To program a digital turtle to trace the outline of a hexagon, the code that should be used is repeat 6 [forward 50 and right 60]. Hence, option B is correct.
What is a program?A computer utilizes a program, which is a collection of instructions, to carry out a particular task. A program is like a computer's recipe, to use an analogy. It has a list of components (known as variables, which can stand for text, graphics, or numeric data) and a list of instructions (known as statements), which instruct the computer on how to carry out a certain operation.
Programming languages with particular syntax, like C++, Python, and Ruby, are used to develop programs. These high-level programming languages are written and readable by humans.
According to the question, the hexagon has six sides so the turtle has to repeat 6 times. Hence, option B is correct.
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A concave mirror has a focal length of 0.50 m. If an object produces a virtual image 0.19 m from the mirror, where is it located?
In order to calculate the object's position, we can use the formula below:
[tex]\frac{1}{f}+\frac{1}{d_o}+\frac{1}{d_i}[/tex]Where f is the focal length, do is the object's position and di is the image's position.
Using f = 0.5 m and di = -0.19 m (we use a negative value because the image is virtual), we have:
[tex]\begin{gathered} \frac{1}{0.5}=\frac{1}{d_o}+\frac{1}{-0.19}\\ \\ 2=\frac{1}{d_o}-5.263\\ \\ \frac{1}{d_o}=2+5.263\\ \\ \frac{1}{d_o}=7.263\\ \\ d_o=\frac{1}{7.263}\\ \\ d_o=0.14\text{ m} \end{gathered}[/tex]Therefore the object is at 0.14 meters from the mirror.
The amount of thermal energy inside a protostar increases with time, even though the protostar is losing radiative energy from its surface. How can we tell how much radiative energy the protostar is losing and how much thermal energy remains in the star? which type of energy can we measure and which type do we infer from the law of conservation of energy?.
The amount of thermal energy inside a protostar increases with time, We will degree radiative power without delay and infer thermal energy from models: type will we infer from the law of conservation of energy
A protostar is a very younger big name that is nonetheless accumulating mass from its parent molecular cloud. The protostellar phase is the earliest one within the system of stellar evolution. for a low-mass superstar, it lasts about 500,000 years.
A protostar is formed as gravity starts offevolved to pull the gases collectively into a ball. This technique is referred to as accretion. As gravity pulls the gasses toward the center of the ball, gravitational strength begins to warmness them, inflicting the gasses to emit radiation. in the beginning, the radiation simply escapes into space.
Protostar is an early degree in the evolution of a celebrity that generally grows to the factor of beginning nuclear fusion and becoming a star through gathering mass. it's miles made from a contracting cloud of bloodless and dark interstellar medium (often hydrogen gas).
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b. A bus travelling in a straight seconds. road covers a distance of 220m in 20 between mass and weight. calculate its velocity
Answer:
11 m/s?
Explanation:
The question is not clear, but I am assuming that it is asking for the velocity of a bus traveling for 220 meters in 20 seconds.
You are Given:
Displacement (d) : 220m (in this question, since there is no change in direction, the distance and displacement is the same)
Time (t) : 20 seconds
You want to Find:
Velocity: ???
Equation to use:
velocity = displacement/time
aka. v = d/t
Now just Solve:
v = d/t
v = 220m/20s
v = 11 m/s
the simplest design for a radio/microwave transmitter antenna is a vertical conducting rod about one quarter-wavelength long. a mobile-phone manufacturer wants to build such an antenna into a gsm-900 (900 mhz) phone. how small can the phone be?
The phone can be 8.35cm small.
Frequency is the variety of occurrences of a repeating occasion in line with a unit of time. it is also from time to time referred to as temporal frequency to emphasize the assessment of spatial frequency, and regular frequency to emphasize the evaluation of angular frequency.
Frequency is an expression of ways regularly a periodic wave shape or signal repeats itself at a given amplitude. it is able to be expressed in hertz (Hz), kilohertz (kHz), megahertz (MHz), and so on.
Frequency is a vital parameter used in technological know-how and engineering to specify the temporal rate of trade discovered in oscillatory and periodic phenomena, such as mechanical vibrations, audio alerts (sound), radio waves, and mild.
GSM 900 band corresponds to a wavelength (\lambda) of
λ= c/f
where c is the speed of light and frequency is f which is 900 MHz
This gives \lambda=33.4cm
Antenna length L=\lambda/4=8.35cm
So the phone should at least be 8.35cm long.
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To prevent water pollution, a factory proposes pumping its wastes into the ground instead of into a river. Would you support this proposal? Why or why not?
A manufacturer of echo effects for music recording studios has built a prototype of an echo device that sounds exactly like a beloved echo machine from the 1970s. even though its computer chips are cheap, they are hard to find because the compounds originally used to make the chips are no longer being manufactured. what type of issue does the effects manufacturer need to overcome in order to be able to mass produce this echo device? availability cost effectiveness safety time effectiveness
Since sound expands and loses intensity as it moves farther away, the echo is weaker than the initial sound and has a smaller amplitude. For instance, when a speaker's voice is heard and its echo is compared.
One kind of audio effect that involves gradually delaying a signal is an echo effect. In this instance, after some amount of time, listeners hear the same signal repeated audibly. When the time delay is relatively long (more than 30 milliseconds), listeners can distinguish between the echoes.
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Answer:
D: Availability
Explanation:
correct on edge 2023
jumping up before the elevator hits. after the cable snaps and the safety system fails, an elevator cab free-falls from a height of 36 m. during the collision at the bottom of the elevator shaft, a 90 kg passenger is stopped in 5.0 ms. (assume that neither the passenger nor the cab rebounds.) what are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision? if the passenger were to jump upward with a speed of 7.0 m/s relative to the cab floor just before the cab hits the bottom of the shaft,
a) The magnitude of the impulse is 2.39 × [tex]10^{3}[/tex] N. b) Average force on the passenger during the collision is 4.78 × [tex]10^{5}[/tex] N. c) impulse is 1.6 × [tex]10^{3}[/tex] N.s d) The corresponding average force would be 3.52 × [tex]10^{5}[/tex] N.
a) By energy conservation, the speed of the passenger when the elevator hits the floor is
1/2 mv² = mgh
v = √2gh = [tex]\sqrt{2(9.8)(36)}[/tex]
= 26.6 m/s
The magnitude of the impulse is
J = |Δp| = m|Δv| = mv = (90kg) (26.6m/s) ≈ 2.39 × 10³ N.s
b) With duration of Δt = 5.0 × [tex]10^{-3}[/tex] s for the collision, the average force is
[tex]F_{avg}[/tex] = J/ Δt = 2.39×10³ N.s / 5.0 × [tex]10^{-3}[/tex] s
≈ 4.78 × [tex]10^{5}[/tex] N
c) If the passenger were to jump upward with a speed of [tex]v^{'}[/tex]= 7.0 m/s, then the resulting downward velocity would be
[tex]v^{n}[/tex] = v- [tex]v^{'}[/tex]
= 26.6m/s - 7.0 m/s
= 19.6 m/s
The magnitude of the impulse becomes
[tex]J^{n}[/tex] = |Δ[tex]p^{n}[/tex]| = m|Δ[tex]v^{n}[/tex]| = m[tex]v^{n}[/tex] = (90kg) (19.6 m/s)
≈ 1.76 × 10³³N.s
d) The corresponding average force which is
[tex]F_{avg } ^{n}[/tex] = [tex]J^{n}[/tex]/ Δt = 1.76 × 10³ N.s/ 5.0 × [tex]10^{-3}[/tex] s
≈ 3.52 × [tex]10^{5}[/tex]N
Therefore the magnitude of the impulse is 26.6 m/s, average force on the passenger during the collision is 4.78 × [tex]10^{5}[/tex]N, the magnitude of the impulse is 1.76×[tex]10^{3}[/tex]N.s, average force is 3.52 × [tex]10^{5}[/tex].
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the one component of the material of the milky way galaxy that prevents us from seeing and photographing the galactic center at optical wavelengths is the one component of the material of the milky way galaxy that prevents us from seeing and photographing the galactic center at optical wavelengths is very cold hydrogen gas. hot hydrogen gas. the glare of light from nearby stars. interstellar dust.
The one component of the material of the milky way galaxy that prevents us from seeing and photographing the galactic center at optical wavelengths is interstellar dust.
The Milky Way is a vast collection of stars dust and gas. It is called a spiral galaxy because it looks like a spinning windmill when viewed from above or below. The Sun is in one of the spiral arms about 25,000 light-years away from the center of the galaxy. Astronomers have released the first images of a supermassive black hole at the center of our Milky Way.
The results provide overwhelming evidence that this object is indeed a black hole and provide valuable clues about the workings of the giant galaxies thought to reside at the center of most galaxies. This extreme environment exposes you to intense UV and X-rays. But much of this activity is obscured from view by a huge plume of interstellar dust. The center of our Milky Way is hidden from the prying eyes of optical telescopes by clouds of dust and gas.
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an ice hockey forward with mass 70.0 kg is skating due north with a speed of 5.5 m/s. as the forward approaches the net for a slap shot, a defensive player (mass 110 kg) skates toward him in order to apply a body-check. the defensive player is traveling south at 4.0 m/s just before they collide. if the two players become intertwined and move together after they collide, in what direction and at what speed do they move after the collision? friction between the two players and the ice can be neglected.
The speed of the two players after the collision is 0.31 m/s in a direction of due south.
What is the final speed of the two players?The final speed of the two players after the collision is calculated as follows.
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
m₁ is the mass of the first playerm₂ is the mass of the second playeru₁ is the initial velocity of the first playeru₂ is the initial velocity of the second playerv is the common speed of both players after the collision.(70 x 5.5) + (-110 x 4) = v(70 + 110)
385 - 440 = 180v
-55 = 180v
v = -55/180
v = -0.31 m/s
Thus, the direction of the speed of the two players after the collision is due south.
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A tennis player can accelerate a tennis ball at 295m/s² using an average force of
58 N with his racquet. What is the mass of the tennis ball?
answer of the question is into the picture ️
Slim Jim, continually maintaining his svelte body, lifts a 70 kg barbell 1.4m above the grounda) How much energy did the barbell have when it was on the ground?b) How much energy does it have after being lifted 1.4m? What kind of energy does it have afterbeing lifted? Where did it come from?c) How much work did Jim do to the lift the object?dIf he lifted it in 1.5s how much power did he use?
Given,
The mass of the barbell, m=70 kg
The height to which Slim Jim lifts the barbell, h=1.4 m
a)
When the barbell was on the ground, it will have zero kinetic energy as it has no velocity. And if assume the height of ground as zero meters, then its potential energy is also zero.
Thus when the barbell was on the ground, its energy was zero joules.
b)
The energy of the barbell when it is at a height of h is given by,
[tex]E=\text{mgh}[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} E=70\times9.8\times1.4 \\ =960.4\text{ J} \end{gathered}[/tex]Thus the energy that the barbell has after being lifted 1.4 m is 960.4 J
This energy is the energy stored in the barbell due to its position. Thus the energy stored in the barbell is the potential energy.
Slim Jim has to do some work to lift the barbell to the given height. This work done will be stored in the barbell in the form of potential energy. That is, the energy of the barbell is supplied to it from Slim Jim through the work.
c)
All the work done by Jim will be stored in the barbell in the form of potential energy. Thus, the work done by Jim is equal to the potential energy of the barbel.
Therefore, the work done by Jim is 960.4 J
d)
Given,
The time interval, t=1.5 s
The power is given by,
[tex]P=\frac{W}{t}[/tex]Where W is the work done by Jim.
On substituting the known values,
[tex]\begin{gathered} P=\frac{960.4}{1.5} \\ =640.27\text{ W} \end{gathered}[/tex]Thus the power used by Jim is 640.27 W
A bar magnet is cut into two pieces from the middle as shown in the figure. Mark the poles on these magnets if it is further broken into smaller pieces?
Giving brainliest!
If we cut the magnet into smaller pieces as we can see in the image attached to the question then each of the ends of the smaller magnets would form a north and south pole respectively.
What is a magnet?The term magnet can be used to describe the material that is able to pick up any magnetizable material. We have to note that a magnet is able to create a magnetic field around it. The magnetic field is the vicinity where the impact of the magnet can be felt.
We know that a magnet has two poles, the north pole and the south pole. These are the points where the influence of the magnetic field is greatest In this discussion, we must keep it at the back of our minds that the influence of the magnetic force is stronger at the ends of the magnet than it is at the center of the magnet as it were.
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An apple is resting on a table. Claire says that there are no forces acting on the apple because it is not moving. Is she right?
Answer:
no
Explanation:
there are forces , even when the apple isnt moving , for example gravity which is keeping the apple on the table
Two identical clocks are set to 12 noon time when on the ground at rest relative to the ground. One clock is placed in a spacecraft, sent into space and accelerated to a speed of v = 0.93 x 108 m/s. What will the moving clock read if the first clock reads 6 p.m.? Express your answer as the number of minutes from 6 pm.
First, for this problem, we will need to know the formula for time dilation
[tex]t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]Where t is the time in the reference frame on earth
t0 is the time on the ship
v is the velocity relative to lightspeed
and c is light speed in a vacuum
For this problem, we will need to convert .93x10^8 to a fraction of light speed.
.93x10^8 = .31c
Now, we can plug in the numbers we get from the given
t0 = 6 hours since the ship has been in the air for 6 hours
[tex]t=\frac{6}{\sqrt{1-\frac{(.31c)^2}{c^2}}}[/tex]t = 6.31089
For the answer to be valid, we will convert .31089 to minutes by multiplying it by 60 (60 minutes in a hour)
18.654 minutes
Select the correct answer. when a small star dies, which of these celestial objects is it most likely to help create? a. a planet b. a star c. a moon d. a meteor e. a comet
It is best to choose option B. When a star runs out of fuel, it can no longer undergo the merger process and dies. The primary elements of stars are helium and hydrogen.
A white dwarf star's maximum mass is determined by Chandrasekhar's limit, a sort of maximum restricted value. The Chandrasekhar limit is often valued at 1.44 times the mass of the sun, or around 2.765 1030 kg, on a global scale.
In other words, the Chandrasekhar limit is the highest figure that may possibly reach 1.44 times the mass of the sun.
When a high-mass star runs out of gas to burn, it enlarges and changes into a red supergiant. While the bulk of stars gradually fade away, the supergiant stars, often known as supernovae, demolish themselves in a massive explosion. New stars can develop as a result of massive stellar deaths.
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Two litres of water, initially at 20 °C, is heated to 40 °C. Determine thevolume of water at 40 °C if the coefficient of volumetric expansion ofwater within this range is 30 x 10-9 °C.
Given data
*The given initial volume of water is V_1 = 2 L
*The given initial temperature of the water is T_i = 20 °C
*The given final temperature of the water is T_f = 40 °C
*The coefficient of volumetric expansion of water is
[tex]\alpha=30\times10^{-9}^{}\text{C}[/tex]The formula for the volumetric expansion is given as
[tex]\frac{\Delta V}{V}=\alpha(T_f-T_i)[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} \frac{\Delta V}{V_1}=30\times10^{-9}\times(40^0-20^0) \\ V_2-V_1=6\times10^{-7}V_1 \\ V_2=6\times10^{-7}V_1+V_1^{} \\ =V_1(6\times10^{-7}+1) \\ =2.00\text{ L} \end{gathered}[/tex]you walk for 6.44 km with at a constant velocity of 2.60 m/s due west, turn around, and then walk with an average velocity of 0.465 m/s due east. from your starting point to your destination, your overall average velocity is 1.37 m/s due west. during the trip, how far east did you walk?
The distance travelled due east during the trip is 875.831 meters.
The distance due west is 6.44 km (6440m) and the velocity due west is 2.6m/s. The velocity due east is 0.465m/s.
We know,
Average velocity = Total displacement/Total time
It is given that, the total average velocity is 0.465m/s.
If distance in west is W and the distance in east is E, then the displacement D is,
D = √((6440)²+(E)²)
The time T₁ in west,
T₁ = 6440/2.6 seconds
Time time T₂ in east,
T₂ = E/0.465
The total time T is,
T = T₁ + T₂
T = 6440/2.6 + E/0.465
So, putting all the values now,
Average velocity = D/T
1.37 = √((6440)²+(E)²)/(6440/2.6 + E/0.465)
On solving it further,
We get,
E = 875.831 meters.
So, the distance in east is 875.831 meters.
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A ball is dropped from a height of 23 meters. Find the velocity of the ball just prior to
striking the ground.
round to nearest hundreds
21.168 m/s velocity of the ball just prior to striking the ground.
Round to nearest hundreds is 21.17
What is the meaning of velocity?The direction of the movement of the body or the object is defined by its velocity. Most of the time, speed is a scalar quantity. In its purest form, velocity is a vector quantity. It measures how quickly a distance changes. It is the rate at which displacement is changing.
“equations of motion”
How long does it take?
s = ut + ½ at²
23 m = 0 + ½ *9.81m/s² * t²
23 m = 4.905 t²
t² = 23 / 4.905
t² = 4.689
t =2.165s
What velocity will it hit the ground?
Take ↓ direction as +
v = u + at
v = 0 + 9.81 m/s² * 2.16 s
v =21.168 m/s
Velocity = 21.168 m/s downwards. ( have left you to decide on significant digits in the answer.
round to nearest hundreds is 21.17
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Write 2 uses of centripetal force.
024 (part 1 of 3) 10.0 points
A 4.0 kg block is pushed 1.0 m at a constant
velocity up a vertical wall by a constant force
applied at an angle of 30.0
◦ with the horizontal, as shown in the figure.
The acceleration of gravity is 9.81 m/s
2
.
1 m
30◦
4 kg
F
Drawing not to scale.
If the coefficient of kinetic friction between
the block and the wall is 0.40, find
a) the work done by the force on the block.
Answer in units of J.
The work done on the force on the block is 23.516J
We are given that,
The mass of the block = m= 4.0 kg
The vertical component of the wall at an angle = sinθ = sin 30°
The horizontal component of the wall at an angle = cosθ = cos30°
The acceleration due to gravity = g = 9.81 m/s²
The coefficient of kinetic friction between the block and the wall= μ=0.40
So that to calculate the work done by the force on the block can be given by the expression, for vertical and horizontal component
Wf = Fycosθ
The Wf is the work done by the force Fy (on y-axis) on the block by the horizontal component,
Wf = Fxsinθ
The Wf is the work done by the force Fx (on x-axis) on the block by the vertical component,
Net Fy = sinθ N -mg =0
Net Fx = 0
Therefore, the work done by the force on the block can be assumed by above equations,
Wf = (mgdsinθ)/(sinθ - μcosθ)
Wf = (4.0kg)(1.0m)(9.81m/s²)/sin30° -(0.40)cos30°
Wf = 33.981/1.445 J
Wf = 23.516J
Thus , the work done by the force on the block would be 23.516J
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use the values from practice it to help you work this exercise. what if an additional mass is attached to the ball? how large must this mass be to increase the downward acceleration by 35%?
If the additional mass is attached to the ball, it must be large enough to increase the downward acceleration by 35%.
What is acceleration?
An object is said to have been accelerated if its velocity changes. An object's velocity can change depending on whether it moves faster or slower or in a different direction. A falling apple, the moon orbiting the earth, or a car stopped at a stop sign are a few instances of acceleration. Through these illustrations, we can see that acceleration happens whenever a moving object changes its direction or speed, or both. Since acceleration has both a magnitude and a direction, it is a vector quantity. Additionally, it is either the first derivative of velocity in relation to time or even the second derivative of position in relation to time.
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does increasing the magnitude of a uniform magnetic field through which a charge is traveling necessarily mean increasing the magnetic force on the charge? does changing the direction of the field necessarily mean a change in the force on the charge?
Since the magnetic force and velocity are orthogonal, the velocity only changes in direction but not in magnitude . Circular motion is the end consequence. The force is in the opposite direction from the right-hand rule and has a negative charge.
Uniform In circular motion, an object moves in two dimensions with a constant speed in a fixed circular direction; however, because the object's direction changes at every point, the velocity also changes, and the direction at each point is always the tangent.
In physics, magnetic force is referred to as an object's maximal size and direction. Both vector and scalar values use magnetic force as a common factor. We understand that by definition
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Answer: The velocity simply changes in direction, not magnitude, as the magnetic force and velocity are orthogonal.
Explanation: The velocity simply changes in direction, not magnitude, as the magnetic force and velocity are orthogonal. The resultant motion is circular. The force has a negative charge and is directed in the opposite direction from the right-hand rule.
Uniform In a circular motion, an item moves in two dimensions with a constant speed in a fixed circular direction; yet, because the object's direction varies at each point, the velocity also alters, and the direction at each point is consistently the tangent.
As an object's maximum size and direction, the magnetic force is referred to in physics. Magnetic force is a factor that is present in both scalar and vector values. We comprehend that by nature.
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A rocket is launched from the surface of the earth with a speed of 9.0x103 m/s. What is the maximum altitude reached by the rocket? (MEarth=5.98x1024 kg, REarth=6.37x106 m)
From the Law of conservation of energy, we know that the sum of the kinetic and potential energy of the rocket is the same at the surface of the Earth and at the maximum altitude. Nevertheless, the kinetic energy of the rocket when it is at the maximum altitude is 0:
[tex]\begin{gathered} K_1+U_1=K_2+U_2 \\ K_2=0 \\ \Rightarrow K_1+U_1=U_2 \end{gathered}[/tex]The kinetic energy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]On the other hand, the gravitational potential energy for big changes in altitude (comparable to the radius of the Earth) is given by the expression:
[tex]U=-\frac{GMm}{r}[/tex]Where M is the mass of the Earth, m is the mass of the rocket, r is the distance from the center of the Earth to the rocket and G is the gravitational constant:
[tex]G=6.67\times10^{-11}N\cdot\frac{m^2}{\operatorname{kg}}[/tex]At the beggining of the movement, the value of r corresponds to the radius of the Earth:
[tex]U_1=-\frac{GMm}{R_E}[/tex]At the end of the movement, the value of r corresponds to the radius of the Earth plus the maximum altitude h:
[tex]U_2=-\frac{GMm}{R_E+h_{}}[/tex]Substitute the expressions for U_1, K_1 and U_2 and simplify the equation by eliminating the factor m:
[tex]\begin{gathered} \frac{1}{2}mv^2-\frac{GMm}{R_E}=-\frac{GMm}{R_E+h} \\ \Rightarrow\frac{1}{2}v^2-\frac{GM}{R_E}=-\frac{GM}{R_E+h} \end{gathered}[/tex]Isolate the term GM/(R_E+h):
[tex]\Rightarrow\frac{GM}{R_E+h}=\frac{GM}{R_E_{}}-\frac{1}{2}v^2[/tex]Divide both sides by the factor GM:
[tex]\Rightarrow\frac{1}{R_E+h}=\frac{1}{R_E}-\frac{v^2}{2GM}[/tex]Take the reciprocal to both sides of the equation:
[tex]\Rightarrow R_E+h=\frac{1}{\frac{1}{R_E}-\frac{v^2}{2GM}}[/tex]Isolate h:
[tex]h=\frac{1}{\frac{1}{R_E}-\frac{v^2}{2GM}}-R_E[/tex]Substitute the values of each variable: R_E=6.37x10^6m, M=5.98x10^24kg, G=6.67x10^-11 N*m^2/kg^2, and v=9.0x10^-3 m/s:
[tex]\begin{gathered} h=\frac{1}{\frac{1}{6.37\times10^6m}-\frac{(9.0\times10^3\cdot\frac{m}{s})^2}{2(6.67\times10^{-11}N\cdot\frac{m^2}{kg^2})(5.98\times10^{24}kg)}}-6.37\times10^6m \\ =18.03\times10^6m-6.37\times10^6m \\ =11.7\times10^6m \end{gathered}[/tex]Therefore, the maximum altitude reached by a rocket with an initial speed of 9.0x10^3m is:
[tex]11.7\times10^6m[/tex]how far from the front of the car will the cm be when two people sit in the front seat 2.30 mm from the front of the car, and three people sit in the back seat 3.40 mm from the front? assume that each person has a mass of 60.0 kgkg . express your answer to three significant figures and include the appropriate units.
The center of mass exists a position defined relative to an object or system of objects.
The proper significant figures and proper rounding of digits exists 2.603 m.
What is meant by center of mass?The unique location where the weighted relative position of the scattered mass adds to zero is known as the center of mass in physics. Here is where a force can be applied to produce a linear acceleration without also producing an angular acceleration.
A position established in relation to an object or set of objects is the center of mass. It represents the system's average location as weighted by each component's mass. The center of mass for straightforward stiff objects with homogeneous density is found at the centroid.
Given:
Mass of car, m = 1150 kg
Distance, d = 2.35 m
The new position of center of mass will be:
[tex]$=\frac{1150 \times 2.35+160 \times 2.7+240 \times 3.75}{240+260+1150} \\[/tex]
simplifying the above equation, we get
= (2702.5 + 432 + 900) / 1650
= 4034.5 / 1650
= 2.603 m
Therefore, the proper significant figures and proper rounding of digits exists 2.603 m.
The complete question is:
The CM of an empty 1150 kg car is 2.35 m behind the front of the car. How far from the front of the car will the CM be when two people sit in the front seat 2.70 m from the front of the car, and three people sit in the back seat 3.75 m from the front? Assume that each person has a mass of 80.0 kg.
express answers in proper significant figures and proper rounding of digits.
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light enters glass from air. the angle of refraction will be: group of answer choices less than the angle of incidence. equal to the angle of incidence. greater than the angle of incidence
Less than the angle of incidence, the angle of refraction will exist.
What is angle of refraction?Angle of refraction refers to the angle formed by the refracted ray and the normal at the point of incident.
In a mirror, the angle of incidence and angle of reflection are the same. If the incident ray falls along the normal, the angle of incidence for a plane mirror is 0 degrees rather than 90 degrees.
The behavior of light is described by Fermat's principle, which states that a light ray always chooses the shortest route to its destination. On reflection, the exact same behavior is seen. The angle of incidence and the angle of reflection are equal when a ray is reflected from a plane.
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when the pendulum bob is at its maximum height (farthest from the center of the earth), is the gravitational potential energy of the bob-earth system at a maximum, minimum or some value in between? briefly explain.
Both gravitational potential energy (PE) and kinetic energy (KE) are present on a swinging pendulum bob. The KE is at its highest and the PE is zero when the bob is in its lowest position. The KE is zero and the PE is at its highest value when the bob is in its highest position on both sides.
When does the pendulum's speed reach its highest point?The pendulum moves at its fastest while it is passing through the middle position. We could interpret this in terms of energy conservation. Bob is at his shortest height when he is in the average position. When Bob is positioned at an extreme, such as the right, it is slightly higher than when it is in the middle.
What happens to a pendulum's length when it is full?The bob's center of mass serves as the standard for determining the pendulum's "length." The center of gravity of the bob would fluctuate in a convoluted manner as the mercury drained out of a simple spherical bob with a small hole in the bottom. The center of gravity would be in the middle of the bob when it is totally full.
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Two people are playing soccer. The kinetic energy of each soccer player depends on the soccer players mass and speed. Compare the two soccer players
The two soccer players kinetic energy are 1 / 2 m1 v1² and 1/ 2 m2 v2² for soccer player 1 and 2 respectively
KE = 1 / 2 m v²
KE = Kinetic energy
m = Mass
v = Velocity
Let the mass and velocity be m1 and v1 respectively for soccer player 1. Let the mass and velocity be m2 and v2 respectively for soccer player 2.
For soccer player 1,
m = m1
v = v1
KE1 = 1 / 2 m1 v1²
For soccer player 2,
m = m2
v = v2
KE2 = 1 / 2 m2 v2²
KE ∝ m
KE ∝ v²
The kinetic energy is directly proportional to mass and velocity squared. So increase in any of these factors will increase the kinetic energy and similarly decrease in any of these factors will result in decrease of kinetic energy.
Therefore, the two soccer players kinetic energy are 1 / 2 m1 v1² and 1 / 2 m2 v2² for soccer player 1 and 2 respectively
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Answer:
Cristiano Ronaldo SUII
Explanation:
A 10 kg box is pulled across a level floor, where the coefficient of kinetic friction is 0. 35. What horizontal force is required for an acceleration of 2. 0 m/s2?
The horizontal force required for an acceleration of 2.0 m/s² is equal to 54.3 N.
What is the force of kinetic friction?The force of kinetic friction is a type of frictions that acts on a moving object on a surface. In order to calculate the force of kinetic friction, we use the formula F = μN where F is the force of kinetic friction, μ is the coefficient of the kinetic friction, and N is the normal force. However, we need to take into account the force along the x-axis. Therefore, the formula ∑F = μN + ma will help us find the horizontal force.
Let ∑F represent the net force, N represent the normal force, m represent the mass of the box, g represent the acceleration of gravity, µ represent the kinetic friction, and a represent an acceleration of 2 m/s².
∑F = µN + ma
∑F = µmg + ma
∑F = m (µg + a)
∑F = 10 kg (0.35 × 9.8 m/s² + 2 m/s²)
∑F = 10 kg (3.43 m/s² + 2 m/s²)
∑F = 10 kg (5.43 m/s²)
∑F = 54.3 kg × m/s²
∑F = 54.3 N
The net force is equal to 54.3 N. That is the horizontal force required!
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a swan on a lake gets airborne by flapping its wings and running on top of the water. (a) if the swan must reach a velocity of 5.80 m/s to take off and it accelerates from rest at an average rate of 0.305 m/s2, how far (in m) will it travel before becoming airborne?
The swan will travel 55.15 m during uniform motion before becoming airbone.
We need to know about the uniform motion to solve this problem. The uniform motion is an object's motion under acceleration. It should follow the rule
vt = vo + a . t
vt² = vo² + 2a . s
s = vo . t + 1/2 . a . t²
where vt is final velocity, vo is initial velocity, a is acceleration, t is time and s is displacement.
From the question above, we know that
vo = 0 m/s
vt = 5.8 m/s
a = 0.305 m/s²
By using the second equation, we can calculate the distance before becoming airbone
vt² = vo² + 2a . s
5.8² = 0² + 2 . 0.305 . s
0.61s = 33.64
s = 55.15 m
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__ is the rate at which an object changes position
Velocity is the rate at which an object changes position.
___________________________________________
"What is velocity?"Velocity is a vector quantity that refers to "the rate at which an object changes its position."
Hope this helps!
Want to learn more? Visit below for more information about velocity.
Look up "Speed versus Velocity" at The Physics Classroom.com
