If you push a 2.1 kg object on a table with 42N of force. The box then slowly skids to a stop over 2.2 meters, then the force of the friction would be 42 Newtons as per the concept of limiting friction.
What is friction?Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.
As given in the problem if push a 2.1 kg object on a table with 42N of force. The box then slowly skids to a stop over 2.2 m, then we have to find the force of the friction.
The force of the friction = The limiting friction force
If you apply 42N of force to a 2.1-kilogram item on a table. According to the theory of limiting friction, when the box gently slides to a halt over a distance of 2.2 meters, the force of friction would be 42 Newtons.
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A Curve in a road has 60m radius The angle of bank of the road is 4.7° Find the maximum speed of car can have without skidding. If the Coefficient of Static Friction between tyres and road is 0.8.
Answer:
An unbanked curve has a radius of 60m. The maximum speed at which a car can make a turn if the coefficient of static friction is 0.75, is.
Explanation:
Three bulbs of resistance 100. Ω, 200, Ω and 300 Ω are connected in parallel to a 120. V DC power supply. Draw the diagram and find thea) current in each bulb b) current drawn from the power supplyc) total power drawn power supply d) the net resistance of all bulbs
Let's use the formula for electric current.
[tex]I=\frac{V}{R}[/tex]Where V is the power supply 120 V, and R is the resistance. Let's find the current in each bulb.
[tex]\begin{gathered} I=\frac{120V}{100\Omega}=1.20A \\ I=\frac{120V}{200\Omega}=0.6A \\ I=\frac{120V}{300\Omega}=0.4A \end{gathered}[/tex](a) The current in each bulb is 1.20A, 0.6A, and 0.4A, respectively.(b) (c) The diagram of the circuit isTo find the net resistance, we use the following formula.
[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}[/tex]Let's use the given magnitudes.
[tex]\begin{gathered} \frac{1}{R}=\frac{1}{100\Omega}+\frac{1}{200\Omega}+\frac{1}{300\Omega} \\ \frac{1}{R}=\frac{6+3+2}{600\Omega} \\ \frac{1}{R}=\frac{11}{600\Omega} \\ R=\frac{600}{11}\Omega \\ R\approx54.55\Omega \end{gathered}[/tex]Therefore, the net resistance of all bulbs is 54.55 ohms.Two long parallel wires 0.552 meter apart are each carrying 1.75 amperes of current, as shown(a) Find the magnitude and direction of the magnetic field at point A due to the current in the top wire. (b) Find the magnitude and direction of the force per unit length this field exerts on the bottom wire.
,Given,
Distance between two-wire, d=0.552 m
Current through the wires, I=1.75 A
(a) The magnetic field at a point on the bottom wire due to top wire is given by,
[tex]B_a=\frac{\mu_0I_1}{2\pi d}[/tex]Where μ₀ is the permeability of the free space.
The direction is given by the right-hand thumb rule. According to this, the direction of the magnetic field produced by top wire will into the plane of the two wires.
On substituting the known values in the above equation,
[tex]B_a=\frac{4\pi\times10^{-7}\times1.75}{2\pi\times0.552}=6.34\times10^{-7}\text{ T}[/tex]The force per unit length is given by,
[tex]F=I_2\times B_a[/tex]The direction is given by the right-hand rule. According to this the force is directed towards the top wire.
On substituting the known values in the above equation,
[tex]F=1.75\times6.34\times10^{-7}=1.11\times10^{-6}\text{ N}[/tex]Therefore the magnetic field acting on the bottom wire due to the current in the top wire is 6.34×10⁻⁷ T and the magnetic force due to this field is 1.11×10⁻⁶ N
Find the force (N) on a 7.36cm radius piston on the other end of a hydraulic system that is driven by 23.37N of force from a 2.98cm radius piston.
The force (N) on a 7.36cm radius piston would be 142.55 Newtons, if on the other end of a hydraulic system that is driven by 23.37 N of force from a 2.98cm radius piston.
What is pressure?The total applied force per unit of area is known as the pressure.
The pressure depends both on externally applied force as well the area on which it is applied.
Pressure = Force / Area
As given in the problem we have to find out the force (N) on a 7.36cm radius piston on the other end of a hydraulic system that is driven by 23.37N of force from a 2.98cm radius piston,
By using the Pascal Law,
F₁ / A ₁ = F₂ / A₂
23.37 / 2.98² = F₂ / 7.36²
F₂ = 23.37 × 7.36² / 2.98 ²
= 142.55 Newtons
Thus, the force (N) on a 7.36cm radius piston would be 142.55 Newtons ,
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If a copper wire with a 0.300 mm diameter is to have a resistance of 0.500 Ω at 20.0ºC, how long should it be? The resistivity of copper is 1.68 * 10^-8 Ω*m
Given:
• Diameter = 0.300 mm
,• Resistance = 0.500 Ω
,• Temperature = 20.0ºC
,• Resistivity = 1.68 * 10⁻⁸ Ω*m
Let's find the length of the copper wire.
To find the length of the wire, apply the formula:
[tex]\begin{gathered} R=\frac{\rho L}{A} \\ \\ R=\frac{\rho L}{\pi r^2} \end{gathered}[/tex]Where:
R is the resistance = 0.500 Ω
p is the resistivity = 1.68 * 10⁻⁸ Ω*m
L is the length
r is the radius = diameter/2 = 0.300mm/2 = 0.15 mm
Convert the radius to meters, where:
1 m = 1000 mm
0.15mm = 0.15 x 10⁻³ m
Now, input values into the formula and solve fo L:
[tex]0.500=\frac{1.68\times10^{-8}\times L}{\pi\times(0.15\times10^{-3})^2}[/tex]Solving further, rewrite the equation for the length L:
[tex]\begin{gathered} L=\frac{0.500\pi\times(0.15\times10^{-3})^2}{1.68\times10^{-8}} \\ \\ L=\frac{3.534\times10^{-8}}{1.68\times10^{-8}} \\ \\ L=2.1\text{ m} \end{gathered}[/tex]Therefore, the length of the copper wire is 2.1 meters.
ANSWER:
2.1 m
Pls quick will mark brainliest.
Lance is working in a library using a trolley to carry books. As he stacks books from the trolley on shelves, it gets easier to push or pull the trolley. Which part of Newton's laws of motion explains the increased ease of moving the trolley?
A. Objects at rest tend to stay at rest.
B. Objects in motion tend to stay in motion.
C. Every action has an equal and opposite reaction.
D. Larger objects require greater amounts of force to move.
Answer: D
Explanation:
Larger objects require greater amounts of force to move.
As he stacks books from the trolley on shelves, it gets easier to push or pull the trolley because the trolley gets lighter when the books are shifted from the trolley to the shelves. So, larger objects require greater amounts of force to move. Hence, option D is correct.
What is inertia?Inertia is a property of bodies that prevents them from moving or, if they are already moving, causes them to change the speed or direction of their motion. The inertia of a body is a passive quality that prohibits it from acting in any way besides opposing active agents like forces and torques.
A moving body continues to move not because of its inertia but rather because no force exists to stop it, alter its direction, or accelerate it.
According to the question, when there are books on the trolley they require a greater amount of force to move but when the books are shifted on the shelves it's easier to move the trolley. So inertia is proportional to mass.
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A child playing by the side of a well is throwing a stone vertically into the well with an initial velocity of 5 m/s. if the stone falls into the water 3 s later, find the height, h, of the point which the stone is thrown from the surface of the water. (g=10 m/s2)
a) 15m
b)30m
c)45m
d)60m
The height of the well is 30m
We are given that ,
The initial velocity of the stone = u = 5m/s
The stone fall into well in time = t = 3s
The acceleration due to gravity = g = 10 m/s²
To get the height of the stone which is thrown from the surface of the water in the well by equation,
h = ut + 1/2 gt²
Where, u is initial velocity in m/s, h is height in m, t is time in secs, g is acceleration due to gravity which is vertically upward direction i.e. (-10m/s²)
h = (5m/s) (3s) + 1/2 (-10m/s²)(3s)²
h = (15m) - (5m/s²)(9s²)
h = -30 m
Therefore, the height of the stone which is thrown upward is 30 m . And here, we can neglect the negative sign due to height is never be in negative.
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S=3-2t+3t^2
What is the instantaneous velocity and it’s acceleration at t=3s
At what time is the particle at rest
Answer:
Explanation:
Given:
X(t) = 3 - 2*t + 3*t²
t = 3 s
_______________
V(t) - ?
a(t) - ?
Speed is the first derivative of the coordinate, acceleration is the second.
1)
V(t) = X' = (3 - 2*t + 3*t²)' = 0 - 2 +6*t = 6*t - 2
V(3) = 6*3 - 2 = 16 m/s
2)
a(t) = X'' = V' = (6*t - 2)' = 6 m/s²
a(3) = 6 m/s²
3)
The body will stop (V = 0 ) in (t) seconds:
V(t) = 6*t - 2
0 = 6*t - 2
6*t = 2
t = 2/6 = 1/3 ≈ 0,33 s
a piece of steel expands 10 cm when it is heated from 20 to 40 deg C. How much would it expand if it was heated from 20 to 60 deg.C?
ANSWER
20 cm
EXPLANATION
Given:
• The change in length of a piece of steel, ΔL = 10 cm, when its temperature change is ΔT = 20°C
Find:
• The change in length of the same piece of steel, ΔL, when its temperature is changed ΔT = 40°C
The change in length of a solid material is,
[tex]\Delta L=\alpha\cdot L_o\cdot\Delta T[/tex]Where L₀ is the original length, α is the linear expansion coefficient and ΔT is the change in temperature.
In this case, we know how much is the piece of steel expanded when the temperature change is 20°C, so we can find the product αL₀,
[tex]\alpha L_o=\frac{\Delta L}{\Delta T}[/tex]Replace the known values and solve,
[tex]\alpha L_o=\frac{10\text{ }cm}{20\degree C}=0.5cm/\degree C[/tex]If the temperature change now is 40°C,
[tex]\Delta L=\alpha L_o\Delta T=0.5cm/\degree C\cdot40\degree C=20cm[/tex]Hence, when the piece of steel is heated from 20°C to 60°C it will expand 20 centimeters.
An interesting question is “In what direction is the dot (representing a particle in the medium)moving at the instant shown above?” The correct answer is “_______”. The way to understandthat is to imagine the pulse an instant later. The wave will have moved a bit to the right.Therefore, since the particle is still on the wave, and can only move up or down, it must be______.
We will have the following:
The correct answer is vertically.
The way to understand that is to image the pulse an instant later. The wave will have moved a bit to the right. Therefore, since the particle is still on the wav, and can only move up or down, it must be lower.
The crazed physic's student's lab partner decides to throw another pumpkin off the
roof with an initial velocity of 19.4 m/s. What is the velocity when the pumpkin
strikes the ground if it takes 3.2 seconds for it to fall?
Please Help :(
The velocity when the pumpkin strikes the ground if it takes 3.2 seconds for it to fall: 50.78 m/s
From the definition of velocity, we can find the velocity of a falling object is:
v = v₀ + gt
Here
v₀ - 19.4 m/s , t - 3.2 seconds , g - 9.80665 m/[tex]s^2\\[/tex]
v = 19.4 + (9.86 x 3.2)
v = 50.78 m/s
What is velocity?
Velocity and speed describe how quickly or slowly an object is moving. We frequently encounter circumstances when we must determine which of two or more moving objects is going faster. If the two are travelling on the same route in the same direction, it is simple to determine which is quicker. It is challenging to identify who is moving the fastest when their motion is in the other direction. The concept of velocity is useful in these circumstances. Learn about the definition of velocity in this article as well as the distinction between speed and velocity.
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If the input force on the lever in problem 5 is 135 N, what is the net force on the rock?
The net force on the rock is determined as 76.2 N.
What is the net force on the rock?The net force on the rock is obtained by subtracting the weight of the rock from the input force on the lever.
F(net) = Input force - weight of the rock
The weight of the rock is calculated as follows;
W = mg
where;
m is mass of the rock = 6 kgg is acceleration due to gravity = 9.8 m/s²W = 6 x 9.8
W = 58.8 N
F(net) = 135 N - 58.8 N
F(net) = 76.2 N
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The complete question is below:
If the input force on the lever in problem 5 is 135 N, what is the net force on the rock? if the mass of the rock is 6kg.
Three vectors are shown in this figure. Their respective moduli are A = 4.00m.B = 3, 20m and C = 2.70mCalculate 2.00 A - B + 1.30 CExpress your answer according toa) Unit vectorsb) The modulus and orientation with respect to the positive part of the x-axis
Given that,
Modulus of vector A=4.00
The angle made by the vector A with the y axis, θ₁=33.0°
The modulus of vector B=3.20 m
The angle made by the vector B with the x-axis is θ₂=40.0+90.0=130°
The modulus of the vector C=2.70 m
The angle made by the vector C with x-axis θ₃=-90°
The x and y components of the vector can be written as
[tex]\begin{gathered} x=r\cos \theta \\ y=r\sin \theta \end{gathered}[/tex]Where r is the magnitude (or modulus) of the vector and θ is the angle made by the vector.
Or a vector, in cartesian coordinates, can be written as,
[tex]R=r\cos \theta\hat{\text{i}}+r\sin \theta\hat{j}[/tex]Therefore, vector A is cartesian coordinates is
[tex]\begin{gathered} \vec{A}=4.00\cos 33^{\circ}\hat{i}+4.00\sin 33.0^{\circ}\hat{j} \\ =3.35\hat{i}+2.18\hat{j} \end{gathered}[/tex]And the vector B is
[tex]\begin{gathered} \vec{B}=3.20\cos (130^{\circ})\hat{i}+3.20\sin (130^{\circ})\hat{j} \\ =-2.06\hat{i}+2.45\hat{j} \end{gathered}[/tex]And vector C is given by,
[tex]\begin{gathered} \vec{C}=2.70\cos (-90^{\circ})\hat{i}+2.70\sin (-90^{\circ})\hat{j} \\ =-2.7\hat{j} \end{gathered}[/tex]The given equation is
[tex]2.00\vec{A}-\vec{B}+1.30\vec{C}[/tex]Let this represents a vector V
On substituting the known values,
[tex]\begin{gathered} \vec{V}=2.00\vec{A}-\vec{B}+1.30\vec{C} \\ =2.00\times(3.35\hat{i}+2.18\hat{j})-(-2.06\hat{i}+2.45\hat{j)}+1.30(-2.7\hat{j}) \\ =8.76\hat{i}-1.6\hat{j} \end{gathered}[/tex](a) This is the representation with the unit vectors, where i and j are the unit vectors along the x-axis and y-axis respectively.
[tex]\vec{V}=8.76\hat{i}-1.6\hat{j}[/tex]b) The modulus of any vector is the square root of the sum of the squares of its components.
That is, the magnitude of the vector V is
[tex]\begin{gathered} V=\sqrt[]{8.76^2+(-1.6)^2} \\ =8.90\text{ m} \end{gathered}[/tex]The angle of this vector with the x-axis is given by
[tex]\begin{gathered} \phi=\tan ^{-1}(\frac{-1.6}{8.75}) \\ =-10.36^{\circ}^{} \end{gathered}[/tex]The negative sign indicates that the vector is below the positive x-axis
Therefore the modulus of the resultant of the above equation is 8.90 m and its angle with the positive x-axis is -10.36°
Calculate the depth in the ocean at which thepressure is three times atmospheric pressure.Atmospheric pressure is 1.013 x 10^5 Pa. Theacceleration of gravity is 9.81 m/s^2and them/sdensity of sea water is 1025 kg/m^3Answer in units of m.
In order to determine the depth in the ocean, use the following equation:
[tex]h=\frac{P-P_o}{\rho g}[/tex]h: depth
P: pressure = 3*Po
Po: atmospheric pressure = 1.013*10^5Pa
g: gravitational acceleration constant = 9.8m/s^2
p: density of water = 1025 kg/m^3
Replace the previoua values into the formula for h and simplify:
[tex]\begin{gathered} h=\frac{3P_o-P_o}{\rho g}=\frac{2P_o}{\rho g} \\ h=\frac{2(1.013\cdot10^5Pa)}{(1025\frac{kg}{m^3})(9.8\frac{m}{s^2})}\approx20.17m \end{gathered}[/tex]Hence, the depth in the ocean is approximately 20.17m
Part of a light ray striking an interface between air and water is refracted, and part is reflected, as shown. The index of refraction of air is 1.00 and the index of refraction of water is 1.33. The frequency of the light ray is 7.85 x 10^16 Hz.(a) If angle 1 measures 40°, find the value of angle 2.(b) If angle 1 measures 40°, find the value of angle 3.(c) Calculate the speed of the light ray in the water.(d) Calculate the wavelength of the light ray in the water.(e) What is the largest value of angle 1, that will result in a refracted ray?
We will use Snell's law, which states:
[tex]n_1\sin \theta_1=n_2\sin \theta_2[/tex]Where n1 and n2 are the refraction indexes and their respective angles are "theta1" and "theta2".
For part A we replace the values:
[tex]1\sin 40=1.33\sin \theta_2[/tex]Now we solve for "theta2" first by dividing both sides by 1.33:
[tex]\frac{\sin40}{1.33}=\sin \theta_2[/tex]Now we use the inverse function for sine:
[tex]\arcsin (\frac{\sin 40}{1.33})=\theta_2[/tex]Solving the operation:
[tex]28.9=\theta_2[/tex]For part B, since "theta1" and "theta3" are angles of reflection, according to the reflection law, these angles are equal, therefore:
[tex]\theta_3=\theta_1=40[/tex]For part C. The index of refraction is defined as:
[tex]n=\frac{c}{v}[/tex]Where "c" is the speed of light in a vacuum and "v" is the speed of light in the medium. Replacing the values:
[tex]1.33=\frac{3\times10^8\text{ m/s}}{v}[/tex]Now we solve for "v":
[tex]v=\frac{3\times10^8\text{ m/s}}{1.33}[/tex]Solving the operation:
[tex]v=2.26\times10^8\text{ m/s}[/tex]For part d. We will use the following formula:
[tex]\lambda=\frac{v}{f}[/tex]Where "v" is the speed and "f" is the frequency. Replacing we get:
[tex]\lambda=\frac{2.26\times10^8\text{ m/s}}{7.85\times10^{16}s^{-1}}[/tex]Solving the operations:
[tex]\lambda=0.288\times10^{-8}m[/tex]For part e. The largest value of the angle of incidence that will result in refraction is 90 degrees.
What is the electric field amplitude of an electromagnetic wave whose magnetic field amplitude is 7.9 mT ?
Given:
The amplitude of the magnetic field of the electromagnetic wave is,
[tex]B_0=7.9\text{ mT}[/tex]To find:
The amplitude of the electric field
Explanation:
Let, the amplitude of the electric field is
[tex]E_0[/tex]As we know,
[tex]\begin{gathered} \frac{E_0}{B_0}=c \\ c=3\times10^8\text{ m/s} \end{gathered}[/tex]Substituting the values we get,
[tex]\begin{gathered} \frac{E_0}{7.9\times10^{-3}}=3\times10^8 \\ E_0=3\times10^8\times7.9\times10^{-3} \\ E_0=2.37\times10^6\text{ N/C} \end{gathered}[/tex]Hence, the amplitude of the electric field is,
[tex]2.37\times10^6\text{ N/C}[/tex]why is the hubble space telescope in space instead of on the ground?
Answer:EARTH'S ATMOSPHERE ALTERS AND BLOCKS THE LIGHT THAT COMES FROM SPANCE.
Explanation:
Hubble orbits above Earth's atmosphere, which gives it a better view of the universe than telescopes have at ground level.
Hunter pushed a couch across the room. He did 800 J of work in 20 seconds.The couch weighed 500 N. How much power did he have?A. 16,000 WB. 1.6WC. 800 WD. 40 W
Power = 40 W
Option D
Explanation:The workdone by the Hunter = 800 Joules
The weight of the couch = 500 N
Time, t = 20 seconds
Power = Workdone/Time
Power = 800/20
Power = 40 W
A +10.31 nC charge is located at (0,8.47) cm and a -2.09 nC charge is located (3.91, 0) cm. Where would a -14.84 nC charge need to be located in order that the electric field at the origin be zero? Express your answer, in cm, as the magnitude of the distance of q3 from the origin.
To make the E-field at the origin become 0, we need to find the E-field at the origin before
E = kq/r^2
E1 = 12934 V/m
E2 = 12303.69 V/m
Etotal = 17851.34 V/m
17851.34 = kq/r^2
r = 8.6526 cm
A golf ball is initially on a tee when it is
struck by a golfer. The ball is given an
initial velocity of 50 m/s at a 37° angle. The
ball hits the side of a building that is 200
meters horizontally away from the golfer.
(a) What are the horizontal and vertical
components of the ball's initial
velocity?
(b) How much time elapses before the
ball strikes the side of the building?
(c) How far from the ground does the ball
strike the building?
Answer:
a.
[tex]horizontal=39.9[/tex] m/s
[tex]vertical=30.1[/tex] m/s
b.
[tex]t=5.009[/tex]
c.
[tex]y=27.7[/tex]
Explanation:
Lets write down what we were given.
Angle = 37°
Initial Velocity = 50 m/s
Displacement in x direction = 200 m
Take note:
I am having some trouble with the theta symbol so let theta = [tex]N[/tex]
Lets do question C first.
We know that time is equal to [tex]\frac{displacement}{velocity}[/tex] aka [tex]t=\frac{x}{v}[/tex].
[tex]x=v[/tex]₀ₓ [tex]t[/tex] ⇒ [tex]\frac{x}{v_{0x} }[/tex] ⇒ [tex]\frac{x}{v_{0} *cos(N)}[/tex]
Now substitute the expression for t into the equation for the position.
[tex]y=(v_{0}sin(N))*(\frac{x}{v_{0}cos(N) })-\frac{1}{2}g(\frac{x}{v_{0}cos(N) }) ^{2}[/tex]
Rearranging terms, we have
[tex]y=(tan(N)*x)-[\frac{g}{2(v_{0}cos(N))^{2} } ]x^{2}[/tex]
Now lets substitute our numbers in for the variables. Then simplify.
[tex]y=(tan37*200)-[\frac{9.81}{2(50*cos37)^{2} } ]200^{2}[/tex]
[tex]y=150.7108-[\frac{9.81}{2(50*cos37)^{2} } ]200^{2}[/tex]
[tex]y=150.7108-[0.0030761]200^{2}[/tex]
[tex]y=150.7108-(0.0030761*40000)[/tex]
[tex]y=150.7108-123.0444[/tex]
[tex]y=27.7[/tex]
Now lets do question B.
Lets steal this from the last question.
We know that time is equal to [tex]\frac{displacement}{velocity}[/tex] aka [tex]t=\frac{x}{v}[/tex].
[tex]x=v[/tex]₀ₓ [tex]t[/tex] ⇒ [tex]\frac{x}{v_{0x} }[/tex] ⇒ [tex]\frac{x}{v_{0} *cos(N)}[/tex]
Now substitute the expression for t into the equation for the position.
[tex]y=(v_{0}sin(N))*(\frac{x}{v_{0}cos(N) })-\frac{1}{2}g(\frac{x}{v_{0}cos(N) }) ^{2}[/tex]
We can substitute [tex]t[/tex] for [tex]\frac{x}{v_{0}cos(N) }[/tex]
[tex]y=(v_{0}sin(N))*(t)-\frac{1}{2}g(t) ^{2}[/tex]
We can rewrite the equation as
[tex](v_{0}sin(N)(t)-\frac{1}{2}*(g(t)^{2})=y[/tex]
Now lets substitute our numbers in for the variables.
[tex](50sin(37)(t)-\frac{1}{2}*(9.81(t)^{2})=27.7[/tex]
After some painful algebra and factoring we get
[tex]30.09075115t-4.905t^{2}=27.6664[/tex]
Subtract [tex]27.6664[/tex] from both sides.
[tex]30.09075115t-4.905t^{2}-27.6664=0[/tex]
Use the quadratic formula to find the solutions.
[tex]\frac{-b+-\sqrt{b^{2}-4ac } }{2a}[/tex]
After some more painful algebra we get
[tex]t=5.00854263, 1.12616708[/tex]
1.126 does not make any sense so.
[tex]t=5.009[/tex]
Finally lets do question A.
Lets draw a triangle. We have the velocity which is the hypotenuse and we have the angle. From there we can solve for the opposite and adjacent sides.
Let [tex]A=horizontal[/tex] and [tex]O=vertical[/tex]
[tex]cos(37)=\frac{A}{50}[/tex]
[tex]A=39.9[/tex]
[tex]sin37=\frac{O}{50}[/tex]
[tex]O=30.1[/tex]
What conditions must be met in order for work to be done?
A. The applied force must make the object move.
B. The output force must be greater than the input force.
C. At least part of the applied force must be in the same direction as the movement of the object.
D. The work must be greater than the momentum.
At least part of the applied force must be in the same direction as the movement of the object must be met in order for work to be done.
What are the conditions to work?The following are the two prerequisites for working: To do the work, the body must be subjected to a force, or F 0. The body must move in the direction of the applied force, or S 0, as a result of the applied force.There must be a force used. The displacement is the distance over which the force must act. The displacement must be a component of the force.A legal term known as a condition precedent refers to an event that must occur before a certain contract is regarded as being in effect or before either party is obliged to fulfill any obligations.
Therefore, the correct answer is option C. At least part of the applied force must be in the same direction as the movement of the object.
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Identify the kinematic equation which relates the velocity and time.
The kinematic equation which relates velocity and time is
[tex]v=v_0+at[/tex]As when the acceleratio
Given v = 145 sin (200πt + 3π/2), what is the period?Question 12 options:10 msec5 sec.03 sec.005 sec
For a sine or cosine function of the form:
[tex]A\sin (B(x+C))+D[/tex]the perfiod is given by:
[tex]\frac{2\pi}{B}[/tex]In this case we have that:
[tex]B=200\pi[/tex]Then we have:
[tex]\frac{2\pi}{200\pi}=\frac{1}{100}=0.01=10\times10^{-3}[/tex]Therefore, the period is 10 ms
A cheetah is running at a velocity of 8 m/s when it accelerptes at 1.5 m/s^2 for 6 seconds. If the
cheetah started at a position of 20m what is the final Position of the cheetah?
Answer:
[tex]95[/tex]
Explanation:
[tex]x= x. + vt+\frac{1}{2} at^{2}[/tex]
[tex]s=20 + (8*6)+ \frac{1}{2} (1.5 * 6^{2} )[/tex]
[tex]s= 95[/tex]
the answer to this question
The car and the delivery truck both start from rest and accelerate at the same rate. So, the final speed of the car as compared to the truck is four times as much. So, option D is correct.
What is acceleration?Acceleration is the term used in mechanics to describe how quickly an object's velocity changes in relation to time. The magnitude of accelerations as a vector. An object will accelerate in the direction that it is being pulled in by the net force.
Acceleration is a vector quantity since it has a magnitude and a direction. Velocity is another aspect of vessel quantities. The change in the vector of velocity over a time interval divided by the time interval is the definition of acceleration.
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Use Newton’s Law of Universal Gravitation and Newton’s Second Law to
Find g = acceleration due to gravity
Show g is independent of mass
By equating the two forces, acceleration due to gravity g is obtained which is independent of mass
What is Newton’s Second Law ?The law state that the rate of change of momentum is directly proportional to the force applied.
From Newton’s Second Law, Force F = mg. And from Newton’s Law of Universal Gravitation, Force F = GMm/r²
Where m is the mass of the satellite or the body revolving round the earth.
Equate the two forces.
mg = GMm/r²
The two m cancelled out leaving
g = GM/r²
Where
g = Acceleration due to gravityM = Mass of the earthG = Universal gravitational constantr = Distance between them.Therefore, since the mass of the satellite m has cancelled out, acceleration due to gravity g is independent of mass.
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How much power is used by a contact lens heating unit that draws 0.05 A of current from a 197 V line?
Given,
The current drawn by the contact lens heating unit, I=0.05 A
The supply voltage, V=197 V
The electric power is given by the product of the current drawn and the supply voltage.
Thus the power used by the given device is given by,
[tex]P=VI[/tex]On substituting the known values,
[tex]\begin{gathered} P=197\times0.05 \\ =9.85\text{ W} \end{gathered}[/tex]Thus the power used by the contact lens heating unit is 9.85 W
Which of the following is an example of Newton's third law of motion?A. A skydiver slows down when her parachute opens.B. A grocery cart moves forward when it is pushed.C. A cannon recoils backwards when it is fired.D. A rolling rock slows down due to friction.
Explanation:
The third law of Newton says that when an object exerts a force on a second object, the first object experiences an equal and opposite force that is exerted by the second object.
So, the example that shows this law is:
C. A cannon recoils backward when it is fired.
Because the cannon e
A child and sled with a combined mass of 25.2 kg slide down a frictionless hill, starting at the top. The sled starts from rest and acquires a speed of 6 m/s by the time it reaches the bottom of the hill. What is the height (in m) of the hill?Answer: _________ m (round to the nearest hundredth)
Given:
The mass of the child and the sled is,
[tex]m=25.2\text{ kg}[/tex]The child acquires a speed at the bottom of the hill is,
[tex]v=6\text{ m/s}[/tex]To find:
The height of the hill
Explanation:
The potential energy at the top of the hill will be converted to the potential energy at the bottom of the hill. So for the hill of height 'h', we can write, the potential energy as
[tex]PE=mgh[/tex]The kinetic energy is
[tex]KE=\frac{1}{2}mv^2[/tex]So,
[tex]\begin{gathered} mgh=\frac{1}{2}mv^2 \\ h=\frac{v^2}{2g} \\ h=\frac{6^2}{2\times9.8} \\ h=1.84\text{ m} \end{gathered}[/tex]Hence, the height of the hill is 1.84 m.
Kinetic energy differs from potential energy inA. Kinetic energy can be created or destroyed, while potential energy can not be created and destroyedB. Kinetic energy can be converted into various forms of energy, whereas potential energy can only be transformed into heat energy.C. Kinetic energy is energy of a moving object, whereas potential energy is energy possessed by matter as a result of its location or structure.D. Kinetic energy is stored energy that has the capacity to do work, and potential energy is the energy of motion.
The kinetic enerfy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]And the potential energy is given by:
[tex]U=mgh[/tex]Where:
v = Velocity
h = height
m = mass
g = gravitational accceleration of earth
As we can see kinetic energy is associated to the movement and the potential energy is associated to the location, therefore the answer is:
C. Kinetic energy is energy of a moving object, whereas potential energy is energy possessed by matter as a result of its location or structure.
