Answer:
The strength of the electric field is [tex]E=15.66\: N/C[/tex]
Explanation:
Here the electric force is equal to Newton's second law.
[tex]F_{e}=ma[/tex]
Let's recall that electric force is the electric field times the charge, so we have:
[tex]qE=ma[/tex]
[tex]E=\frac{ma}{q}[/tex] (1)
Where:
m is the proton mass
q is the proton charge
a is the acceleration
Using the equation (1) we have:
[tex]E=\frac{1.67*10^{-27}1.5x10^{9}}{1.6*10^{-19}}[/tex]
Therefore, the strength of the electric field is [tex]E=15.66\: N/C[/tex]
I hope it helps you!
A bowling ball weighing 71.7 N is attached to the ceiling by a rope of length 3.73 m . The ball is pulled to one side and released; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.60 m/s. At this instant, what are:
a. the acceleration of the bowling ball, in magnitude and direction
b. the tension in the rope?
Answer:
A) a = 5.673 m/s²
The direction will be upwards vertically towards the point where it is suspended.
B) T = 113.2 N
Explanation:
A) We are given;
Weight of bowling ball; W = 71.7 N
Speed; v = 4.6 m/s
Rope length; r = 3.73 m
Now, formula for the centripetal acceleration is;
a = v²/r
Thus; a = 4.6²/3.73
a = 5.673 m/s²
The direction will be upwards vertically towards the point where it is suspended.
B) since weight is 71.7 N, it means that;
Mass = weight/acceleration = 71.7/9.8
Mass(m) = 7.316 kg
Thus,
Centripetal force is;
F_cent = 7.316 × 5.673
F_cent = 41.5 N
Thus, Tension in the rope is;
T = W + F_cent
T = 71.7 + 41.5
T = 113.2 N
A boy standing at one end of a floating raft that is stationary relative to the shore walks to the opposite end of the raft, away from the shore. As a consequence, the raft does which of the following?
a. remains stationary
b. moves away from the shore
c. moves toward the shore.
Answer:
The raft moves towards the shore
Explanation:
In which of Earth’s systems would we find cloud droplets, wind, & weather?
Cryosphere
Geosphere
Biosphere
Atmosphere
Answer:
atmosphere
Explanation:
.........,....
Name the three types of kin etic energy. Define each.
Answer: Out of the 5, the first 3 that came to my mind first were thermal energy, electrical energy, and sound energy.
Explanation:
Thermal-Generates due to the quick motion of atoms and molecules, especially when they collide with each other. It is also called heat energy. The matters in the universe consist of atoms or molecules which are always in motion. However, we can’t see the movement of this energy with our naked eyes. We can feel it at the time it touches our skin.
Sound-The vibration of an object causes sound energy. Sound is the movement of energy generated by vibrations through some substance, such as air or water or solid. Sound energy can travel through any medium to transfer energy from one particle to another, and you can hear it. However, it cannot travel through a vacuum as a vacuum does not contain any particles that can act as a medium. When an object vibrates, it makes the surrounding particles vibrate by transferring its energy. These particles, when colliding with other particles, make them vibrate. In this way, the sound energy gets transferred from one particle to another.
Electrical- Every object in the universe is made up of small particles called atoms. Atoms consist of tiny particles, namely electrons, protons, and neutrons. The electrons in the atom always move around the nucleus of an atom. While applying voltage, the electrons present in the atom get energy and break the bonding with the parent atom and thus become free. The energy of this free electron is called electrical energy or electricity. Therefore, it is the energy of these moving free electrons. Electrons are negatively and positively charged and usually move through a wire.
A satellite of mass m orbits a moon of mass M in uniform circular motion with a constant tangential speed of v. The satellite orbits at a distance R from the center of the moon. Write down the correct expression for the time T it takes the satellite to make one complete revolution around the moon?
The gravitational force exerted by the moon on the satellite is such that
F = G M m / R ² = m a → a = G M / R ²
where a is the satellite's centripetal acceleration, given by
a = v ² / R
The satellite travels a distance of 2πR about the moon in complete revolution in time T, so that its tangential speed is such that
v = 2πR / T → a = 4π ² R / T ²
Substitute this into the first equation and solve for T :
4π ² R / T ² = G M / R ²
4π ² R ³ = G M T ²
T ² = 4π ² R ³ / (G M )
T = √(4π ² R ³ / (G M ))
T = 2πR √(R / (G M ))
The correct expression for the time T it takes the satellite to make one complete revolution around the moon is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].
We can find the period T (the time it takes the satellite to make one complete revolution around the moon) from the gravitational force:
[tex] F = \frac{GmM}{R^{2}} [/tex] (1)
Where:
G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²
R: is the distance between the satellite and the center of the moon
m: is the satellite's mass
M: is the moon's mass
The gravitational force is also equal to the centripetal force:
[tex] F = ma_{c} [/tex] (2)
The centripetal acceleration ([tex]a_{c}[/tex]) is equal to the tangential velocity (v):
[tex] a_{c} = \frac{v^{2}}{R} [/tex] (3)
And from the tangential velocity we can find the period:
[tex] v = \omega R = \frac{2\pi R}{T} [/tex] (4)
Where:
ω: is the angular speed = 2π/T
By entering equations (4) and (3) into (2), we have:
[tex] F = m\frac{v^{2}}{R} = m\frac{(\frac{2\pi R}{T})^{2}}{R} = \frac{mR(2\pi)^{2}}{T^{2}} [/tex] (5)
By equating (5) and (1), we get:
[tex] \frac{mR(2\pi)^{2}}{T^{2}} = \frac{GmM}{R^{2}} [/tex]
[tex] T^{2} = \frac{R^{3}(2\pi)^{2})}{GM} [/tex]
[tex] T = \sqrt{\frac{R^{3}(2\pi)^{2})}{GM}} [/tex]
[tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex]
Therefore, the expression for the time T is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].
Find more here:
https://brainly.com/question/13340745?referrer=searchResultshttps://brainly.com/question/13451473?referrer=searchResultsI hope it helps you!
The effects of force on gravity is more noticeable when the object is_____
Answer:
Gravity only becomes noticeable when there is a really massive object like a moon, planet or star. We are pulled down towards the ground because of gravity. The gravitational force pulls in the direction towards the centre of any object.
P and S waves from an earthquake travel at different speeds and this difference helps in locating the epicenter (point of origin) of the earthquake. (a) Assuming P waves travel at 10.3 km/s and S waves travel at 4.2 km/s, how far away did the earthquake occur if a particular seismic station detects the arrival of these two types of waves 3.25 minutes apart
Answer:
x = 1382.9 km
Explanation:
The speed of the wave is constant, so we can use the uniform motion relationships
p wave
[tex]v_p[/tex] = x / t₁
t₁ = x /v_p
S wave
v_s = x / t₂
t₂ = x / v_s
indicate that the time difference between the two waves is
t₂ - t₁ = 3.25 min (60 s / 1 min)
t₂ -t₁ = 195 s
let's substitute
[tex]\frac{x}{v_s} - \frac{x}{v_p}[/tex] = 195
x ([tex]\frac{1}{v_s} - \frac{1}{v_p}[/tex] = 195
let's calculate
x [tex]( \frac{1}{4.2} - \frac{1}{10.3} )[/tex] = 195
x (0.1410) = 195
x = 195 /0.141
x = 1382.9 km
A 50 kg crate slides down a 5.0 m loading ramp that is inclined at an angle of 25 to the horizontal. A worker pushes on the crate parallel to the surface of the ramp so that the crate slides down with a constant velocity. If the coefficient of kinetic friction between the crate and the ramp is 0.33, how much work is done by (a) the worker
Answer:
The magnitude of the work done by the worker is 303 J.
Explanation:
The work done by the worker can be found as follows:
[tex] W = |F|\cdot |d| cos(\alpha) [/tex]
Where:
F: is the force applied by the worker
d: is the displacement = 5.0 m
α: is the angle between the force applied and the displacement = 180°
We need to find the force applied by the worker:
[tex]\Sigma F = ma[/tex]
Taking as positive the movement direction of the crate we have:
[tex] -F - F_{\mu} + P_{x} = 0 [/tex]
Where:
m: is the crate's mass
a: is the acceleration = 0 (It is moving at constant speed)
F: is the force applied by the worker
Pₓ: is the weight in the horizontal direction
[tex]F_{\mu}[/tex]: is the frictional force
Hence, the force applied by the worker is:
[tex]F = P_{x} - F_{\mu} = mgsin(\theta) - \mu mgcos(\theta)[/tex]
[tex] F = 50 kg*9.81 m/s^{2}*(sin(25) - 0.33cos(25)) = 60.6 N [/tex]
Then, the work done by the worker is:
[tex] W = |F|\cdot |d| cos(\alpha) = 60.6 N*5.0 m*cos(180) = -303 J [/tex]
Therefore, the magnitude of the work done by the worker is 303 J.
I hope it helps you!
The work is done by the worker will be 303 J. Work done is described as the multiplication of applied force and the amount of displacement.
What is work done?Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.
Work may be zero, positive and negative.it depends on the direction of the body displaced. if the body is displaced in the same direction of the force it will be positive.
The given data in the problem is;
F is the force applied by the worker
d is the displacement = 5.0 m
α is the angle between the force applied and the displacement = 180°
m is the crate's mass= 50 kg
a is the acceleration = 0
Pₓ: is the weight in the horizontal direction
The net force on the crate is found as;
[tex]\rm F_{net}= P_X - F_{\mu} \\\\ \rm F_{net}= mg sin \theta - \mu mg cos \theta \\\\ \rm F_{net}=50 \times \times 9.81 (sin 25^0 -0.33 cos(25) \\\\ \rm F_{net}= 60.6 N \\\\[/tex]
The work done by the worker will be;
[tex]\rm W= Fd cos \alpha \\\\ \rm W= 60.6 \times 5.0 cos 180^0 \\\\\rm W=-303 \ J[/tex]
Hence the work is done by the worker will be 303 J.
To learn more about the work done refer to the link ;
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You have two small spheres, each with a mass of 2.40 grams, separated by a distance of 10.0 cm. You remove the same number of electrons from each sphere.
1) What is the charge on each sphere if their gravitational attraction is exactly equal to their electrical repulsion?
2) How many electrons did you remove from each sphere?
Answer:
q = 2.066* 10⁻¹³ C.
n = 1,291,250 electrons.
Explanation:
1)
If the gravitational attraction is equal to their electrical repulsion, we can write the following equation:[tex]F_{g} = F_{c} (1)[/tex]
where Fg is the gravitational attraction, that can be written as follows according Newton's Universal Law of Gravitation:[tex]F_{g} = G*\frac{m_{1}*m_{2}}{r_{12}^{2}} (2)[/tex]
Fc, due to it is the electrical repulsion between both charged spheres, must obey Coulomb's Law (assuming we can treat both spheres as point charges), as follows:[tex]F_{c} = k*\frac{q_{1}*q_{2}}{r_{12}^{2}} (3)[/tex]
since m₁ = m₂ = 0.0024 kg, and r₁₂ = 0.1m, G and k universal constants, and q₁ = q₂ = Q, we can replace the values in (2) and (3), so we can rewrite (1) as follows:[tex]G*\frac{(0.0024kg)^{2}}{r_{12}^{2}} = k*\frac{Q^{2}}{r_{12}^{2}} (4)[/tex]
Since obviously the distance is the same on both sides, we can cancel them out, and solve (4) for Q² first, as follows:[tex]Q^{2} = \frac{6.67e-11*(0.0024kg)^{2}}{9e9Nm2/C2} = 4.27*e-26 C2 (5)[/tex]
Since both charges are the same, the charge on each sphere is just the square root of (5):Q = 2.066* 10⁻¹³ C.2)
Assuming that both spheres were electrically neutral before being charged, the negative charge removed must be equal to the positive charge on the spheres.Now, since each electron carries an elementary charge equal to -1.6*10⁻¹⁹ C, in order to get the number of electrons removed from each sphere, we need to divide the charge removed from each sphere (the outcome of part 1) with negative sign) by the elementary charge, as follows:[tex]n_{e} =\frac{-2.066e-13C}{-1.6e-19C} = 1,291,250 electrons. (6)[/tex]Two cars, a Porsche Boxster convertible and a Toyota Scion xB, are traveling at constant speeds in the same direction. Suppose, instead, that the Boxster is initially 170 m behind the Scion. The speed of the Boxster is 24.4 m/s and the speed of the Scion is 18.6 m/s. How much time does it take for the Boxster to catch the Scion
Answer:
It will take 29.31 seconds for the Boxster to catch the Scion
Explanation:
Given the data in the question;
lets say Toyota Scion xB is car A and Porsche Boxster convertible is B and Toyota Scion xB is car A
the distance travelled by car A is
x = [tex]V_{A}[/tex] × t
where [tex]V_{A}[/tex] is the speed of the car and t is time
the distance travelled by car B before reaching car A will be;
x + x₀ = [tex]V_{B}[/tex] × t
Now lets replace x by [tex]V_{A}[/tex] × t
so
([tex]V_{A}[/tex] × t) + x₀ = [tex]V_{B}[/tex] × t
x₀ = ([tex]V_{B}[/tex] × t) - ([tex]V_{A}[/tex] × t)
x₀ = t ([tex]V_{B}[/tex] - [tex]V_{A}[/tex])
t = x₀ / ([tex]V_{B}[/tex] - [tex]V_{A}[/tex])
so we substitute
t = 170 m / (24.4 - 18.6)
t = 170 / 5.8
t = 29.31 s
Therefore; it will take 29.31 s for the Boxster to catch the Scion
There is always one way to calculate the magnitude of the net vector of any direction vectors.
True
False
Answer and I will give you brainiliest
Answer:
False
Explanation:
Hope it help mark as BrainlistAnswer:
False
Explanation:
On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a constant rate from 20 m/s to a complete stop over a 10 second interval. How does the distance traveled by the truck compare to that of the car? So assuming that I've sketched a graph, do I just do distance x time?
a. The truck travels twice as far as the car.
b.There is not enough information to answer the question.
c .The truck travels the same distance as the car.
d The truck travels half as far as the car.
Answer:
a)
Explanation:
Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:[tex]x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} (1)[/tex]
Since the car starts from rest, v₀ =0.We know the value of t = 5 sec., but we need to find the value of a.Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:[tex]a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2 (2)[/tex]
Replacing a and t in (1):[tex]x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m. (3)[/tex]
Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:[tex]a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2 (4)[/tex]
Replacing v₀, at and t in (1), we have:[tex]x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m (5)[/tex]
Therefore, as the truck travels twice as far as the car, the right answer is a).A quarterback, Patrick, throws a football down the field in a long arching trajectory to wide receiver, Tyreek. The football and Tyreek are traveling in the same direction, started at the same spot, and the football was thrown at the same instant that Tyreek began running. Furthermore, both Tyreek and the football have horizontal components of speed of 22.6 mph. Under these circumstances, no matter what angle the football is thrown at, it will land on Tyreek (whether he catches it or not). True or false
Answer:
the statement is False [tex]\frac{x_{ball} }{ x_{rec} }[/tex] = cos θ
Explanation:
Let's analyze this problem, the ball and the receiver leave the same point and we want to know if at the same moment they reach the same point, for this we must have both the ball and the receiver travel the same distance.
Let's start by finding the time it takes for the ball to reach the ground
y = [tex]v_{oy}[/tex] t - ½ g t²
when it reaches the ground its height is y = 0
0 = vo sin θ - ½ g t²
0 = t (vo sin θ - ½ g t)
The results are
t = 0 exit point
t = 2 v₀ sin θ/g arrival point
at this point the ball traveled
[tex]x_{ball}[/tex]= v₀ₓ t
x_{ball} = v₀ cos θ 2v₀ sin θ / g
x_{ball}= 2 v₀² cos θ sin θ/ g
Now let's find that distantica traveled the receiver in time
[tex]x_{rec}[/tex] = v₀ t
x_{rec} = v₀ (2 v₀ sin θ / g)
x_{rec} = 2 v₀² sin θ / g
without dividing this into two distances
[tex]\frac{x_{ball} }{ x_{rec} }[/tex] = cos θ
therefore the distances are not equal to the ball as long as behind the receiver
Therefore the statement is False
Magnetic dipole X is fixed and magnetic dipole Y is free to move. Dipole Y will initially:
A. Move toward X but not rotate
B. Move away from X but not rotate
C. Move toward X and rotate Move away from X and rotate
D. Rotate but not translate
Answer: The correct option is A.
Move toward X but not rotate
Explanation:
This is because from the question, X is fixed and y is free to move. Since magnetic dipole of X is fixed, that is it can't move and that of y is free to move, therefore y will move toward x because it's forces of attraction is linear and not rotational and besides X and Y are on a linear path, therefore Y will move towards X that is fixed and it will not rotate since it's linear.
If two clay masses produce a gravitational force of 340N, what will be the force if the distance is divided by 3 and the mass of one is divided by 2?
75.56 N
1530 N
510 N
56.67 N
Answer:
B: 1530 N
Explanation:
We know ghat formula for force of gravity is;
F_grav = G•m1•m2/d²
We are told that two clay masses produce a gravitational force of 340N.
Thus;
G•m1•m2/d² = 340
Now, if the distance is divided by 3 and the mass of one is divided by 2, we have;
F_g = (G × m1/2 × m2)/(d/3)²
Thus gives;
F_g = ½(G•m1•m2)/((1/9)d²)
Simplifying this gives;
F_g = (9/2)G•m1•m2/d²
From earlier, we saw that;
G•m1•m2/d² = 340
Thus;
F_g = (9/2) × 340
F_g = 1530 N
In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its speed is decreased linearly from 60 mph to 30 mph in 10 seconds. Calculate the theoretical maximum energy in kWh that can be recovered during this interval. Ignore all losses.
Answer:
the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
Explanation:
Given that;
weight of vehicle = 4000 lbs
we know that 1 kg = 2.20462
so
m = 4000 / 2.20462 = 1814.37 kg
Initial velocity [tex]V_{i}[/tex] = 60 mph = 26.8224 m/s
Final velocity [tex]V_{f}[/tex] = 30 mph = 13.4112 m/s
now we determine change in kinetic energy
Δk = [tex]\frac{1}{2}[/tex]m( [tex]V_{i}[/tex]² - [tex]V_{f}[/tex]² )
we substitute
Δk = [tex]\frac{1}{2}[/tex]×1814.37( (26.8224)² - (13.4112)² )
Δk = [tex]\frac{1}{2}[/tex] × 1814.37 × 539.5808
Δk = 489500 Joules
we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule
so
Δk = 489500 / 3.6 × 10⁶
Δk = 0.13597 ≈ 0.136 kWh
Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
What are some iron objects magnetic and some are not
Iron is magnetic, so any metal with iron in it will be attracted to a magnet. Steel contains iron, so a steel paperclip will be attracted to a magnet too. Most other metals, for example, aluminum, copper, and gold are NOT magnetic. Two metals that aren't magnetic are gold and silver.
An object with a mass of 8 kg moves at a speed of 5 m/s. How much kinetic energy does the object have?
Question 2 options:
100 J
200 J
40 J
20 J
Answer:
Kinetic energy =1/2mv^2
Where m is the mass of the object. V is the velocity
K.E=0.5(8*5^2)
K.E=100J
Explanation:
please help fast due next period!!!
write the chemical formula for the following ionic compounds
Zinc (III) Phosphide
Answer:
The chemical formula for zinc (III) phosphide is Zn3P2
To wait until the oncoming vehicle passes before completing a left turn is known as:
a)
b)
c)
IPDE strategy
Risk acceptance
Risk rejection
Inappropriate maneuver
Answer:
Risk rejection
Explanation:
There are several factors that contribute to the degree of driving risks and they include but not limited to the ability of the driver and the condition of a vehicle. Other factors are condition of the environment and the condition of the highway. When driving, a driver may wait until an oncoming vehicle passes before making a complete left turn as a risk rejection strategy. Left turns are more dangerous when making them because drivers tend to accelerate on to a left turn. The wider radius of a left turn is know to led to higher speeds and greater pedestrian exposure. A driver is advised to have more mental and physical efforts when making a left turn.
After an initial race George determines that his car loses 35 percent of its acceleration due to air resistance travelling at 38 m/s on flat ground. Assuming that his car travels with a constant acceleration, calculate the maximum speed (Vm), in meters per second, his car can reach on flat ground?\
Answer:
64.2 m/s
Explanation:
We are given that
Speed ,v=38 m/s
We have to find the maximum speed when his car reach on flat ground.
Using dimensional analysis
[tex]F_{res}\propto v^2[/tex]
If 35% acceleration reduced by F(res) at 38 m/s
Then, 100% acceleration can be reduced by F(res) at v' m/s
[tex]\frac{F_1}{F_2}=\frac{v^2}{v'^2}[/tex]
[tex]v'^2=\frac{F_2}{F_1}v^2[/tex]
[tex]v'=v\sqrt{\frac{F_2}{F_1}}[/tex]
Substitute the values
[tex]v'=38\times \sqrt{\frac{100}{35}}[/tex]
[tex]v'=64.2 m/s[/tex]
Hence, the maximum speed when his car can reach on flat ground=64.2 m/s
The maximum velocity of the car at constant acceleration on the road will be 62.2 m/s.
What is the relation between resistance and speed?
The air resistance is directly proportional to the square of the velocity of an object.
R ∝ v²
The speed of the car was reduced by 35 % at 38 m/s.
So, the speed of the car was reduced by 100% at v' m/s.
The relationship can be given by,
[tex]\dfrac {R_1}{R_2} = \dfrac {v^2}{v'^2}[/tex]
Put the values in the formula and calculate for [tex]v'[/tex],
[tex]v' = \sqrt {\dfrac {R_2 v^2}{R_1 }}[/tex]
[tex]v' = \sqrt {\dfrac { 100\times 38 ^2}{35 }}\\\\v' = 62.2 \rm \ m/s[/tex]
Therefore, the maximum velocity of the car at constant acceleration on the road will be 62.2 m/s.
Learn more about resistance and speed?
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a 75 kg object traveling at 4 m/s collides with and sticks to a 125 kg object initially at rest, what is the final velocity of the two objects?
Answer:
1.5m/s
Explanation:
Given parameters:
M1 = 75kg
V1 = 4m/s
M2 = 125kg
V2 = 0m/s
Unknown:
Final velocity of the two objects = ?
Solution:
To solve this problem, we must understand that this is an inelastic collision. To conserve momentum;
M1 V1 + M2 V2 = V (M1 + M2)
( 75 x 4 ) + ( 125 x 0) = V (75 + 125)
300 = 200V
V = 1.5m/s
How are wavelength and frequency related?
As wavelength increases frequency increases.
b. As wavelength decreases frequency decreases.
As wavelength increases frequency decreases.
d. None of the above
Answer:
Hey dear
Explanation:
Its option C
As Wavelength increases frequency decreases
In other case,
When Wavelength decreases frequency increases
its opposite
Tq
Need help in the question that is down
Answer:
4th ans
Explanation:
The chain of DNA consists of
ILL MARK BRAINLIST !! get it right
how much work is required to make a 1400 kg car increase its speed from 10 m/s to 20 m/s?
what average force is required if the car travels 15 m during this speed change?
What name is given to "an object or group of objects that we wish to consider for a physics problem?
Answer:
System
Explanation:
A system is an object or group of objects that we wish to consider for a physics problem. Examples of systems are universe, house and its surroundings etc.
There are three different types of systems which are:
Isolated system: In this type of system there is no energy or matter exchange with the surroundings. Closed system: In this type of system, only energy is exchanged with the surroundings, matter is not exchanged. Open system: in this system both matter and energy are exchanged with the surroundings..A person pushes an object of mass 5.0 kg along the floor by applying a
force. If the object experiences a friction force of 10 N and accelerates at
18 m/s^2, what is the magnitude of the force exerted by the person?
Answer:
The magnitude of the force exerted by the person is 100 N
Explanation:
Net Force
According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:
Fn = ma
Where a is the acceleration of the object.
The net force is the sum of all forces exerted over a body. When an object is moved along a rough surface it experiences two horizontal forces and two vertical forces (provided there is no vertical component of the applied force).
The vertical forces are the Normal and the Weight and they are balanced, i.e.: N = W = mg.
The horizontal forces are The applied force (Fa) and the friction force (Fr). They are not balanced because the object is accelerated in that direction. The net force is:
Fn = Fa - Fr
Applying the first equation:
Fa - Fr = ma
Solving for Fa:
Fa = Fr + ma
Substituting the given values m=5 kg, Fr=10 N, [tex]a=18\ m/s^2[/tex].
Fa = 10 + 5*18 = 10 + 90 = 100
Fa = 100 N
The magnitude of the force exerted by the person is 100 N
Basically.
Given: F + m x a
Equation: 10 N + 5.0 kg x 18 m/s2
Solve: 100 N
A bat, flying at 5.00 m/s, emits a chirp at 40.0 kHz. If this sound pulse is reflected by a wall, what is the frequency of the echo received by the bat
Answer:
The answer is below
Explanation:
Firstly, the frequency is received by the wall and then it is reflected and received by the bat.
The frequency received by the wall (f') is given by:
f' = [tex]f(v\pm v_o)/v\\\\[/tex]
The - sign is used when the observer is moving away from the source and + sign when the observer is moving towards the source.
Since the bat is moving towards the wall, we use a positive sign. Hence:
f' = [tex]f(v+ v_o)/v\\\\[/tex]
The frequency reflected and received by the bat f" is:
f'' = [tex]f'\frac{v}{ (v\pm v_s)}\\\\[/tex]
- sign is used when the source moves toward the observer and + is used when the source moves away
since the bat moves towards the wall, then::
f'' = [tex]f'\frac{v}{(v-v_s)} =\frac{f(v+v_o)}{v}*\frac{v}{(v-v_s)} =f\frac{(v+v_o)}{(v-v_s)} \\\\[/tex]
v = speed of sound in air = 331 m/s, vo = velocity of observer = 5 m/s, vs = velocity of source = 5 m/s. Therefore:
[tex]f"=f\frac{(v+v_o)}{(v-v_s)} =40\ kHz\frac{(331\ m/s+5\ m/s)}{(331\ m/s+5\ m/s)} \\\\f"=41\ kHz[/tex]
The frequency of the echo received by the bat is 38.8 kHz.
The given parameters:
Speed of the bat, V = 5 m/sActual frequency of the chirp, Fs = 40 kHzSpeed of sound, Vs = 331 m/sThe observed frequency or frequency received by the bat is calculated by applying Doppler effect as follows;
[tex]f_0 = f_s(\frac{v \ +/-\ v_0}{v\ +/- \ v_s} )[/tex]
Since the bat is flying away from the wall the frequency received will be smaller than the actual frequency;
[tex]f_o = f_s (\frac{v \ - \ v_0}{v \ + \ v_s} )\\\\f_o = 40 \ kHz \times (\frac{331 - 5}{331 + 5} ) \\\\f_0 = 38.8 \ kHz[/tex]
Thus, the frequency of the echo received by the bat is 38.8 kHz.
Learn more about Doppler effect here: https://brainly.com/question/3841958
An electron moving along the x axis has a position given by x 16te$t m, where t is in seconds. How far is the electron from the origin when it momentarily stops
This question is incomplete, the complete question is;
An electron moving along the x axis has a position given by x = 16 t[tex]e^{-t}[/tex] m, where t is in seconds. How far is the electron from the origin when it momentarily stops
Answer:
the electron is 5.88 m far from the origin when it momentarily stops
Explanation:
Given that;
the position of the electron is x = 16 t[tex]e^{-t}[/tex] m
now, if the electron stopped after a time t, then its velocity is zero
so
V = dx/dt = 0
d/dt( 16 t[tex]e^{-t}[/tex] m) = 0
16( -t[tex]e^{-t}[/tex] + [tex]e^{-t}[/tex] ) = 0
16(-t + 1) [tex]e^{-t}[/tex] = 0
16(1 - t) [tex]e^{-t}[/tex] = 0
1 - t = 0
t = 1 sec
so
x = 16 × 1 × [tex]e^{-1}[/tex] m
x = 16 × 1 × 0.36787 m
x = 5.88 m
Therefore, the electron is 5.88 m far from the origin when it momentarily stops
