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Aleks [24]
1 year ago
13

The velocity of a car increases from 2.0 m/s to 16.0 m/s in a time period of 3.5 s. What was the average acceleration?

Physics
2 answers:
dangina [55]1 year ago
8 0

Answer:

the acceleration is a=3ms^-2

Explanation:

lara [203]1 year ago
5 0

Answer:4.0m/s^2

Explanation:

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Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov
frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v_{1} be the velocity of ball A along positive x axis

v_{2} be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v_{1}

∴  v_{1} = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v_{2}

2 = u +  v_{2} → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 =  v_{2} - u → equation 2

By adding both equations 1 and 2 we get

v_{2} = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

                         = 8·85×m J

Total momentum = m×√(2² + 3·7²)

                             = m× 4·21 kgm/s

3 0
2 years ago
"The predictions of Einstein’s Theory of General Relativity were tested on a double pulsar system in January of 2004. His equati
Rasek [7]

Answer:

99.95%

Explanation:

A double pulsar system named PSR J0737-3039A/B  in Puppis constellation was discovered in the year 2003. Pulsars are second most densest object in the universe after black holes and they emit radio waves at regular intervals. This pair presented a great and natural setup to test the Theory of General Relativity presented by Einstein in 1915. In this theory Einstein had presented a set of equations on how the space-time fabric will be curved because of the very dense objects such as Neutron stars. It also predicted how the gravitational waves are created because of stars orbiting each other.

A team of astrophysicists led by Michael Kramer, conducted a study on how these gravitational waves will impact the time in which the radio waves emitted by pulsars will reach Earth. The result of the study proved the theory of General Relativity to be accurate up to 99.95%.

8 0
2 years ago
A helium ion of mass 4m and charge 2e is accelerated from rest through a potential difference V in vacuum. Its final speed will
Pavel [41]

Answer:

Final Velocity = √(eV/m)

Explanation:

The Workdone, W, in accelerating a charge, 2e, through a potential difference, V is given as a product of the charge and the potential difference

W = (2e) × V = 2eV

And this work is equal to change in kinetic energy

W = Δ(kinetic energy) = ΔK.E

But since the charge starts from rest, initial velocity = 0 and initial kinetic energy = 0

ΔK.E = ½ × (mass) × (final velocity)²

(Velocity)² = (2×ΔK.E)/(mass)

Velocity = √[(2×ΔK.E)/(mass)]

ΔK.E = W = 2eV

mass = 4m

Final Velocity = √[(2×W)/(4m)]

Final Velocity = √[(2×2eV)/4m]

Final Velocity = √(4eV/4m)

Final Velocity = √(eV/m)

Hope this Helps!!!

8 0
1 year ago
An astronaut takes what he measures to be a 10-min nap in a space station orbiting Earth at 8000 m/s. A signal is sent from the
svet-max [94.6K]

Answer:

longer than

Explanation:

given,

time of nap = 10 min

speed of orbiting earth = 8000 m/s

c is the speed of light

using the equation of time dilation

t' = \dfrac{t}{\sqrt{1-\dfrac{v^2}{c^2}}}

now inserting all the values

t' = \dfrac{10}{\sqrt{1-\dfrac{8000^2}{3\times 10^8)^2}}}

t' = \dfrac{10}{0.9999}

t' = 10.001 s

on solving the above equation we will get a value greater than 10minutes.

hence, On earth time of nap measured will be longer than 10 min

3 0
2 years ago
In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that
mars1129 [50]

Answer:

The maximum transverse speed of the bead is 0.4 m/s

Explanation:

As we know that the Amplitude of the travelling wave is

A = 3.65 mm

Now the speed of the travelling wave is

v_x = 13.5 m/s

now we know that distance of first antinode from one end is 27.5 cm

so length of the loop of the standing wave is given as

\frac{\lambda}{4} = 27.5 cm

\lambda = 110 cm

now we have

N = \frac{2L}{\lambda}

N = \frac{2(1.65)}{1.10}

N = 3

now we have

R = 2A sin(kx)

R = 2(3.65) sin(\frac{2\pi}{1.10}x)

R = 7.3 sin(1.82 \pi x)

now at x = 13.8 cm

R = 7.3 sin(1.82 \pi (0.138))

R = 5.18 mm

now we have

f = \frac{v}{\lambda}

f = \frac{13.5}{1.1}

f = 12.27 Hz

now maximum speed is given as

v_y = R\omega

v_y = (5.18 \times 10^{-3})(2\pi(12.27))

v_y = 0.4 m/s

4 0
1 year ago
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