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adelina 88 [10]

3 years ago

7

If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision? Is it possible f

or one to be at rest after the collision.

1 answer:

Ulleksa [173]3 years ago

8 0

Answer:

yes

Explanation:

if the objects a magnetic with a certain amount of force they should be at a stop

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In one of the original Doppler experiments, a tuba was played at a frequency of 64.0 Hz on a moving flat train car, and a second

wolverine [178]

Answer:

$f_{beat} = 1.64\ Hz$

Explanation:

given,

frequency of tuba.f = 64 Hz

Speed of train approaching, v = 8.50 m/s

beat frequency = ?

using Doppler's effect formula

$f' = f(\dfrac{v}{v-v_s})$

v_s is the velocity of the source

v is the speed of sound, v = 340 m/s

now,

$f' = 64\times (\dfrac{340}{340 - 8.50})$

f' = 65.64 Hz

now, beat frequency is equal to

$f_{beat} = f' - f$

$f_{beat} = 65.64 - 64$

$f_{beat} = 1.64\ Hz$

hence, beat frequency is equal to 1.64 Hz

3 0

3 years ago

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

So

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

So

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

6 0

3 years ago

The acceleration of an object is constant when its velocity is :

emmasim [6.3K]

Answer:

B. changing by a constant amount each second

Explanation:

thats my answer

5 0

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What is the term that describes waves that require a medium through which to travel?

gladu [14]

Answer:

Mechanical waves are waves that require a medium. This means that they have to have some sort of matter to travel through. These waves travel when molecules in the medium collide with each other passing on energy. One example of a mechanical wave is sound.

8 0

3 years ago

Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1150 kg and was approaching at

devlian [24]

Force acting during collision is internal so momentum is conserve so (initial momentum = final momentum) in both directions Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1150 kg and was approaching at 5.00 m/s due south. The second car has a mass of 750 kg and was approaching at 25.0 m/s due west. Let Vx is and Vy are final velocities of car in +x and +y direction respectively. initial momentum in +ve x (east) direction = final momentum in +ve x direction (east)

- 750*25 + 1150*0 = (750+1150)
Vx initial momentum in +ve y (north) direction = final momentum in +ve y direction (north)

750*0 - 1150*5 = (750+1150)
Vy from here you can calculate Vx and Vy so final velocity V is

<span>V=<span>(√</span><span>V2x</span>+<span>V2y</span>)
</span>
and angle make from +ve x axis is

<span>θ=<span>tan<span>−1</span></span>(<span><span>Vy</span><span>Vx</span></span>)

</span><span> kinetic energy loss in the collision = final KE - initial KE</span>

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3 years ago

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