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mestny [16]
3 years ago
8

A 2.9-kg cart is rolling across a frictionless, horizontal track toward a 1.4-kg cart that is held initially at rest. The carts

are loaded with strong magnets that cause them to attract one another. Thus, the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is +4.7 m/s, and the second cart's velocity is −1.9 m/s. (Indicate the direction with the sign of your answer.)
Physics
1 answer:
VikaD [51]3 years ago
6 0

Answer:

The question is incomplete. I will assume you intend to find the total momentum of the two carts during collision. Therefore, we can use the conservation of momentum principle to get the total momentum at a certain instant before collision.

Explanation:

The conservation of momentum principle states that the initial net momentum of two bodies before collision is equal to the final net momentum after collision.

In this case, let's denote the rolling cart as <em>'a'</em> and the stationary cart as <em>'b'</em>.

Momentum, P = MaVa +MbVb

P=(2.9)(+4.7) + (1.4)(-1.9)\\P= 13.68 + (-2.66)\\P= 13.68-2.66\\P=+11.02 Kg.m/s

Therefore, the total momentum before collision is 11.02 Kg.m/s.

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A 27.0-m steel wire and a 48.0-m copper wire are attached end to end and stretched to a tension of 145 N. Both wires have a radi
algol13

Answer:

The time taken by the wave to travel  along the combination of two wires is 458 ms.

Explanation:

Given that,

Length of steel wire= 27.0 m

Length of copper wire = 48.0 m

Tension = 145 N

Radius of both wires = 0.450 mm

Density of steel wire \rho_{s}= 7.86\times10^{3}\ kg/m^{3}

Density of copper wire \rho_{c}=8.92\times10^{3}\ kg/m^3

We need to calculate the linear density of steel wire

Using formula of linear density

\mu_{s}=\rho_{s}A

\mu_{s}=\rho_{s}\times\pi r^2

Put the value into the formula

\mu_{s}=7.86\times10^{3}\times\pi\times(0.450\times10^{-3})^2

\mu_{s}=5.00\times10^{-3}\ kg/m

We need to calculate the linear density of copper wire

Using formula of linear density

\mu_{c}=\rho_{s}A

\mu_{c}=\rho_{s}\times\pi r^2

Put the value into the formula

\mu_{c}=8.92\times10^{3}\times\pi\times(0.450\times10^{-3})^2

\mu_{c}=5.67\times10^{-3}\ kg/m

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

v_{s}=\sqrt{\dfrac{T}{\mu_{s}}}

v_{s}=\sqrt{\dfrac{145}{5.00\times10^{-3}}}

v_{s}=170.3\ m/s

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

v_{c}=\sqrt{\dfrac{T}{\mu_{c}}}

v_{c}=\sqrt{\dfrac{145}{5.67\times10^{-3}}}

v_{c}=159.9\ m/s

We need to calculate the time taken by the wave to travel  along the combination of two wires

t=t_{s}+t_{c}

t=\dfrac{l_{s}}{v_{s}}+\dfrac{l_{c}}{v_{c}}

Put the value into the formula

t=\dfrac{27.0}{170.3}+\dfrac{48.0}{159.9}

t=0.458\ sec

t=458\ ms

Hence, The time taken by the wave to travel  along the combination of two wires is 458 ms.

4 0
4 years ago
How does the time at which you see things happen at a baseball game compare to the time when you hear things happen? Explain you
hoa [83]

Answer:

Because of the speed of the sound.

Explanation:

The first thing that happens in such cases is to take into account the speed of the sound. First, we see that the player hits the ball with the bat, if we are in the stands far enough we will hear the sound of the batting time later, this time depends on the speed of the sound which is equal to 345 [m/s].

Another visible and practical example is a fireworks display, where people nearby immediately hear the explosion. while those at a great distance will be able to see first the explosion followed by the sound.

With the following equation, we can calculate how long it takes to hear a hit or explosion

t = x / v

where:

x = distance [m]

v = sound velocity = 345 [m/s]

t = time [s]

7 0
3 years ago
5. The wire in consists of two segments of different diameters but made from the same metal. The current in segment 1 is I1. a.
Volgvan

Answer:

hello your question is incomplete attached below is the complete question

answer :

a) I1 = I2

b) J1 > J2

c) E 1 > E2

d) ( vd1 ) > ( vd2 )

Explanation:

a) The currents in the two segments are the same  i.e. I1 = I2  and this is because the segments are connected in series

b) Comparing the current densities J1 and J2 in the two segments

note : current density ∝ 1 / area

The area of the second segment is > the area of first segment  therefore

J1 > J2

J1 ( current density of first segment )

J2 ( current density of second segment )

c) Comparing the electric field strengths E1 and E2

 note : electric field strength ∝ current density

since current density of first segment is > current density of second segment  and conductivity of the materials are the same hence

E 1 > E2

d) Comparing the drift speeds Vd1 and Vd2

( vd1 ) > ( vd2 )

this because  ; vd ∝ current density

7 0
3 years ago
A ball on a frictionless plane is swung around in a circle at constant speed. The acceleration points in the same direction as t
Kitty [74]
I believe it is False, only because the plane is Frictionless. Hope this helps, good luck.
4 0
3 years ago
Read 2 more answers
Suppose that you are standing on a train accelerating at 0.39g. What minimum coefficient of static friction must exist between y
timurjin [86]

Answer:

0.39

Explanation:

In order not to slide, you must have exactly the same acceleration of the train:

a=0.39 g

where

g = 9.81 m/s^2

There is only one force acting on you: the static frictional force that "pulls" you forward, and it is given by

F_s = \mu_s mg

According to Newton's second law, the net force acting on you (so, the frictional force) must be equal to your mass times the acceleration, so we have

F= ma = \mu_s mg

from which we find

\mu_s = \frac{a}{g}=\frac{0.39 g}{g}=0.39

so, the minimum coefficient of static friction must be 0.39.

5 0
3 years ago
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