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4 years ago

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Objects 1 has half the mass of object 2 and the objects move toward each other and experience an inelastic collision. If both ob

jects do not move after the collision compare the velocity of both objects before the collision.

1 answer:

Vika [28.1K]4 years ago

3 0

Answer:

Explanation:

Given

Object 1 is of mass $\frac{m}{2}$

Object 2 has mass of $m$

They undergone inelastic collision and do not move after collision i.e. the collision is perfectly inelastic

Final momentum of both the object is zero

suppose $v_1$ and $v_2$ are the velocities of object 1 and 2 respectively then

$m_1v_1+m_2v_2=0$

$\frac{v_1}{v_2}=-\frac{m_2}{m_1}$

$\frac{v_1}{v_2}=-2$

$v_1=-2v_2$

i.e. velocity of object 1 is twice the velocity of object 2 but opposite to the direction of object 2

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3. A 92 kg Tarzan is holding on to a level 22m vine. He swings on the vine. What will his speed at the bottom of the swing be?

insens350 [35]

To solve this problem, we must always remember that energy is conserved. In this case, since he is falling down, he has highest potential energy at the top and zero at bottom. While his kinetic energy is zero at the top since he started from rest and highest at the bottom. We can also say that Potential Energy lost is Kinetic Energy gained thus,

- ΔPE = ΔKE ---> one is negative since PE is losing energy

- m g (h2 – h1) = 0.5 m (v2^2 – v1^2)

Where,

m = mass of tarzan (cancel that out)

g = gravitational acceleration

h2 = height at the bottom= 0

h1 = height at top = 22 m

v2 = velocity at the bottom

v1 = velocity at top = 0 (started from rest)

Therefore substituting all values:

- 9.8 (- 22) = 0.5 (v2^2)

v2 = 20.77 m / s (ANSWER)

6 0

3 years ago

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

Caculate the total charge transferred through a wire carrying a current of 0.5 Ameter for a period of 20 seconds.

Marina CMI [18]

charge = current x time = 0.5x 20=10Coulombs

5 0

3 years ago

A vector is<br> In English

zzz [600]

A vector is a quantity or phenomenon that has two independent properties: magnitude and direction.

3 0

3 years ago

A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o

iVinArrow [24]

There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: $v=15 m/s$
- radius of the hill: $r=100 m$

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car $W=mg$ (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
$mg=m \frac{v^2}{r}$
from which we find
$v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s$

7 0

4 years ago