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2 years ago

A 35 kg object has -450 kgm/s of momentum. Calculate its velocity.

Physics

1 answer:

shtirl [24]2 years ago

7 0

Hi there!

Recall the equation for momentum:

$\large\boxed{p = mv}\\$

p = momentum (kgm/s)

m = mass (kg)

v = velocity (m/s)

We can plug in the given values to solve for velocity:

$-450 = 35v\\\\v = \frac{-450}{35} = \boxed{12.86 m/s}$

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3 years ago

A square loop of wire is held in a uniform 0.24 T magnetic field directed perpendicular to the plane of the loop. The length of

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Answer:

Explanation:

Given that,

Magnetic field of 0.24T

B = 0.24T

Field perpendicular to plane i.e 90°

Rate of decrease of length of side of square is 5.4cm/s

dL/dt = 5.4cm/s = 0.054m/s

Since it is decreasing

Then, dL/dt = -0.054m/s

When L is 14cm, what is the EMF induced?

L = 14cm = 0.14m

EMF is give as

ε = - dΦ/dt

Where flux is given as

Φ = BA

Where A is the area of the square

A = L²

Then, Φ = BL²

Substituting this into the EMF

ε = - dΦ/dt

ε = - d(BL²)/dt

B is constant

ε = - Bd(L²)/dt

ε = -2BL dL/dr

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4 years ago

After polishing his 2-kg wrestling trophy, Mike sets it down on the ground and walks away to find more polish. Meanwhile, Julie

klio [65]

1) The initial momentum of the trophy is zero

2) The initial momentum of the bowling ball is 160 kg m/s

3) The total momentum before the collision is 160 kg m/s

4) The total momentum of the system after the collision is 160 kg m/s

5) The final velocity of the trophy is 32 m/s

Explanation:

The momentum of an object is given by

$p=mv$

where

m is the mass of the object

v is its velocity

In this problem, the data for the trophy before the collision are:

m = 2 kg is the mass

v = 0 is its initial velocity

Therefore, the initial momentum of the trophy is

$p_1=(2)(0)=0$

Using the same equation used in part 1), the initial momentum of the bowling ball is

$p=mv$

where

m is the mass of the bowling ball

v is its initial velocity

The data of the problem are

m = 8 kg is the mass

v = 20 m/s is the velocity

Substituting,

$p_2=(8)(20)=160 kg m/s$

The total momentum of the system before the collision is given by the sum between the initial momentum of the trophy and the initial momentum of the bowling ball:

$p_i = p_1 + p_2$

where

$p_1$ is the initial momentum of the trophy

$p_2$ is the initial momentum of the ball

Here we have

$p_1 = 0$

$p_2 = 160 kg m/s$

Therefore, the total momentum is

$p_i = 0 + 160 = 160 kg m/s$

According to the law of conservation of momentum, for an isolated system (=no external unbalanced forces acting on the system), the total momentum of the system is conserved before and after the collision:

$p_i = p_f$

where

$p_i$ is the total momentum before the collision

$p_f$ is the total momentum after the collision

If we consider the system in the problem to be isolated (i.e. no frictional forces acting on the ball or the trophy), we can therefore say that the total momentum after the collision must be equal to the total momentum before the collision: therefore,

$p_f = 160 kg m/s$