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3 years ago

How do I solve the following problem?

Physics

1 answer:

lukranit [14]3 years ago

7 0

Use the impulse-momentum theorem.

$Ft = mv$

Substitute your known values:

$(120kg)(20m/s) = F(1.5)

F= 1600N$

Hope this helps!

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Two sacks contain the same number of identical apples and are separated by a distance r. The two

andrezito [222]

Answer:

The appropriate response will be " $F_2=\frac{3}{4}F$ ". A further explanation is given below.

Explanation:

According to the question,

⇒ $F_1=\frac{G(m_1 m_2)}{r^2}$ ....(equation 1)

and,

⇒ $F_2=\frac{G(m_1 m_2)}{r^2}$ ....(equation 2)

Now,

On dividing both the equations, we get

⇒ $\frac{F_1}{F_2}=\frac{\frac{G(2)(2)}{(1)^2}}{\frac{G(1)(3)}{(1)^2}}$

⇒ $\frac{F_1}{F_2}= \frac{4}{3}$

⇒ $F_2=\frac{3}{4}F$

4 0

3 years ago

A 0.22kg mousetrap car has 3.1 J of potential energy due to the spring in the mouse

Stella [2.4K]

Answer:

r56

Explanation:

7 0

3 years ago

A student has to work the following problem: A block is being pulled along at constant speed on a horizontal surface a distance

brilliants [131]

Answer:

The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.

Explanation:

Since block moves with constant speed

So, frictional force

f = FCosq

Work done by friction

W = - fd

W = - fd Cos q

The answer may be greater or less than - fdSinq. It depends on the value of q which is less than, or equal to 45°.

6 0

3 years ago

The ball will oscillate along the z axis between z=dz=d and z=−dz=−d in simple harmonic motion. What will be the angular frequen

Eddi Din [679]

Answer:

$\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }$

Explanation:

Additional information:

The ball has charge $-q_0$ , and the ring has positive charge $+Q$ distributed uniformly along its circumference.

The electric field at distance $z$ along the z-axis due to the charged ring is

$E_z= \dfrac{kQz}{(z^2+a^2)^{3/2}}.$

Therefore, the force on the ball with charge $-q_0$ is

$F=-q_oE_z$

$F=- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}$

and according to Newton's second law

$F=ma=m\dfrac{d^2z}{dz^2}$

substituting $F$ we get:

$- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=m\dfrac{d^2z}{dz^2}$

rearranging we get:

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=0$

Now we use the approximation that

$z^2+a^2\approx a^2$ (we use this approximation instead of the original $d^2+a^2\approx a^2$ since $z$ , our assumption still holds )

and get

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(a^2)^{3/2}}=0$

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{a^{3}}=0$

Now the last equation looks like a Simple Harmonic Equation

$m\dfrac{d^2z}{dz^2}+kz=0$

where

$\omega=\sqrt{ \dfrac{k}{m} }$

is the frequency of oscillation. Applying this to our equation we get:

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Q}{a^{3}}z=0\\\\m=m\\\\k= \dfrac{kq_0Q}{a^{3}}$

$\boxed{\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }}$

5 0

4 years ago

A 1-kg chunk of putty moving at 1 m/s collides with and sticks to a 5-kg bowling ball initially at rest. The bowling ball and pu

mars1129 [50]

Momentum P is conserved because there are no external forces acting on the system.

P before = P after = m₁ v₁ + m₂ v₂

m₁ = 1
m₂ = 5

before:
v₁ = 1
v₂ = 0

Pᵇ = 1·1 + 5 · 0

after:
v₁ = v₂

Pᵃ = Pᵇ

4 0

3 years ago

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