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3 years ago

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A force in the negative direction of an x-axis is applied for 17ms to a 0.18 kg ball initially moving at 16.0 m/s in the positiv

e direction of the axis. the force varies in magnitude and results in an impulse of magnitude 4.28 ns. what is the velocity of the ball just after the force is applied? indicate the direction of the velocity by its sign.

1 answer:

Marizza181 [45]3 years ago

5 0

Impulse = Ft=mΔv => Δv = Ft/m = 4.28/0.18 = 23.78 m/s

But,
Δv = v1-v2, where v1 = initial velocity = 16 m/s, v2 = final velocity

Therefore,
v1 - v2 = 23.78 => v2 = v1 - 23.78 => v2 = 16 - 23.78 = -7.78 m/s

The velocity of ball after the force is 7.78 m/s in the direction of the force.

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A thin block of soft wood with a mass of 0.072 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is

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Answer:

366.90149 m/s

923.821735 J

324.734 J

Initial Kinetic energy > Final kinetic energy

Explanation:

$m_1$ = Mass of block = 0.072 kg

$m_2$ = Mass of bullet = 4.67 g

$u_1$ = Initial Velocity of block = 0

$u_2$ = Initial Velocity of bullet = 629 m/s

$v_1$ = Final Velocity of block = 17 m/s

$v_2$ = Final Velocity of bullet

In this system the linear momentum is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow v_2=\frac{m_{1}u_{1}+m_{2}u_{2}-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{0.072\times 0+4.67\times 10^{-3}\times 629-0.072\times 17}{4.67\times 10^{-3}}\\\Rightarrow v_2=366.90149\ m/s$

Final Velocity of bullet is 366.90149 m/s

The initial kinetic energy

$K_i=\frac{1}{2}m_2u_2^2\\\Rightarrow K_i=\frac{1}{2}4.67\times 10^{-3}\times 629^2\\\Rightarrow K_i=923.821735\ J$

The final kinetic energy

$K_f=\frac{1}{2}m_2v_2^2+\frac{1}{2}m_1v_1^2\\\Rightarrow K_f=\frac{1}{2}4.67\times 10^{-3}\times 366.90149^2+\frac{1}{2}0.072\times 17^2\\\Rightarrow K_f=324.734\ J$

Initial Kinetic energy > Final kinetic energy

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When Jackson throws a baseball in a straight path what two forces causes the ball to eventually stop moving

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Gravity and wind resistence

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A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m bef

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Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

μ_k = 0.1773

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Answer:

$T=66262.4s$

Explanation:

From the question we are told that:

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Mass $m=5.97 * 10^{24} kg$

Radius $r=6.38 *10^6 m.$

Generally the equation for Satellite Speed is mathematically given by

$V=(\frac{GM}{d} )^{0.5}$

$V=(\frac{6.67*10^{-11}*5.97 * 10^{24}}{6.38 *10^6+2.90 *10^7} )^{0.5}$

$V=3354.83m/s$

Therefore

Period T is Given as

$T=\frac{2 \pi *a}{V}$

$T=\frac{2 \pi *(6.38 *10^6+2.90 *10^7}{3354.83}$

$T=66262.4s$

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A star produces 2x10^26 watts. how much energy does it lose every minutes

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Answer:

Energy loss per minute will be $120\times 10^{26}j$

Explanation:

We have given the star produces power of $2\times 10^{26}W$

We know that 1 W = 1 J/sec

So $2\times 10^{26}W=2\times 10^{26}J/sec$

Given time = 1 minute = 60 sec

So the energy loss per minute $=2\times 10^{26}\times 60=120\times 10^{26}j$

We multiply with 60 we have to calculate energy loss per minute

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4 years ago