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3 years ago

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A rifle of mass M is initially at rest. A bullet of mass m is fired from the rifle with a velocity of v relative to the ground.

What is the expression that gives the velocity of the rifle relative to the ground after the bullet is fired?

1 answer:

Iteru [2.4K]3 years ago

7 0

Answer:

$V_{rifle} = -\frac{mV_{bullet} }{M}$ .

Explanation:

According to conservation of momentum, the momentum of rifle before fire is equal to the momentum of bullet after fire plus momentum of rifle after fire.

This in mathematics translates to.

$(M+m)v_{rifle-before-fire}=Mv_{rifle-after-fire}+mv_{bullet}$

rifle is stationary before fire so $v_{rifle-before-fire}=0$ this means that left side is 0.

$0=Mv_{rifle-after-fire}+mv_{bullet}$

solving for $v_{rifle-after-fire}$ in above gives us our answer.

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the coefficient of static friction between a 40 kg picnic table and the ground below is .43. what is the greatest horizontal for

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The force equals the coefficient of static friction times the weight. Use gravity g=9.8 m/s^2
0.43*40*9.8=16.856 N

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A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls

nekit [7.7K]

Given:

density of air at inlet, $\rho_{a} = 1.20 kg/m_{3}$

density of air at inlet, $\rho_{b} = 1.05 kg/m_{3}$

Solution:

Now,

$\dot{m} = \dot{m_{a}} = \dot{m_{b}}$

$\rho_{a} A v_{a} = \rho _{b} Av_{b}$ (1)

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$v_{a}$ = velocity of air at inlet

$v_{b}$ = velocity of air at outlet

Now, using eqn (1), we get:

$\frac{v_{b}}{v_{a}} = \frac{\rho_{a}}{\rho_{b}}$

$\frac{v_{b}}{v_{a}} = \frac{1.20}{1.05}$ = 1.14

% increase in velocity = $1.14\times 100$ =114%

which is 14% more

Therefore % increase in velocity is 14%

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3 years ago

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The CERN particle accelerator is circular with a circumference of 7.0 km.

Contact [7]

Answer:

$a_c=2.0196\times 10^{13}\ m/s^2$

$F=3.37273\times 10^{-14}\ N$

Explanation:

m = Mass of proton = $1.67\times 10^{-27}\ kg$

v = Speed of proton = 0.5c = $0.5\times 3\times 10^8=1.5\times 10^8\ m/s$

Circumference of the colider is 7 km

$P=2\pi r\\\Rightarrow r=\frac{P}{2\pi}\\\Rightarrow r=\frac{7000}{2\pi}\ m$

$a_c=\frac{v^2}{r}\\\Rightarrow a_c=\frac{\left(1.5\times 10^8\right)^2}{\frac{7000}{2\pi}}\\\Rightarrow a_c=2.0196\times 10^{13}\ m/s^2$

Centripetal acceleration is $2.0196\times 10^{13}\ m/s^2$

$F_c=ma_c\\\Rightarrow F_c=1.67\times 10^{-27}\times 2.0196\times 10^{13}\\\Rightarrow F=3.37273\times 10^{-14}\ N$

Force on protons is $3.37273\times 10^{-14}\ N$

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3 years ago

If the resistance of dry skin is 200 times larger than the resistance of wet skin, how do the maximum voltages without shock com

Lelu [443]

Answer:

<em> B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.</em>

<em></em>

Explanation:

This is the complete question

A person will feel a shock when a current of greater than approximately 100μ A flows between his index finger and thumb. If the resistance of dry skin is 200 times larger than the resistance of wet skin, how do the maximum voltages without shock compare in each scenario?

A) The voltage on dry skin needs to be 200 times smaller than the voltage on wet skin.

B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.

C) The voltage on dry skin is the same as the voltage on wet skin.

D) The voltage on dry skin needs to be 40,000 times larger than the voltage on wet skin.

Ohm's law states that electric current is proportional to voltage and inversely proportional to resistance.

the equation is written as

V = IR

Where V is the voltage

I is the current

R is the resistance

for this case, the current I is 100μ A = 100 x 10^16 A

resistance of wet skin = R

resistance of dry skin = 200R

for the wet skin, voltage will be

V = IR = $100*10^{-6} R$

for dry skin, voltage will be

V = IR = $100*10^{-6}*200R = 0.02R$

Comparing both voltages

$0.02R$ ÷ $100*10^{-6} R$ = 200

<em>this means that the voltage on the wet skin should be 200 times lesser than the voltage on the dry skin or the voltage on the dry skin should be 200 times more than the voltage on the wet skin.</em>

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4 years ago

A 61 kg skater is traveling at 2.5 m/s while carrying a 4.0 kg bowling ball. After he throws the bowling ball forward at twice t

gregori [183]

The final velocity of the skater is 2.34 m/s forward

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system before and after the ball is thrown must be conserved, in absence of external forces.

Before the ball is thrown, the total momentum is:

$p_i = (M+m)u$

where

M = 61 kg is the mass of the skater

m = 4.0 kg is the mass of the ball

u = 2.5 m/s (forward) is the combined velocity of the skater and the ball

After, the ball is thrown at twice the velocity, so the final total momentum is

$p_f = MV+mv$

where

V is the final velocity of the skater

v = 2(2.5) = 5.0 is the final velocity of the ball

Since the total momentum must be conserved, we can write

$p_i = p_f\\(M+m)u = MV+mv\\V=\frac{(M+m)u-mv}{M}=\frac{(61+4.0)(2.5)-(4.0)(5.0)}{61}=2.34 m/s$

So, the skater is moving at 2.34 m/s (forward) after the shot.

Learn more about momentum:

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brainly.com/question/6573742

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3 years ago