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4 years ago

12

Suppose that Paul D. Trigger fires a bullet from a gun. Will the speed of the bullet leaving the muzzle will be the same as the

speed of the recoiling gun? Explain.

1 answer:

Y_Kistochka [10]4 years ago

6 0

Answer:

Explanation:

The speed of the bullet will only be the same if the mass of the bullet is equal to the mass of the gun. This because only their momentum ( product of mass and velocity) is conserved; the momentum before the gun and pullet was fired is equal to the momentum of the recoil of the gun and the momentum of the released bullet. The smallest mass receives the highest speed (the bullet) while the larger mass ( the gun) receives the lowest speed ( speed of recoil).

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Acceleration can be best described as:

zhuklara [117]

Answer:

The ability to speed up

Explanation:

Speed up since the motion is rising , thats acceleration

3 0

4 years ago

A student wants to demonstrate entropy using the songs on her portable

Salsk061 [2.6K]

Answer:

Explanation:

a

8 0

3 years ago

Suppose there are two identical gas cylinders. One contains the monatomic gas krypton (Kr), and the other contains an equal mass

tensa zangetsu [6.8K]

Answer:

Explanation:

Let equal mass of Ne and Kr be m gm

no of moles of Ne and Kr will be m / 20 and m / 84 ( atomic weight of Ne and Kr is 20 and 84 )

Let the pressure and volume of both the gases be P and V respectively .

The temperature of Ne be T₁ and temperature of Kr be T₂.

For Ne

PV = (m / 20) x R T₁

For Kr

PV = (m / 84) x R T₂

T₁ / T₂ = 84 / 20

We know that

average KE of an atom of mono atomic gas = 3 / 2 x k T

k is boltzmann constant and T is temperature .

KEKr/KENe = T₂ / T₁

= 20 / 84

4 0

3 years ago

A projectile of mass 6.8 kg kg is shot horizontally with an initial speed of 14.5 m/s from a height of 26.7 m above a flat deser

sleet_krkn [62]

Answer:

Explanation:

Given

mass of projectile $m=6.8\ kg$

initial horizontal speed $u_x=14.5\ m/s$

height $h=26.7\ m$

Considering vertical motion

velocity gained by projectile during 26.7 m motion

$v^2-u^2=2 as$

v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v^2-(0)^2=2\times (9.8)\times (26.7)$

$v=\sqrt{523.32}$

$v=22.87\ m/s$

Horizontal velocity will remain same as there is no acceleration

final velocity $v_{net}=\sqrt{(v)^2+(u_x)^2}$

$v_{net}=\sqrt{733.57}=27.08\ m/s$

Initial kinetic Energy $K_i=\frac{1}{2}mu_x^2$

$K_i=\frac{1}{2}\times 6.8\times (14.5)^2=714.85\ J$

Final Kinetic Energy $K_f=\frac{1}{2}mv_{net}^2$

$K_f=\frac{1}{2}\times 6.8\times (27.08)^2$

$K_f=2493.30\ J$

Work done by all the force is equal to change in kinetic Energy of object

Work done by gravity is $W_g$

$W_g=\Delta K$

$W_g=2493.30-714.85=1778.45\ J$

8 0

3 years ago

A cylinder with rotational inertia I1=2.0kg·m2 rotates clockwise about a vertical axis through its center with angular speed ω1=

Mrac [35]

Answer:

<em>a) 0.67 rad/sec in the clockwise direction.</em>

<em>b) 98.8% of the kinetic energy is lost.</em>

Explanation:

Let us take clockwise angular speed as +ve

For first cylinder

rotational inertia $I$ = 2.0 kg-m^2

angular speed ω = +5.0 rad/s

For second cylinder

rotational inertia $I$ = 1.0 kg-m^2

angular speed = -8.0 rad/s

The rotational momentum of a rotating body is given as = $I$ ω

where $I$ is the rotational inertia

ω is the angular speed

The rotational momenta of the cylinders are:

for first cylinder = $I$ ω = 2.0 x 5.0 = 10 kg-m^2 rad/s

for second cylinder = $I$ ω = 1.0 x (-8.0) = -8 kg-m^2 rad/s

The total initial angular momentum of this system cylinders before they were coupled together = 10 + (-8) = <em>2 kg-m^2 rad/s</em>

When they are coupled coupled together, their total rotational inertia $I_{t}$ = 1.0 + 2.0 = 3 kg-m^2

Their final angular rotational momentum after coupling = $I_{t}$ $w_{f}$

where $I_{t}$ is their total rotational inertia

$w_{f}$ = their final angular speed together

Final angular momentum = 3 x $w_{f}$ = 3 $w_{f}$

According to the conservation of angular momentum, the initial rotational momentum must be equal to the final rotational momentum

this means that

2 = 3 $w_{f}$

$w_{f}$ = final total angular speed of the coupled cylinders = 2/3 = <em>0.67 rad/s</em>

From the first statement, <em>the direction is clockwise</em>

b) Rotational kinetic energy = $\frac{1}{2} Iw^{2}$

where $I$ is the rotational inertia

$w$ is the angular speed

The kinetic energy of the cylinders are:

for first cylinder = $\frac{1}{2} Iw^{2}$ = $\frac{1}{2}*2*5^{2}$ = 25 J

for second cylinder = $\frac{1}{2}*1*8^{2}$ = 32 J

Total initial energy of the system = 25 + 32 = 57 J

The final kinetic energy of the cylinders after coupling = $\frac{1}{2}I_{t}w^{2} _{f}$

where

where $I_{t}$ is the total rotational inertia of the cylinders

$w_{f}$ is final total angular speed of the coupled cylinders

Final kinetic energy = $\frac{1}{2}*3*0.67^{2}$ = 0.67 J

kinetic energy lost = 57 - 0.67 = 56.33 J

percentage = 56.33/57 x 100% = <em>98.8%</em>

6 0

3 years ago