The Moon (mass = 7.35 x 1022 kg) is 3.84 x 108 m from the Earth (mass = 5.98 x 1024 kg).

Consider a rifle, which has a mass of 2.44 kg and a bullet which has a mass of 150 grams and is loaded in the firing chamber.Whe

svetlana [45]

Answer:

a. 0 kgm/s

b. 0 kgm/s

c. 66 kgm/s

d. -66 kgm/s

e. 0 kgm/s

f. -27.05 m/s

g. 173.68 N

h. 12.58 m/s

i. 0.772 m

j. 14487 J

Explanation:

150 g = 0.15 kg

a. Before the bullet is fired, both rifle and the bullet has no motion. Therefore, their velocity are 0, and so are their momentum.

b. The combination is also 0 here since the bullet and the rifle have no velocity before firing.

c. After the bullet is fired, the momentum is:

0.15*440 = 66 kgm/s

d. As there's no external force, momentum should be conserved. That means the total momentum of both the bullet and the rifle is 0 after firing. That means the momentum of the rifle is equal and opposite of the bullet, which is -66kgm/s

e. 0 according to law of momentum conservation.

f. Velocity of the rifle is its momentum divided by mass

v = -66 / 2.44 = -27.05 m/s

g. The average force would be the momentum divided by the time

f = -66 / 0.38 = 173.68 N

h. According the momentum conservation the bullet-block system would have the same momentum as before, which is 66kgm/s. As the total mass of the bullet-block is 5.1 + 0.15 = 5.25. The velocity of the combination right after the impact is

66 / 5.25 = 12.58 m/s

i. The normal force and also friction force due to sliding is

$F_f = N\mu = Mg\mu = 5.25*9.81*0.83 = 42.75N$

According to law of energy conservation, the initial kinetic energy will soon be transformed to work done by this friction force along a distance d:

$W = K_e$

$dF_f = 0.5Mv_0^2$

$d = \frac{Mv_0^2}{2F_f} = \frac{5.25*12.58^2}{2*42.75} = 0.772 m$

j.Kinetic energy of the bullet before the impact:

$K_b = 0.5*m_bv_b^2 = 0.5*0.15*440^2 = 14520 J$

Kinetic energy of the block-bullet system after the impact:

$K_e = 0.5Mv_0^2 = 0.5 * 5.25*12.58^2 = 33 J$

So 14520 - 33 = 14487 J was lost during the lodging process.

3 0

3 years ago

A ball is thrown horizontally at a speed of 24 m/s from the top of a cliff. if the ball hits the ground 4.0 s later, approximate

marishachu [46]

144 m 24 divided by 4 = 6 24 x 6 = 144

6 0

3 years ago

A 0.0208 m diameter coin rolls up a 18.0◦ inclined plane. The coin starts with an initial angular speed of 56.0 rad/s and rolls

anastassius [24]

Answer:

h = 0.0259 m

Explanation:

given,

diameter of the cone = 0.0208 m

radius,r = 0.0104 m

angle of inclination,θ = 18°

initial angular velocity, ω_i = 56 rad/s

final angular velocity ,ω_f = 0 rad/s

height, h = ?

Rotational kinetic energy

$KE_r = \dfrac{1}{2}I\omega^2$

Moment of inertia of coin

$I = \dfrac{1}{2}MR^2$

so,

$KE_r = \dfrac{1}{4}MR^2\omega^2$

Transnational Kinetic energy

$KE_t = \dfrac{1}{2}Mv^2$

v = r ω

$KE_t = \dfrac{1}{2}MR^2\omega^2$

now,

using conservation energy

Kinetic energy of the coin is converted into the potential energy

KE_r + KE_t = PE

$\dfrac{1}{4}MR^2\omega^2 + \dfrac{1}{2}MR^2\omega^2 = Mgh$

$\dfrac{3}{4}R^2\omega^2=gh$

$\dfrac{3}{4}\times 0.0104^2\times 56^2=9.8\times h$

h = 0.0259 m

Vertical height gain by the coin is equal to 0.0259 m

7 0

4 years ago

A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th

ankoles [38]

<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Time, t = 3 s

Substituting

s = ut + 0.5 at²

s = 0 x 3 + 0.5 x 9.81 x 3²

s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

4 0

4 years ago

Scientists have investigated how quickly hoverflies start beating their wings when dropped both in complete darkness and in a li

Vesna [10]

Answers:

A) 0.204 m

B) 0.285 s

Explanation:

<h2>Answer A:</h2>

This described situation is free fall, this means the initial velocity of the fly is zero, and the equation that will be used is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y$ is the final height of the fly

$y_{o}=40 cm=0.4 m$ is the initial height of the fly

$V_{o}=0$ is the initial velocity of the fly

$t=200(10)^{-3} s$ is the time

$g=9.8 m/s^{2}$ is the acceleration due to gravity

$y=0.4 m+0-\frac{1}{2}(9.8 m/s^{2})(200(10)^{-3} s)^{2}$ (2)

$y=0.204 m$ (3) This is the distance at which the fly would begin to beat its wings

<h2>Answer B:</h2>

In this part we will also use equation (1), but we will find the time:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0$ is the final height of the fly

$y_{o}=40 cm=0.4 m$ is the initial height of the fly

$V_{o}=0$ is the initial velocity of the fly

$t$ is the time we need to find

$g=9.8 m/s^{2}$ is the acceleration due to gravity

$0=0.4 m+0-\frac{1}{2}(9.8 m/s^{2})t^{2}$ (4)

Isolating $t$ :

$t^{2}=\frac{(-2)(-0.4 m)}{9.8 m/s^{2}}$ (5)

$t=0.285 s$ (6) This is the time it would take for a fly to hit the bottom of the box

6 0

4 years ago