0.75 km expressed in centimeters

Ratling [72]

Answer:

D) momentum of cannon + momentum of projectile= 0

Explanation:

The law of conservation of momentum states that the total momentum of an isolated system is constant.

In this case, the system cannon+projectile can be considered as isolated, because no external forces act on it (in fact, the surface is frictionless, so there is no friction acting on the cannon). Therefore, the total momentum of the two objects (cannon+projectile) must be equal before and after the firing:

$p_i = p_f$

But the initial momentum is zero, because at the beginning both the cannon and the projectile are at rest:

$p_i = 0$

So the final momentum, which is sum of the momentum of the cannon and of the projectile, must also be zero:

$p_f = p_{cannon}+p_{projectile} =0$

6 0

4 years ago

You are a nurse in a hospital. You are told to move a patient through three corridors each measuring 14.33 meters. Because of fr

dimulka [17.4K]

Work done by force is given by formula

$W = F.d$

now here work done against friction is given so force of friction here is

$F_f = 848 N$

It moved by three corridors with each measures

$d = 14.33 m$

so total distance will be

$s = 14.33 \times 3 = 43 m$

now from above formula of work done we will say

$W = 848 \times 43$

$W = 36464 J$

so above is the work done to move three corridors

3 0

3 years ago

A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.

enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, $KE_{0} = 0.5 mu^{2}$

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, $KE_{f} = 0.5mv^{2}$

a) Using the principle of energy conservation,

$KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s$

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0

4 years ago

Scientists are making plans to put a probe in orbit around Earth. They want the probe to enter the orbit shown below.

iris [78.8K]

An arrow which shows the direction that the probe should be moving in order for it to enter the orbit is X.

<h3>What is an orbit?</h3>

An orbit can be defined as the curved path through which a astronomical (celestial) object such as planet Earth, in space move around a Moon, Sun, planet or star.

In this scenario, if the scientists want the probe to enter the orbit they should ensure that probe moves in direction X. This ultimately implies that, the probe must move in the same direction as the orbit, in order to enter it.

Read more on orbit here: brainly.com/question/18496962

#SPJ1

6 0

3 years ago

Read 2 more answers

a body is projected at an angle of 30 degrees to the horizontal with a velocity of 15m/s, calculate the time it takes to reach t

andreev551 [17]

Explanation:

The vertical component the velocity of the projectile is 15 m/s x sin 30 = 7.5 m/s.

The body is accelarating downwards at 10 m/s^2.

This means that every second its upward velocity reduces by 10 m/s.

So if the body is travelling upwards at 7.5 m/s then how long does it take for the velocity to become 0?