Which of the following is a device

Projectile motion 1 and 2

shtirl [24]

1. Vf =62.57778m/s

h=62.5778

theta final=45°????

8 0

3 years ago

After polishing his 2-kg wrestling trophy, Mike sets it down on the ground and walks away to find more polish. Meanwhile, Julie

klio [65]

1) The initial momentum of the trophy is zero

2) The initial momentum of the bowling ball is 160 kg m/s

3) The total momentum before the collision is 160 kg m/s

4) The total momentum of the system after the collision is 160 kg m/s

5) The final velocity of the trophy is 32 m/s

Explanation:

1)

The momentum of an object is given by

$p=mv$

where

m is the mass of the object

v is its velocity

In this problem, the data for the trophy before the collision are:

m = 2 kg is the mass

v = 0 is its initial velocity

Therefore, the initial momentum of the trophy is

$p_1=(2)(0)=0$

2)

Using the same equation used in part 1), the initial momentum of the bowling ball is

$p=mv$

where

m is the mass of the bowling ball

v is its initial velocity

The data of the problem are

m = 8 kg is the mass

v = 20 m/s is the velocity

Substituting,

$p_2=(8)(20)=160 kg m/s$

3)

The total momentum of the system before the collision is given by the sum between the initial momentum of the trophy and the initial momentum of the bowling ball:

$p_i = p_1 + p_2$

where

$p_1$ is the initial momentum of the trophy

$p_2$ is the initial momentum of the ball

Here we have

$p_1 = 0$

$p_2 = 160 kg m/s$

Therefore, the total momentum is

$p_i = 0 + 160 = 160 kg m/s$

4)

According to the law of conservation of momentum, for an isolated system (=no external unbalanced forces acting on the system), the total momentum of the system is conserved before and after the collision:

$p_i = p_f$

where

$p_i$ is the total momentum before the collision

$p_f$ is the total momentum after the collision

If we consider the system in the problem to be isolated (i.e. no frictional forces acting on the ball or the trophy), we can therefore say that the total momentum after the collision must be equal to the total momentum before the collision: therefore,

$p_f = 160 kg m/s$

5)

We can write the total momentum after the collision as

$p_f = m_1 v_1 + m_2 v_2$

where:

$m_1 = 2 kg$ is the mass of the trophy

$v_1$ is the final velocity of the trophy

$m_2 = 8 kg$ is the mass of the bowling ball

$v_2 = 12 m/s$ is the final velocity of the ball

Since we also know the value of the final total momentum,

$p_f = 160 kg m/s$

we can solve the equation to find the velocity of the trophy:

$v_1 = \frac{p_f - m_2 v_2 }{m_1}=\frac{160-(8)(12)}{2}=32 m/s$

Learn more about momentum:

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#LearnwithBrainly

4 0

3 years ago

Write two differences between practical machine and ideal machine

slava [35]

Answer:

Ideal machines are said to have parts which are weightless , frictionless and strings if any are inextensible. Practical machines have parts which have weights, friction and it causes some loss of work done. So these are the differences between Ideal Machine and Practical Machines.

6 0

2 years ago

An astronomer observes that the wavelength of light from a distant star is shifted toward the red part of the visible spectrum.

balandron [24]

Answer:

The distance between the earth and the star is increasing.

Explanation:

When we observe an object and its electromagnetic radiation has been displaced to blue, it means that it is getting closer to us, causing the light waves it emits to get closer together and its wavelength to decrease towards blue, this is knowm as blueshift.

On the contrary, when an object is rapidly moving away from us, the light waves or electromagnetic radiation it emits have been stretched from their normal wavelength to a longer wavelength, towards the red part of the spectrum. This is known as redshift.

This phenomenon of changes in wavelength and frequency due to movement (whether the source approaches or moves away) is described by the Doppler effect.

So for this case because the light we perceive from the star has moved to the red part of the visible spectrum, we can conclude that it is moving away from the earth, and that the distance between the star and the earth is increasing.

7 0

3 years ago

A small fish is dropped by a pelican that is rising steadily at 0.500 m/s. How far below the pelican is the fish after 2.50 s?

sesenic [268]

Refer to the diagram shown below.

h = original height of the pelican when the fish is dropped (not relevant).
S = distance traveled by the fish as a function of time, measured upward.
u = 0.5 m/s, the upward velocity with which the fish is dropped.
g = 9.8 m/s², the acceleration due to gravity.

Use the following equation:
S = ut + (1/2)gt²

S = (0.5 m/s)*(2.5 s) + 0.5*(-9.8 m/s²)*(2.5 s)²
= -29.375 m

The negative sign means that the fish drops by 29.375 m from the original height of h.

Answer: The fish is 29.375 m below where the pelican dropped it after 2.5 s.

7 0

3 years ago