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geniusboy [140]
3 years ago
7

Object A of mass 0.70 kg travels horizontally on a frictionless surface at 20 m/s. It collides with object B, which is initially

stationary and rebounds with 70% of its initial kinetic energy. What is the magnitude of the change in momentum of object A?
Physics
1 answer:
AnnZ [28]3 years ago
6 0

Answer:

25.71 kgm/s

Explanation:

Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.

Given that K₂ = 0.7K₁

1/2mv₂² = 0.7(1/2mv₁²)

v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s

Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.

Th magnitude of object A's momentum change is thus 25.71 kgm/s

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In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i
RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is 2.7 rad/s^2

Explanation:

a)

In a uniform circular motion, the centripetal acceleration is given by

a_c=\omega^2 r

where:

\omega is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

\omega=1.7 rad/s is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity (g=9.8 m/s^2), so:

a_c=3.2 g = 3.2(9.8)=31.4 m/s^2

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m

b)

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

a=4.4 g

So, 4.4 times the acceleration due to gravity.

The total acceleration is the resultant of the centripetal acceleration (a_c) and the tangential acceleration (a_t):

a=\sqrt{a_c^2+a_t^2}

We know that:

a = 4.4g

a_c = 3.2 g

So, we can find the tangential acceleration:

a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2

The angular acceleration is related to the tangential acceleration by

\alpha = \frac{a_t}{r}

where r = 10.9 m is the length of the centrifuge. Substituting,

\alpha = \frac{29.6}{10.9}=2.7 rad/s^2

Learn more about centripetal and angular acceleration here:

brainly.com/question/2562955

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

8 0
3 years ago
When a Lunar Module landed on the Moon, it used thrusters to slow its descent to the surface. When other spacecraft are returned
mote1985 [20]
A parachute is a device designed to generate a LOT of air resistance.
Parachutes do that very well in places where there's any air to work with.
There is no air on the Moon.

7 0
3 years ago
Technician A says that the freezing point of a substance is the temperature at which the substance changes state from a solid to
pashok25 [27]

Answer:

Technician B is correct

Explanation:

Freezing is a method of conversion of substance in its liquid state to solid state. It is the process by which a liquid substance changes to a solid at a particular temperature.

Increasing the pressure and decreasing the temperature of a liquid increases its freezing point. For example, in other to freeze water i.e to change water to ice, it has to be kept in a fridge at a temperature lower than the temperature of the water. The essence of covering the fridge after placing the water in the fridge is to increase the pressure of the liquid hence increasing its freezing rate.

Based on the above explanation, it can be concluded that technician B is correct.

5 0
3 years ago
In order to attain orbit around earth, the ATLAS-V rocket must accelerate up to a
IrinaK [193]
A) average acceleration = final velocity - initial velocity / time

= 7700 - 0 / 11

= 700ms^-2


B) force = mass x acceleration

= (3.05 x 105) x 700

= 320.25 x 700

= 224,175N
7 0
3 years ago
A weight of 1400 pounds is suspended from two cables as shown in the figure. What is the tension in the left cable? _________ po
juin [17]

Answer:

Following are the solution to this question:

Explanation:

Law:

\to \theta= 180^{\circ}- 50^{\circ}- 25^{\circ}

      = 180^{\circ}- 75^{\circ}\\\\= 105^{\circ}

\to \frac{T_{L}}{\sin (90+50)}= \frac{T_{R}}{\sin (25+90)}=\frac{1400}{\sin (105)}

\to T_L=931.65 \ pounds \\\\ \to T_R=1313.59 \ pounds \\\\

4 0
3 years ago
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