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aniked [119]
3 years ago
11

1.imagine you were able to throw a ball in a frictionless environment such as outer space.once you let go of the ball,what will

happen to the ball according to the law of inertia?
a.it will travel in a straight forever line
b.it will travel striaight line but eventually slow down
c.it will continue to speed up forever
d.it will slow down and turn
2.a car accelerates down the road .what is the reaction to the tires pushing on the road?
a.the car in the tires moving the car
b.the masses of the car pushing on the tires
c.the road pushing on the tires
d.the road moving away from the tires
Physics
1 answer:
Scorpion4ik [409]3 years ago
5 0
Imagine you were able to throw a ball in a frictionless environment
such as outer space.  Once you let go of the ball, it will travel forever
in a straight line, and at a constant speed.  (At least until it bumps into
something.)

A car accelerates down the road.  The reaction to the tires pushing
on the road is the road pushing on the tires.

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A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsul
jeka94

The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

<h3>Given:</h3>

Mass of the astronaut, m_a = 126 kg

Speed he acquires, v_{a}  = 2.70 m/s

Mass of the space capsule, m_{c} = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

P_f = m_av_a + m_cv_c

P_I = 0

m_av_a + m_cv_c = 0

v_c =\frac{- m_a v_a}{m_c}}\\\\

   = \frac{126* 2.70}{1800}

   = - 0.189 m/s

Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

    = 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

  = 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = \frac{1}{2} m v^2

     = \frac{1}{2} × 126 × (2.70) ^2

     = 459.27 J

The Kinetic Energy of the capsule;

K.E = \frac{1}{2} m v^2

     = \frac{1}{2}×1800×(0.189) ^2

     = 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Learn more about kinetic energy here:

brainly.com/question/26520543

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3 0
2 years ago
A coach shouts "Go!" from the finish line of a 100.0-m track to the runners.
irina [24]

Answer:

a. Speed = 342.5 meters per seconds.

b. Wavelength = 2.0 meters

Explanation:

Given the following data;

Distance = 100m

Time = 292 milliseconds to seconds = 292/1000 = 0.292 seconds

Frequency = 171 Hz

a. To find the speed of sound in air;

Speed = distance/time

Speed = 100/0.292

Speed = 342.5 m/s

b. To find the wavelength;

Wavelength = speed/frequency

Wavelength = 342.5/171

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A student of mass 40kg takes 10s to run up a flight of 50steps. If each step is 15cm high, calculate the Potential Energy of the
astra-53 [7]

Answer:

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Explanation:

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Substitute the given numbers: (I take acceleration due to gravity as 9.81 m s^-2)

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Calculate the difference between<br>the points B (-2, 3) and C (-5,8)​
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Explanation:

the answer to this question would be mj.us 1 and minus 3

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