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dexar [7]
2 years ago
7

A student pushes a block from rest across a frictionless surface while the block is in front of a motion detector for 3 seconds.

The block first speeds up for the 1st second, then moves at constant velocity for the 2nd second, and finally slows down during the final second.
Which statement best explains the motion of the block using Newton's 1st Law?
There is a net force on the block during the 2nd second as a net force keeps the motion constant.
There is a net force on the block during the 3rd second as a net force keeps the motion constant.
There is no net force on the block during the 2nd second as zero net force keeps the motion constant.
There is no net force on the block during the 3rd second as zero net force causes the motion to accelerate.​
Physics
1 answer:
Furkat [3]2 years ago
5 0

The true statement according to Newton's laws of motion is; "there is no net force on the block during the 2nd second as zero net force keeps the motion constant."

According to the Newton first law of motion, an object will continue in its state of rest or uniform motion unless it is acted upon by a net force. The action of a net force leads to an acceleration.

In the 2nd second of the journey, there is no net force on the block hence the block continues to move at constant velocity.

Learn more: brainly.com/question/13678295

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At t = 0 the components of a radio-controlled car's velocity are vx = 0.5 m/s and vy = 1.2 m/s. Find the components of the car's
FinnZ [79.3K]

Answer:

x-component of velocity = 5.7 m/s

y-component of velocity = -1.4 m/s          

Explanation:

Use first equation of motion to find components of velocity at a given time:

v = u + at

where, v is  the final velocity, u is the initial velocity, a is the acceleration and t is the time.

Given:

u_x= 0.5 m/s\\u_y=1.2 m/s\\a_x=2 m/s^2\\a_y=-1m/s^2\\t = (2.6-0)s =2.6 s

v_x=u_x+a_xt\\\Rightarrow v_x=0.5+2\times 2.6\\\Rightarrow v_x=5.7 m/s

v_y=u_y+a_yt\\\Rightarrow v_y=1.2-1\times 2.6\\\Rightarrow v_y=-1.4 m/s

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PLZ HELP I WILL GIVE BRAINLIEST
Vaselesa [24]

Answer:

12.7m/s

Explanation:

Given parameters:

Mass of diver  = 77kg

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Unknown:

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             v²   = u²  + 2gH

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

H is the height

 Now insert the parameters and solve;

       v² = 0²  +  2 x 9.8 x 8.18

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A rocket travels at a speed of 14,000 m/s. What is the distance travelled by the rocket in 150s?
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1) The distance travelled by the rocket can be found by using the basic relationship between speed (v), time (t) and distance (S):

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Rearranging the equation, we can write

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In this problem, v=14000 m/s and t=150 s, so the distance travelled by the rocket is

S=vt=(14000 m/s)(150 s)=2.1 \cdot 10^6 m


2) We can solve the second part of the problem by using the same formula we used previously. This time, t=300 s, so we have:

S=vt=(14000 m/s)(300 s)=4.2 \cdot 10^6 m/s

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