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Vaselesa [24]
1 year ago
8

Newton's laws of motion are true on earth and space.

Physics
1 answer:
frez [133]1 year ago
8 0

Answer:

False

Explanation:

It is a common misunderstanding that objects in space have no weight. If that were true, they would just float away from the Earth, the Sun and the other planets. Objects in low Earth orbit experience about 90% of the weight that they feel on the surface of the Earth.

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Artemon [7]

Answer:

constant volicty of the pumper when they hit ground 7.03-/s

8 0
3 years ago
Bádminton is played to a score of
Temka [501]
Badminton is played to a score of 21 points
4 0
2 years ago
Read 2 more answers
A graph of the net force F exerted on an object as a function of x position is shown for the object of mass M as it travels a ho
saul85 [17]

The change in kinetic energy is \Delta K = 3Fd

Explanation:

According to the work-energy theorem, the work done on an object is equal to the change in kinetic energy of the object. Mathematically:

W=K_f -K_i= \Delta K

where :

W is the work done on the object

K_f is the final kinetic energy of the object

K_i is the initial kinetic energy

Also, the work done on an object is (assuming that the force is applied parallel to the motion of the object):

W=F\Delta x

where

F is the magnitude of the force

\Delta x is the displacement of the object

In this problem, the force acting on the object is

F

While the displacement is the horizontal distance travelled, so

\Delta x = 3d

Therefore, the work done is

W=(F)(3d)=3Fd

And so the change in kinetic energy is

\Delta K = 3Fd

Learn more about work and kinetic energy:

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/6536722

#LearnwithBrainly

5 0
2 years ago
Ms. Stafford wants to try racing with a push start. A student pushes her at 5 m/s before the rocket kicks in. The rocket still o
Gnom [1K]
The initial velocity of Ms. Stafford is v_0 = 5 m/s, while her acceleration is 
a=4 m/s^2
This is a uniform accelerated motion, so we can calculate the total distance travelled by Ms. Stafford in a time of t=7.0 s using the law of motion for a uniform accelerated motion:
S=v_0t +  \frac{1}{2} at^2 = (5 m/s)(7.0 s)+ \frac{1}{2}(2 m/s^2)(7.0 s)^2 = 84 m
8 0
2 years ago
1. Is the sequence geometric? If so, identify the common ratio.
frozen [14]
1) <span>yes;2 6*2=12 12*2=24 24*2=48
2)</span><span>Next Term (or nth term) = ar^n-1 
</span>
a = first term, i.e. 5 
<span>r = common ratio i.e. 3 (as 15/5=3 and 45/15=3 </span>
<span>n = .. </span>
<span>as you already have 1st , 2nd and 3rd terms</span>
<span>substituting now </span>
<span>T4= ar^n-1 </span>
<span>= 5*3^4-1 </span>
<span>= 5*3^3 </span>
<span>= 5*27 </span>
<span>T4 = 135 
</span>T5= ar^n-1 
<span>= 5*3^5-1 </span>
<span>= 5*3^4 </span>
<span>= 5*81 </span>
<span>T5 = 405 </span>


6 0
3 years ago
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