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Anestetic [448]
3 years ago
14

A car is traveling north. can its acceleration vector ever Point South? explain​

Physics
1 answer:
Scrat [10]3 years ago
7 0
Yes.

The acceleration vector WILL point south when the car is slowing down while traveling north

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I give a ball a push on an acclivity. The "start velocity" is on 7m/s. The time it took the ball to get back to me was 10 second
Degger [83]

Answer:

7÷10

Explanation:

initial velocity=7m/s

final velocity=0m/s

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4 years ago
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After the first solid grains formed in our solar system, these particles could then grow by the process of ____, the collision a
labwork [276]

Answer:

Accretion

Explanation:

Accretion is the process by which there is an accumulation of particles into a bigger object by attracting more mass by gravitational force into an accretion disk. This is one of the first steps in the formation of our solar system. There was a collapse of a gas cloud which resulted in most of the mass collecting in the center leading to the formation of the sun and the rest spread out forming the planets.

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3 years ago
The speed for a car that went a distance of 125 miles in 2 hours time is ______. Question 2 options: 250 meters/sec 62.5 miles/h
Goryan [66]

Answer:

62.5 miles per hour

Explanation:

Speed = Distance travelled / Time taken

Speed = 125/2 = 62.5

You derive the units of the speed...by using the speed formula....,

Speed = Distance/Time

Speed = miles/hour

Hence the units for the speed = miles/hour

5 0
3 years ago
What is the study of the mone and stars
maks197457 [2]
The study of the moon as well as the stars is astronomy
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3 years ago
A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,
alexgriva [62]
The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
S= \frac{1}{2} a t^2
where a is the acceleration. Using the data of the problem, we can find a:
a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J

and so the power output is
P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W

b) power output at t=4.0s 
The velocity at t=4.0 s is
v(t)=at=(2.0 m/s^2)(4.0 s)=8.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(8.0 m/s)^2=1792 J

and so the power output is
P= \frac{E}{t} = \frac{1792 J}{4.0 s} =448 W

c) Power output at t=6.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J

and so the power output is
P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W
8 0
3 years ago
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