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zlopas [31]
3 years ago
6

Multiplying or dividing vectors by scalar results in?

Physics
1 answer:
Minchanka [31]3 years ago
6 0
You divide sort of the similar but vector is distance and displacement and scalar is just distance
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If u speed up from rest to 12 m/s in 3 seconds, what is your acceleration
irakobra [83]
4

Just divide 12 by 3, so if it takes 3 seconds, then every second, it goes up 4.
4 0
3 years ago
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Plz answer it...plzzzzzz​
Bad White [126]
I think it’s C) lmk if I’m wrong
7 0
3 years ago
You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s
marta [7]

1) Acceleration of the car in front: -7.89 m/s^2

The only data we need for this part of the problem is:

u = 27.7 m/s --> initial velocity of the car

\mu=0.804 --> coefficient of friction between the car wheels and the road

From the coefficient of friction, we can find the deceleration of the car. In fact, the force of friction is given by

F=-\mu mg

where m is the car's mass and g=9.81 m/s^2 is the acceleration due to gravity. We can find the car's acceleration by using Newton's second law:

a=\frac{F}{m}=\frac{-\mu mg}{m}=\mu g=(0.804)(9.81 m/s^2)=-7.89 m/s^2

And the negative sign means it is a deceleration.


2) Braking distance for the car in front: 48.6 m

This can be found by using the following SUVAT equation:

v^2 - u^2 = 2aS

where

v=0 is the final velocity of the car

u=27.7 m/s is the initial velocity of the car

a=-7.89 m/s^2 is the acceleration of the car

S is the braking distance

By re-arranging the formula, we find S:

S=\frac{v^2-u^2}{2a}=\frac{0-(27.7 m/s)^2}{2(-7.89 m/s^2)}=48.6 m


3) Minimum safe distance at which you can follow the car: 15.0 m

In this case, we must calculate the thinking distance, which is the distance you travel before hitting the brakes. During this time, the speed of your car is constant, so the thinking distance is given by

d_t = ut=(27.7 m/s)(0.543 s)=15.0 m

After hitting the brakes, your car decelerates at the same rate of the car in front of you, so the braking distance is the same of the other car:

d_b=48.6 m

So the total distance your car covers is

S'=d_t+d_b=15.0 m +48.6 m=63.6 m

At the same time, the car in front of you just covered a distance of 48.6 m. So, in order to avoid the collision, you should travel at a distance equal to

d=S'-S=63.6 m-48.6 m=15.0 m


6 0
3 years ago
Consider a stone at rest on the ground. There are two interactions that involve the stone. One is between the stone and the Eart
finlep [7]

5. between the stone and the ground

for a stone at rest on the ground, we have two forces acting on the stone. first is the force on the stone by the earth due to gravity of earth. second is the force applied by the ground in upward direction to balance the force of gravity on the stone.

so first interaction is between earth and stone : earth pulls stone towards it and stone pulls earth towards it by same amount of force.

second interaction is between stone and ground : ground push the stone in upward direction and stone push the ground in down direction by same amount of force.

hence the correct choice is

5. between the stone and the ground

3 0
3 years ago
Find the displacement the body in the following graph​
Len [333]

Answer:

150

Explanation:

15m/s×10s= 15m/a

and that is the correct answer

7 0
2 years ago
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