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Arisa [49]
3 years ago
6

If a vehicle is traveling at constant velocity and then comes to a sudden stop, has it undergone negative acceleration or positi

ve acceleration? Explain your answer
Physics
1 answer:
sertanlavr [38]3 years ago
6 0

Answer:

Negative acceleration

Explanation:

Let us assume a vehicle is moving with a constant velocity u and then it comes to a sudden stop.

The initial velocity of the vehicle is u and finally it comes to rest, it means final velocity is 0.

As we know that,

Acceleration is equal to change in velocity divided by time taken.

a=\dfrac{v-u}{t}\\\\\text{Put v=0}\\\\a=\dfrac{-u}{t}

We can see that the value of acceleration is negative. Hence, it leads to negative acceleration.

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when he returns his hertz rent-a-rocket after one week's cruising in the galaxy, spock is shocked to be billed for three weeks'
Flura [38]

Mr. spoke is travelling with speed v = 9.4c.

Mr. spoke is apparently unknowing of  theory of special relativity by Einstein. The theory states that moving clock tick slower than stationery clock.

Time elapse =∆t = 3 weeks

(In earth stationery frame)

Time ellipse= ∆t' = 1 week

(In the frame of rocket)

∆t' = ∆t'/γ

1/y² = 1 - v²/c²

∆t'²/∆t² =

=1²/3²

=1/9

c= 3×10⁸m/s

on solving above equation we get :

v = 9.4c

He was spreading around at close to the speed of light.

Experience significant time dilatation.

He is travelling with speed  v = 9.4c.

To know more about special theory of relativity :

brainly.com/question/28289663

# SPJ4

4 0
2 years ago
A merry-go-round rotates at the rate of 0.17 rev/s with an 79 kg man standing at a point 1.6 m from the axis of rotation.
dezoksy [38]

Hi there!

We can use the conservation of angular momentum to solve.

L_i = L_f\\\\I\omega_i = I\omega_f

I = moment of inertia (kgm²)

ω = angular velocity (rad/sec)

Recall the following equations for the moment of inertia.

\text{Solid cylinder:} I = \frac{1}{2}MR^2\\\\\text{Object around center:} = MR^2

Begin by converting rev/sec to rad sec:


\frac{0.17rev}{s} * \frac{2\pi rad}{1 rev} = 1.068 \frac{rad}{s}

According to the above and the given information, we can write an equation and solve for ωf.

1.068(\frac{1}{2}(34)(1.6)^2 + (79)(1.6)^2) = \omega_f(\frac{1}{2}(34)(1.6^2) + 79(0^2))\\\\\omega_f = \boxed{6.03 \frac{rad}{sec}}

4 0
3 years ago
The diagram shows two balls before they collide.
Ronch [10]

Answer:

The Answer is B)0.2 kg • m/s

Explanation:

I made a 100 on my test. Sorry if I'm late but hope I helped.

4 0
3 years ago
Read 2 more answers
If the radius of a coin is 1 cm then calculate its area.​
igor_vitrenko [27]

Answer:

3.14*1²

3.14 cm²

I hope this will help

5 0
3 years ago
Read 2 more answers
A light ray in air enters and passes through a block of glass. What can be stated with regard to its speed after it emerges from
Alik [6]

Answer:

Speed is same as that before it entered glass.

Explanation:

Given:

A light ray enters and passes through the glass as shown in the diagram.

We have to analyze its speed.

Speed of light in air is 3\times 10^8\ ms^-^1 and speed of light in glass is 2.25\times 10^8\ ms^-^1

Whenever a light ray enters a glass block or slab there is bending of light at the interface of the two media.

So speed of light will decrease in glass medium and again it passes to the air.

So

Speed of light in air will again increase or will be equivalent to the earlier speed when it was entering the glass block.

Finally

Speed is same as that before it entered glass as it in the same medium (air).

6 0
3 years ago
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