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Vlada [557]
3 years ago
15

Describe the motion of an automobile on an east-west

Physics
1 answer:
Tanya [424]3 years ago
4 0

Answer:

1. The automobile is traveling due east and is speeding up.

2. The car is traveling due east and is is slowing down.

3. The automobile is traveling due east at a constant speed.

4. The car is traveling due west and is slowing down.

5. The automobile is traveling due west and is speeding up.

6. The automobile is traveling due west at a constant speed.

7. The automobile is accelerating due east from rest.

8. The automobile is accelerating due west from rest.

Explanation:

The key to understanding this is:

When the acceleration and initial velocity of the automobile have the same sign (positive or negative) then the automobile is speeding up. Explained further, if acceleration and the initial velocity are both positive or they are both negative the automobile is speeding up but whenever they have opposite signs (that is acceleration is positive and initial velocity is negative or vice versa) the automobile is slowing down. When the acceleration is zero the automobile is maintaining a unform motion at a constant speed (the speed is not changing with time). The + or - sign indicates the direction of travel. In this case east is + and west is -. It is my pleasure answering this question. I hope you find it helpful. Thank you.

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tatuchka [14]
C.

The train first sped up, giving it a positive acceleration in the beginning. This eliminates D since that choice states that it begins with a negative acceleration. This also eliminates B since that choice states that the train only had a negation acceleration.

Next, the train slows down, giving it a negative acceleration. We’re looking for the answer choice that starts with a positive acceleration and ends with a negative one. This makes C the correct answer. Hope this helps!
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A spring gun is made by compressing a spring in a tube and then latching the spring at
inn [45]

Answer:

a)v=13.2171\,m.s^{-1}

b)H=8.9605\,m

Explanation:

Given:

mass of bullet, m=4.97\times 10^{-3}\,kg

compression of the spring, \Delta x=0.0476\,m

force required for the given compression, F=9.12 \,N

(a)

We know

F=m.a

where:

a= acceleration

9.12=4.97\times 10^{-3}\times a

a\approx 1835\,m.s^{-2}\\

we have:

initial velocity,u=0\,m.s^{-1}

Using the eq. of motion:

v^2=u^2+2a.\Delta x

where:

v= final velocity after the separation of spring with the bullet.

v^2= 0^2+2\times 1835\times 0.0476

v=13.2171\,m.s^{-1}

(b)

Now, in vertical direction we take the above velocity as the initial velocity "u"

so,

u=13.2171\,m.s^{-1}

∵At maximum height the final velocity will be zero

v=0\,m.s^{-1}

Using the equation of motion:

v^2=u^2-2g.h

where:

h= height

g= acceleration due to gravity

0^2=13.2171^2-2\times 9.8\times h

h=8.9129\,m

is the height from the release position of the spring.

So, the height from the latched position be:

H=h+\Delta x

H=8.9129+0.0476

H=8.9605\,m

4 0
3 years ago
Which wave must have a medium to travel?
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Letter B

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