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Kryger [21]
3 years ago
6

A car accelerate from a stop to 27 m/sec in 5 seconds. What was the car's acceleration?

Physics
1 answer:
Furkat [3]3 years ago
4 0
The acceleration should 5.4 m/s^2
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Based on the data given, in what direction will the car accelerate?
skad [1K]
Vertical forces:
There is a force of 579N acting upward, and a force of 579N
acting downward.
The vertical forces are balanced ... they add up to zero ...
so there's no vertical acceleration. 
Not up, not down.

Horizontal forces:
There is a force of 487N acting to the left, and a force of 632N
acting to the right.
The net horizontal force is

        (487-left + 632-right)  -  (632-right - 487-right) =  145N to the right.

The net force on the car is all to the right.
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7 0
3 years ago
calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the s
IceJOKER [234]

Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m  * 1609.344 m

  = 149.668 * 10^8 m

next we will express the distance between the earth and the sun

r = \frac{a(1-E^2)}{1+Ecos\beta }   --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

\beta = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

v^2 = \frac{4\pi^2 }{r_{c} }   ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

(V^2 = (\frac{4\pi^{2}  }{149.626*10^9})

therefore : V = 1.624* 10^-5 m/s

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