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Svet_ta [14]
3 years ago
11

If you are the driver or owner of a vehicle which is in a crash that is your fault, and you are not insured in compliance with t

he Financial Responsibility Law
Physics
1 answer:
Oksi-84 [34.3K]3 years ago
3 0
If a driver is at fault in an accident and is not insured by the Financial Responsibility Law then he or she may be required to pay for the damages caused to the other person's vehicle or property due to the accident before the driver at fault is given the privilege to drive again.
The Financial Responsibility Law is a law that requires drivers to prove that they have enough money saved to pay for the damages that may occur in the case of an accident.
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A sprinter starts from rest and reaches a speed of 15 m/s in 4.25 s. Find his acceleration
kakasveta [241]

Answer:

a=3.53 m/s^2

Explanation:

Vo=0 m/s (because he is not moving at the start)

V1=15 m/s

t= 4.25 s

a = (V1-Vo) / t = 15/4.25 = 3.53 m/s^2

5 0
3 years ago
143°C = _____<br><br> 416 K<br> -130 K<br> 0 K<br> 143 K
dezoksy [38]
The answer is 0k because 143c equals nothing
5 0
2 years ago
Read 2 more answers
A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m
ArbitrLikvidat [17]

Answer:

<h2>206.67N</h2>

Explanation:

The sum of force along both components x and y is expressed as;

\sum Fx = ma_x  \ and \ \sum Fy = ma_y

The magnitude of the net force which is also known as the resultant will be expressed as R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

a_x = \frac{d^2 x }{dt^2}

a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4  )\\\\a_x = \frac{d}{dt}(12t  )\\\\a_x = 12m/s^{2}

Similarly,

a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2}   )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2

\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N

R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

7 0
3 years ago
A4 40 kg girl skates at 3.5 m/s one ice toward her 65 kg friend who is standing still, with open arms. As they collide and hold
salantis [7]

Explanation:

Below is an attachment containing the solution.

8 0
4 years ago
Standing on the surface of the earth you have a weight of 100 N. If you were to travel until you were 2
Vika [28.1K]

Answer:

E

Explanation:

8 0
3 years ago
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