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borishaifa [10]
3 years ago
15

Which car is harder to stop and why?

Physics
2 answers:
kondor19780726 [428]3 years ago
8 0

Answer:

The momentum of bath cars is 40000 Ns which make the difficulty to stop each car in aspect of fprce is the same.

Explanation:

Momentum (P) =mass(m) × velocity (v)

For car A,

                    P = m × v = 1000 × 40 = 40000 Ns

For car B,

                    P = m × v = 4000 × 10 = 40000 Ns

Force (F) = Momentum change(ΔΡ)/ time taken(t)

            F = ΔΡ/t

When stopping the car the momentum changes from 40000 Ns to 0

So momentum change in both cars is the same. So to stop the two cars in a  given time (t) you need the same force, which means you will feel same difficulty.

             

Bond [772]3 years ago
8 0

Answer:

The third option

Explanation:

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ou have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3,
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5 0
3 years ago
Two identical strings, of identical lengths of 2.00 m and linear mass density of μ=0.0065kg/m, are fixed on both ends. String A
kolezko [41]

Answer:

beat frequency = 13.87 Hz

Explanation:

given data

lengths l = 2.00 m

linear mass density μ = 0.0065 kg/m

String A is under a tension T1 = 120.00 N

String B is under a tension T2 = 130.00 N

n = 10 mode

to find out

beat frequency

solution

we know here that length L is

L = n × \frac{ \lambda }{2}      ........1

so  λ = \frac{2L}{10}  

and velocity is express as

V = \sqrt{\frac{T}{\mu } }    .................2

so

frequency for string A = f1 = \frac{V1}{\lambda}

f1 = \frac{\sqrt{\frac{T}{\mu } }}{\frac{2L}{10}}

f1 = \frac{10}{2L} \sqrt{\frac{T1}{\mu } }      

and

f2 = \frac{10}{2L} \sqrt{\frac{T2}{\mu } }

so

beat frequency is = f2 - f1

put here value

beat frequency = \frac{10}{2*2} \sqrt{\frac{130}{0.0065}}  - \frac{10}{2*2} \sqrt{\frac{120}{0.0065} }

beat frequency = 13.87 Hz

6 0
3 years ago
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