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jeyben [28]
2 years ago
8

A car with two people traveling down the road has a mass of 100 kg and a velocity of 5 m/s. The car pulls over and picks up two

people. The mass of the car is now 150 kg. What is the new momentum if the velocity stays the same?
Physics
1 answer:
Andrej [43]2 years ago
3 0

Answer:

New_momentum = 750 [kg.m/s]

Explanation:

Momentum is defined as

M  = mass* velocity

Original momentum

M_old = 100 Kg * 5 m/s = 500 [kg.m/s]

If the car pulls over and picks up two people. The mass of the car is now 150 kg, but the velocity stays the same.

New momentum

M_new = 150 Kg * 5 m/s = 750 [kg.m/s]

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Pepsi [2]

Answer:

\boxed{\sf Time \ in \ which \ train \ will \ come \ to \ rest = 20 \ sec}

Given:

Initial velocity (u) = 30 m/s

Final speed (v) = 0 m/s

Acceleration (a) = - 1.5 m/,s²

To Find:

Time in which train will come to rest (t).

Explanation:

\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ a:  \\  \sf \implies 0 = 30 + ( - 1.5)(t) \\   \sf  \implies 0 = 30 - 1.5(t) \\  \sf \implies 30 - 1.5(t) = 0 \\  \\  \sf Subtract  \: 30  \: from  \: both  \: sides: \\  \sf \implies (30 -  \boxed{ \sf 30}) - 1.5(t) =  \boxed{ \sf  - 30} \\  \\  \sf 30 - 30 = 0 :  \\  \sf \implies  - 1.5(t) =  - 30 \\  \\  \sf Divide  \: both  \: sides \:  of \:  - 1.5(t) =  - 30 \: by \:  - 1.5 :  \\  \sf \implies  \frac{  - 1.5(t)}{ \boxed{ \sf - 1.5}}  =  \frac{ - 30}{ \boxed{ \sf -1.5 }}  \\  \\  \sf \frac{ \cancel{ \sf 1.5}}{\cancel{ \sf 1.5}}  = 1 :  \\  \sf \implies t =  \frac{ - 30}{ - 1.5}  \\  \\   \sf  \frac{ - 30}{ - 1.5}  =  \frac{\cancel{ \sf 1.5} \times 20}{\cancel{ \sf 1.5}}  = 20 :  \\  \sf  \implies t = 20 \: sec

So,

Time in which train will come to rest = 20 seconds

4 0
2 years ago
Can someone help? Please?
zubka84 [21]

Answer:

A. Speed

Explanation:

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2 years ago
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vazorg [7]

The magnitude of the induced emf is given by:

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