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Sergeeva-Olga [200]

3 years ago

12

A 55 kg person is in a head-on collision. The car's speed at impact is 12 m/s. Estimate the net force on the person if he or she

is wearing a seat belt and if the air bag deploys.

1 answer:

ollegr [7]3 years ago

4 0

1.244 m per second the person driving will go

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An electron (e = 1.6 x 10-19 C) is traveling at 4.00 x 107 m/s due North in a horizontal plane through a point where the earth’s

Irina18 [472]

Answer:

Explanation:

Force on the electron due to magnetic field 's north component will be zero because both velocity of electron and direction of magnetic field are same .

Force due to downward component of magnetic field

= B q v where B is magnetic field , q is charge moving and v is velocity of charge

F = 5 .00 x 10⁻⁵ x 1.6 x 10⁻¹⁹ x 4 x 10⁷

= 32 x 10⁻¹⁷ N

acceleration = F / m where m is mass of electron

= 32 x 10⁻¹⁷ / 9.11 x 10⁻³¹

= 3.5 x 10¹⁴ m/s²

4 0

3 years ago

A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it

Ilia_Sergeevich [38]

Below is the attachment of the solutions.

5 0

4 years ago

!! Offering 50 points !!

Evgesh-ka [11]

Answeryes they are the same

Explanation:

0.5-0.5=10 is gotta be sorry if its wrong

8 0

3 years ago

Read 2 more answers

Please help me.

vesna_86 [32]

A) the tension in the string.

4 0

4 years ago

Read 2 more answers

A 50.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and the

Dimas [21]

Answer:

98N and 147N

Explanation:

We have the following information:

$m=50kg\\\mu_s =0.4\\\mu_k = 0.2\\F=140N$

We can find the static fricton force as follow,

$F=\mu_s * N$

Where N is the normal force (mg)

$F=0.3*50*9.8\\F=147N$

Static friction force at 147N is greater than the force applied hence body does not move.

$F=\mu_k N = 0.2*50*9.8= 98N$

3 0

4 years ago