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LuckyWell [14K]
2 years ago
13

A driver without a seat belt getting thrown from the car in a collision is an example of Newton’s?

Physics
1 answer:
kupik [55]2 years ago
5 0
Newton’s first law. An object stays in motion unless acted on by an external force or in this case the driver was in motion but the sudden crash stoped the car but not the driver
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Electromagnetic waves all move at the same wave speed. which of these offer solutions to slowing them down select all that apply
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option c...have the waves move thru the vacuum or space

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What is a therapuetic modality in psychology?
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Answer:

There are several modalities, or methods, of treatment: individual therapy, group therapy, couples therapy, and family therapy

Explanation:

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2 years ago
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The magnitude of a component of a vector must be
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Less than or equal to the magnitude of the vector
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PLEASE PLEASE PLEASE A mad scientist places massive amounts of charge on basketball sized aluminum balls. The charge on the ball
bazaltina [42]

Answer:

4.2 x 10⁷N

Explanation:

Given parameters:

Charge on ball:

             q₁  = 3C

              q₂ = 14C

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Unknown:

Force acting on the two balls

Solution:

The force experienced by the two charges is given by coulombs law. It is mathematically expressed as;

                      F  = \frac{k q_{1} q_{2} }{r^{2} }

where k  = 9 x 10⁹Nm²/C²

           q is the charges

             r is the distance

Input the variables and solve;

                 

        F  = \frac{9 x 10^{9} x 3 x 14 }{9000}  = 4.2 x 10⁷N

8 0
3 years ago
A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit a
Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

m_1v_1+m_2v_2 = (m_1+m_2)V_f

Where,

m_{1,2}= Mass of each object

v_{1,2}= Initial Velocity of Each object

V_f= Final Velocity

Our values are given as

m_1 = 2190Kg

v_1 =10.5m/s

m_2 = 3220kg

V_f = 4.74m/s

Replacing we have that

m_1v_1+m_2v_2 = (m_1+m_2)V_f

(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)

v_2 = 0.8224m/s

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 0
3 years ago
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