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Oksi-84 [34.3K]
2 years ago
12

It is safer for a car driver to be wearing a seat belt, compared with not wearing a seat belt, if the car is involved in a colli

sion.
Explain why.​
Physics
1 answer:
Korolek [52]2 years ago
8 0

Explanation: It is because when a car is moving both the car and the driver is in inertia of motion. When a car is involved in collision it comes to a sudden stop and the car comes into inertia of rest whereas the person still in inertia of motion moves forward and might result in major injuries. But this can be prevented by wearing a seatbelt

Hope it helps :)

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If 20 beats are produce with one second which of the following frequency could possibly be held by two sound waves traveling thr
musickatia [10]

ANY two frequencies whose difference is 20 Hz.

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3 years ago
If an object has a kinetic energy of 30 j and mass of 34kg how fast is the object moving ?
cricket20 [7]

Ek = (m*V^2) / 2 where m is mass and V is speed, then we can take this equation and manipulate it a little to isolate the speed.

Ek = mv^2 / 2 — multiply both sides by 2

2Ek = mv^2 — divide both sides by m

2Ek / m = V^2 — switch sides

V^2 = 2Ek / m — plug in values

V^2 = 2*30J / 34kg

V^2 = 60J/34kg

V^2 = 1.76 m/s — sqrt of both sides

V = sqrt(1.76)

V = 1.32m/s (roughly)

3 0
3 years ago
Explain why the gravitational acceleration of any object near earth is the same matter what was the mass of the object
Gre4nikov [31]
It takes greater force to accelerate an object that has more mass. But the gravitational force between the Earth and an object is greater when the object hass more mass. It works out just right to make any object with any mass accelerate at the same rate.
5 0
3 years ago
Where does the engery of an earthquake originate
allochka39001 [22]
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6 0
3 years ago
The next four questions refer to the situation below.
Anna11 [10]

Answer:

 t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in},      t_{out} = \frac{D}{v_s +v_r}

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

         v_{sg 1} = v_{sr} + v_{rg}

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

           v_{sg1} = D / t_{out}

           D = v_{sg1}  t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

        v_{sg 2} =  v_{sr}  - v_{rg}

         v_{sg 2} = D / t_{in}

         D = v_{sg 2}  t_{in}

with the distance is the same we can equalize

           v_{sg1} t_{out} = v_{sg2} t_{in}

          t_{out} =  t_{in}

           t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in}

This must be the answer since the return time is known. If you want to delete this time

            t_{in}= D / v_{sg2}

we substitute

            t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

            t_{out} = \frac{D}{v_s +v_r}

7 0
2 years ago
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