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2 years ago

When you are riding in a car then the driver suddenly steps on the gas pedal, why do you feel a jolt and push back?

Physics

2 answers:

Nana76 [90]2 years ago

6 0

Answer:

Effects of Interia

As a more familiar example of inertia, think about riding in a car. ... If the car comes to a sudden stop, your body tends to keep moving forward. When the car starts moving again, your body tends to stay at rest. You move forward because the car seat exerts an unbalanced force on your body.

inessss [21]2 years ago

4 0

Answer:

due to acceleration lol

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A solid weighs 16.5N on the surface of the moon. The force of gravity on the moon is 1.7N/Kg.

Fiesta28 [93]

Answer:

mass = 9.7 kg

Explanation:

Weight = Mass x Acceleration due to gravity (g)

16.5 = mass x 1.7

mass = $\frac{16.5}{1.7}$ = 9.7 kg

3 0

2 years ago

A spherical tank with radius 4 m is half full of a liquid that has a density of 900 kg/m3. The tank has a 1 m spout at the top.

kompoz [17]

The answer & explanation for this question is given in the attachment below.

7 0

2 years ago

For a particular type of motion, the velocity is zero but the speed is a nonzero quantity. Which statement can you make about th

masha68 [24]

Answer:

because speed is the modulus of velocity which is a vector

the velocity to be zero it must be a round trip

Explanation:

This is because speed is the modulus of velocity which is a vector.

For the velocity to be zero it must be a round trip, therefore the resulting vector zero

On the other hand, the speed of the module is the same in both directions

8 0

2 years ago

Two trains A and B of length 400 m each are moving on two parallel tracks with

Vinvika [58]

Answer:

The initial distance between the trains is 1450 m. 

Explanation:

In the question two trains are of equal length 400 m and moves at a uniform speed of 72 km/h. train A is moving ahead of train B. If the train B has to overtake train A it should accelerate.

Train B’s acceleration is $1m/s^2$ and it accelerated for 50 seconds.

 $a=1 m/s^2$ 

t=50 s 

initial speed u=72km/h 

we have to convert this speed into m/s 

 $u=72 \times 5/18=20 m/s$ 

Distance covered in accelerating phase $S=ut+1/2 at^2$ 

 $=20 \times 50+1/2 \times 1 \times 50^2$ 

 $=1000+1250=2250 m$ 

If a train is just behind another, the distance covered by the train located behind during overtaking phase will be equal to the sum of the lengths of the trains.

Here length of train A+length of train $B=400+400=800 m$ 