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3 years ago

Inertia law what happens when You are standing on the bus and the bus stops abruptly:

Physics

2 answers:

lord [1]3 years ago

6 0

Answer:

You will fall towards the front of the bus i.e towards its front seat

Explanation:

When the bus is moving, you are supposed to be in that motion. Your body is experiencing a forward force as the bus is moving forward So when bus stops suddenly your inertia does not allow you to stay in that standing position it will let you move forrward to maintain your inertia so you will feel a forward push. (If you were sitting on a seat with seatbelt on you that seatbelt will make you come back when car or bus stops.)

I am Lyosha [343]3 years ago

4 0

Answer:

You will fly forward in the bus until you hit something.

Explanation:

While standing there on the bus, you are traveling at the same speed as the bus. If the bus suddenly stops, you will still be traveling at the same speed you started with. That is until you hit something hard enough or big enough to stop you.

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Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin

Zarrin [17]

Answer:

$\rm 9.186\times 10^{-7}\ C.$

Explanation:

<u>Given:</u>

Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
Distance of separation between the plates, d = 1.0 cm = 0.01 m.
Minimum value of electric field that produces spark, $\rm E=3\times 10^6\ N/C.$

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

$\rm E=\dfrac{\sigma}{\epsilon_o}.$

where,

$\rm \sigma$ = surface charge density of the plate of the capacitor = $\dfrac qA$ .

$\rm q$ = magnitude of the charge on each of the plate.
$\rm A$ = surface area of each of the plate = $\rm \pi \times (Radius)^2=\pi \times\left ( \dfrac{D}{2}\right )^2= \pi \times \left ( \dfrac{0.21}{2}\right )^2=3.46\times 10^{-2}\ m^2.$
$\epsilon_o$ = electrical permittivity of free space, having value = $8.85\times 10^{-12}\rm \ C^2N^{-1}m^{-2}.$

For the minimum value of electric field that produces spark,

$\rm E = \dfrac{q}{A\epsilon_o}\\\Rightarrow q = E\ A\epsilon_o\\=3\times 10^6\times 3.46\times 10^{-2}\times 8.85\times 10^{-12}\\=9.186\times 10^{-7}\ C.$

It is the maximum value of the magnitude of charge which can be added up to each of the plates of the capacitor.

4 0

2 years ago

What effect does temperature have on the composition of ocean water?

Burka [1]

The answer is letter B. XD

4 0

3 years ago

Read 2 more answers

An Airbus A350 is initially moving down the runway at 6.0 m/s preparing for takeoff. The pilot pulls on the throttle so that the

alexdok [17]

Answer:

t=67.7s

Explanation:

From this question we know that:

Vo = 6m/s

a = 1.8 m/s2

D = 1500m

And we also know that:

$X=V_{o}*t + \frac{a*t^{2}}{2}$ Replacing the known values:

$1500=6t+0.9*t^{2}$ Solving for t we get 2 possible answers:

t1 = -44.3s and t2 = 67.7s Since negative time represents an instant before the beginning of the movement, t1 is discarded. So, the final answer is:

t = 67.7s

8 0

3 years ago

The cheetah can run a distance of 275 m for 9 seconds. Calculate its speed.

AnnZ [28]

Answer: 110000

Explanation:

26/9=30.5555555556

30.5555555556 x 60=1833.33333333

110000 x 60=110000

4 0

3 years ago

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.

astra-53 [7]

Answer:

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

Y(0) = 0 m

Vy(0) = 15 m/s

ay = -9.81 m/s^2 (negative because it points down)

Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

t = 0.654 s

At this time it will have a height of:

Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m

The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

t1 = 0 This is the moment it jumped into the air

0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

5 0

2 years ago