All categories

3 years ago

Please help with this :(

Physics

1 answer:

Rainbow [258]3 years ago

3 0

Answer:

I don't really know but i have an idea

Explanation:

so acceleration means: increase in the rate or speed of something.

so just a head start.

your welcome!!

You might be interested in

(a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.34 m/s² and subway stations are located

andreev551 [17]

Answer:

The correct answer is 32.9 m/s

Explanation:

To solve this, we list out the known and the unknown variables as follows

Maximum allowable acceleration = 1.34 m/s²

Distance between sttions = 806 m

Therefore from the equation of motion

v = ut + 0.5·×at²

Where v = final velocity

u = initial velocity

S = distance covered

t = time

a = acceleration

Also v² = u² + 2·a·S

where u is the initial velocity, which we can take as u = 0, then

v² = 2·1.34·S = 2.68S m²/s² then

Also the train has to decelerate from maximum speed to stop at the next tran station wherev = 0, thus v² = u² -2·1.34·Z, so u² = 2.68Z

since u² = 2.68S from the previous calculation, then for v = 0

2.68S = 2.68Z thus S = Z which and to reach the next subway station S + Z must be = 806 m, then S = 806 m ÷ 2 = 403 m

and v² = 2.68S m²/s² = 1080.04 m²/s²

v = 32.9 m/s

The maximum speed a subway train can attain between stations is 32.9 m/s

3 0

3 years ago

The maker of an automobile advertises that it takes 8 seconds to accelerate from 15 kilometers per hour to 70 kilometers per hou

jeyben [28]

Answer:

The distance the car traveled during the given time is 94.44 m.

Explanation:

Given;

time of motion t = 8 seconds

initial velocity, u = 15 km/hr = 4.167 m/s

final velocity, v = 70 km/hr = 19.444 m/s

The distance traveled by the car during this time is given by;

$s = (\frac{u+v}{2} )t\\\\s = (\frac{4.167+19.444}{2} )*8\\\\s = 94.44 \ m$

Therefore, the distance the car traveled during the given time is 94.44 m.

8 0

3 years ago

When you drop an object from a certain height, it takes time T to reach the ground with no air resistance. If you dropped it fro

alexandr402 [8]

Answer:

Explanation:

Given

When we drop an object from height , suppose h

it takes time T

using equation of motion

$h=ut+\frac{1}{2}at^2$

where

$h=displacement$

$u=initial\ velocity$

$a=acceleration$

$t=time$

here $u=0$ because it dropped from a certain height

$h=\frac{1}{2}gT^2$

$T=\sqrt{\frac{2h}{g}}$

When height is increases to three times of original height

i.e. $h'=3 h$

then time period becomes

$T'=\sqrt{\frac{2\times 3h}{g}}$

$T'=\sqrt{3}T$

5 0

3 years ago

If you toss a stick into the air, it appears to wobble all over the place. specifically, about what place does it wobble?

Otrada [13]

I would assume the end?

8 0

3 years ago

How do I know what kinematic equations to use when solving a question?

olganol [36]

Explanation:

There are 5 kinematic equations, and 5 variables.

Each question will give you 3 variables and ask you to solve for a fourth.

To determine which equation to use, look at which variable is <em>not</em> included in the problem.

For example, if the question does not include time, then you need to use a kinematic equation that does not have t in it. That would be:

v² = v₀² + 2aΔx

Or, if the question does not include the final velocity, then you need a kinematic equation that does not have v in it. That would be:

Δx = v₀ t + ½ at²

5 0

3 years ago