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2 years ago

For for each situation say how the ideas of force pressure and area can be applied :

Physics

1 answer:

fomenos2 years ago

4 0

<h3 /><h3><u><em>Solution-:</em></u></h3><h3><u><em>more force as expansion is much</em></u></h3>

<u><em>also, less force area is much</em></u>

<h3> mark me Brainliest</h3>

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An object is thrown directly upwards from the ground at a velocity of 9ms. Recalling that the acceleration due to gravity is −gm

olga55 [171]

Answer:

Explanation:

Given

Object is thrown with a velocity of $u=9\ m/s$

Acceleration due to gravity is -g (i.e. acting downward)

Vertical distance traveled by object is given by

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

at maximum height final velocity is zero

$0-(9)^2=2\times (-g)\times (s)$

$s=\frac{81}{2g}=\frac{40.5}{g}\ m$

time taken to reach maximum height

using

v=u+at

0=9-gt

$t=\frac{9}{g}\ s$

6 0

3 years ago

oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student

puteri [66]

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

3 0

2 years ago

An object with a mass M = 250 g is at rest on a plane that makes an angle θ = 30 o above the horizontal. The coefficient of kine

liubo4ka [24]

Answer:

v = 79.2 m/s

Solution:

As per the question:

Mass of the object, m = 250 g = 0.250 kg

Angle, $\theta = 30^{\circ}$

Coefficient of kinetic friction, $\mu_{k} = 0.100$

Mass attached to the string, m = 0.200 kg

Distance, d = 30 cm = 0.03 m

Now,

The tension in the string is given by:

$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = T$ (1)

Also

T = m(g + a)

Thus eqn (1) can be written as:

$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = m(g - a)$

$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = mg + ma$

$mg - Mgsin\theta - \mu_{k}Mgcos\theta = (M - m)a$

$a = \frac{0.2\times 9.8 - 0.250\times 9.8\times sin30^{\circ} - 0.1\times 0.250\times 9.8\times cos30^{\circ}}{0.250 - 0.200}$

$a = 10.45\ m/s^{2}$

Now, the speed is given by the third eqn of motion with initial velocity being zero:

$v^{2} = u^{2} + 2ad$

where

u = initial velocity = 0

Thus

$v = \sqrt{2ad}$

$v = \sqrt{2\times 10.45\times 0.03} = 0.792\ m/s$

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3 years ago

Fossils are generally found inside of what type of rocks?

dusya [7]

Answer:

C sedimentary rocks is the answer.

Explanation:

8 0

3 years ago

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When a nerve cell depolarizes, charge is transferred across the cell membrane, changing the potential difference. For a typical

igomit [66]

Answer:

1.8nA

Explanation:

I = dq/dt

I = 9.0 × 10 ^-12 / 0.5 × 10^-3

I = 1.8 × 10× 10^-8

I = 1.8nA

6 0

3 years ago

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