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2 years ago

6

A car’s brakes decelerate it at a rate of -1.70 m/s2. What is the time, in seconds, required to slow the car from 15 m/s to 9.0

m/s?

1 answer:

jeka942 years ago

3 0

Answer:

t = 3.52 s

Explanation:

Given that,

The deceleration of a car is, a = -1.7 m/s²

The initial velocity of the car, u = 15 m/s

Final velocity of the car, v = 9 m/s

We need to find the time that is required to slow the car from 15 m/s to 9 m/s. The definition of acceleration is :

$a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{9-15}{-1.7}\\\\t=3.52\ s$

So, it will take 3.52 seconds to slow the car from 15 m/s to 9 m/s.

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What is general relativity

Katena32 [7]

Explanation:

General relativity is a theory of space and time. ... The central idea of general relativity is that space and time are two aspects of spacetime. Spacetime is curved when there is matter, energy, and momentum resulting in what we perceive as gravity. The links between these forces are shown in the Einstein field equations.

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3 years ago

Find the pressure exerted by a force 8000 newton's on an area of 25m^2. Give your answer in newton's/n^2

Vanyuwa [196]

The pressure exerted by this force is equal to 320 $N/m^2$ .

<u>Given the following data:</u>

Force = 8000 Newton.

Area = 25 $m^2$ .

To calculate the pressure exerted by this force:

<h3>How to calculate pressure.</h3>

Mathematically, pressure is given by this formula:

$P=\frac{F}{A}$

<u>Where:</u>

P is the pressure.

F is the force.

A is the area.

Substituting the given parameters into the formula, we have:

$P = \frac{8000}{25}$

P = 320 $N/m^2$ .

Read more on pressure here: brainly.com/question/8033367

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2 years ago

A 3-m-high, 7-m-wide rectangular gate is hinged at the top edge and is restrained by a fixed ridge. Determine the hydrostatic fo

Shalnov [3]

Answer:

The Hydrostatic force is $F = 137.2 kN$

The location of pressure center is $Z = 1.333 \ m$

Explanation:

From the question we are told that

The height of the gate is $h = 3 \ m$

The weight of the gate is $w = 7 \ m$

The height of the water is $h_w = 2 \ m$

The density of water is $\rho_w = 1000 \ kg/m^3$

Note used $h_w$ for height of water and height of gate immersed by water since both have the same value

The area of the gate immersed in water is mathematically represented as

$A = h_w * w$

substituting values

$A = 2* 7$

$A = 14 \ m^2$

The hydrostatic force is mathematically represented as

$F = \rho_w * g * h_f * A$

Where

$h_f =h- h_w$

$h_f =3 -2$

$h_f = 1\ m$

So

$F = 1000 * 9.8 * 1 * 14$

$F = 137.2 kN$

The center of pressure is mathematically represented as

$Z = h_f + \frac{I_g}{h_f * A}$

Where $I_g$ is the moment of inertia of the gate which mathematically represented as

$I_g = \frac{w * h_w^2}{12}$

The $h_w$ is the height of gate immersed in water

$I_g = \frac{7 * 2^2 }{12}$

$I_g = 4.667\ kg m^2$

Thus

$Z = 1 + \frac{4.66}{1 * 14}$

$Z = 1.333 \ m$

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A Ping-Pong ball has a mass of 2.3 g and a terminal speed of 9.4 m/s. The drag force is of the form bv^2 What is the value of b?

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Drag Force = bv^2 = ma; a = g = 9.81 m/s^2

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