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Mila [183]
2 years ago
9

A sports car starts from rest it covers a distance of 900 m to attain a speed of 80m s determine the acceleration of the car and

the time required to reach this speed
Physics
1 answer:
Lady bird [3.3K]2 years ago
5 0

The acceleration of the car and the time required to reach this speed will be 8 m/s² and 10 sec.

<h3>What is acceleration?</h3>

The rate of velocity change concerning time is known as acceleration.

Given data;

Initial velocity, u=0  m/s

Final velocity, v= 80 m/sec

Distance travelled,s =900m

From Newton's third equation of motion;

v²=u²+2as

a =(v²-u²)/2s

Substitute the given values;

a = (80²-0)/2 ×900

a = 6400/1800

a=8 m/s²

The time required to reach this speed is found in Newton's first equation of motion as;

v = u+at

Substitute the given values;

80 = 0 + 8t

t=80/8

t = 10 sec

Hence, the acceleration of the car and the time required to reach this speed will be 8 m/s² and 10 sec.

To learn more about acceleration, refer to the link brainly.com/question/2437624

#SPJ1

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When looking at the chemical symbol, the charge of the ion is displayed as the
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Answer:

superscript

Explanation:

When looking at the chemical symbol, the charge of the ion is displayed as the Superscript. This is because the charge of ions is usually written up on the chemical symbol while the atom/molecule is usually written down the chemical symbol. The superscript refers to what is written up on the formula while the subscript is written down on the formula.

An example is H2O . The 2 present represents two molecule of oxygen and its written as the subscript while Fe2+ in which the 2+ is written up is known as the superscript.

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Which of the following is equal to Velocity change per second​
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Acceleration

Explanation:

Acceleration is a physical quantity that expresses the change in the velocity of a body per unit of time.

  Acceleration = \frac{V - U }{T}

V is the initial velocity

U is the final velocity

T is the time

It is has a unit of m/s²

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3 years ago
A 244.0 N block is at rest on a flat, frictionless table. A hooked cable applies an upward force of 24.0 N on the block. What is
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Answer:

268N

Explanation:

The upward force acting on the block are the reaction and the hooked table..

The total normal force acting = normal reaction + 24N

Note that the normal reaction is always equal the weight of the table

Hence the normal force acting in the block is 244.0+24 = 268.0N

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A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th
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Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2 (1)

where

p_1 = 7340 N/m^2 is the pressure in the pipe

p_2 = 5505 N/m^2 is the pressure in the constricted section

\rho = 821 kg/m^3 is the density of the oil

v_1 is the velocity of the oil in the pipe

v_2 is the velocity of the oil in the constricted section

Also, according to the continuity equation,

A_1 v_1 = A_2 v_2

where

A_1 = \pi r_1^2 is the cross-sectional area of the pipe, with

r_1 = \frac{1.03}{2}=0.515 m is the radius

A_2 = \pi r_2^2 is the cross-sectional area of the constricted section, with

r_2=\frac{0.618}{2}=0.309 m is the radius

So the equation becomes

r_1^2 v_1 = r_2^2 v_2

So we can write

v_2=\frac{r_1^2}{r_2^2}v_1

Substituting into eq.(1),

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2

And solving the equation for v_1:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s

And the velocity in the constricted section is

v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s

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