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zysi [14]
3 years ago
11

What is the displacement of a car with an acceleration of 4.00 m/s2 as it increases.

Physics
1 answer:
insens350 [35]3 years ago
4 0

Answer:

100 m.

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 10 m/s

Final velocity (v) = 30 m/s

Acceleration (a) = 4 m/s²

Displacement (D) =?

Next, we shall determine the time.

This can be obtained as follow:

Acceleration is defined as the rate of change of velocity with time. Mathematically, it is expressed as:

a = (v – u) / t

Where

a is the acceleration.

u is the initial velocity.

v is the final velocity.

t is the time.

With the above formula, we can obtain the time as follow:

Initial velocity (u) = 10 m/s

Final velocity (v) = 30 m/s

Acceleration (a) = 4 m/s²

Time (t) =?

a = (v – u) / t

4 = (30 – 10)/t

4 = 20/t

Cross multiply

4 × t = 20

Divide both side by 4

t = 20/4

t = 5 secs.

Finally, we shall calculate the displacement of car as illustrated below:

Velocity is defined as rate of change of displacement with time. Mathematically, it can be expressed as:

(v – u) = D/ t

Where:

u is the initial velocity.

v is the final velocity.

t is the time.

D is the displacement.

With the above formula, we can calculate the displacement of the car as follow:

Initial velocity (u) = 10 m/s

Final velocity (v) = 30 m/s

Time (t) = 5 s.

Displacement (D) =?

(v – u) = D/ t

(30 – 10) = D/5

20 = D/5

Cross multiply

D = 20 × 5

D = 100 m

Therefore, the displacement of the car is 100 m

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Four copper wires of equal length are connected in series. Their cross-sectional areas are 1.6 cm2 , 1.2 cm2 , 4.4 cm2 , and 7 c
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Answer:

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Explanation:

We are given that

A_1=1.6 cm^2=1.6\times 10^{-4} m^2

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A_2=1.2 cm^2=1.2\times 10^{-4} m^2

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A_4=7 cm^2=7\times 10^{-4} m^2

Potential difference,V=140 V

We know that

R=\frac{\rho l}{A}

According to question

l_1=l_2=l_3=l_4=l

In series

R=R_1+R_2+R_3+R_4

R=\rho l(\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4})

R=\rho l(\frac{1}{1.6\times 10^{-4}}+\frac{1}{1.2\times 10^{-4}}+\frac{1}{4.4\times 10^{-4}}+\frac{1}{7\times 10^{-4}})

R=\rho l(18284.6)

I=\frac{V}{R}=\frac{140}{\rho l\times 18284.6}

Potential across 1.2 square cm=V_1=IR_1=\frac{140}{\rho l\times 18284.6}\times \rho l(\frac{1}{1.2\times 10^{-4}}=63.8 V

Hence, the voltage across the 1.2 square cm wire=63.8 V

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Hope this helps!
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Answer:

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Since, it is fixed at one point in the sky in the North direction, explorers used to take help of Polaris to find direction in the night time. Since Polaris is not the brightest star in the night sky (it is 48th brightest star) our ancestors took help of constellations to find it. One of the method was to use Big Dipper of "Ursa Major" constellation. The pointer stars (Merak and Dubhe) of big dipper point towards the Polaris. One can easily spot it in the sky as well.

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