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3 years ago

A driver starts is parked car and within 5 seconds reaches a speed of 60km/h, as he travels east. What his acceleration?

Physics

1 answer:

atroni [7]3 years ago

3 0

The acceleration of the car is 3.3 m/s²

<u>Explanation:</u>

Given-

Speed of the car, (we can represent as s) = 60 km/hr = 60 × $\frac{5}{18}$ = 16.67 m/s

Time,(we can represent as t) = 5 s

Acceleration, (we can represent as a) = ?

We know,

a = $\frac{s}{t}$

a = $\frac{16.67}{5}$

a = 3.334 m/s²

The acceleration of the car is 3.3 m/s²

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Answer:

1.925 μC

Explanation:

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The formula for the charge stored in a capacitor is given as,

Q = CV ................... Equation 1

Where Q = charge, C = Capacitor, V = Voltage.

Note: 1 μF = 10⁻⁶ F

Given: C = 0.55 μF = 0.55×10⁻⁶ F, V = 3.5 V.

Substitute into equation 1

Q = 0.55×10⁻⁶×3.5

Q = 1.925×10⁻⁶ C.

Q = 1.925 μC

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kolezko [41]

Answer:

beat frequency = 13.87 Hz

Explanation:

given data

lengths l = 2.00 m

linear mass density μ = 0.0065 kg/m

String A is under a tension T1 = 120.00 N

String B is under a tension T2 = 130.00 N

n = 10 mode

to find out

beat frequency

solution

we know here that length L is

L = n × $\frac{ \lambda }{2}$ ........1

so λ = $\frac{2L}{10}$

and velocity is express as

V = $\sqrt{\frac{T}{\mu } }$ .................2

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f1 = $\frac{\sqrt{\frac{T}{\mu } }}{\frac{2L}{10}}$

f1 = $\frac{10}{2L} \sqrt{\frac{T1}{\mu } }$

and

f2 = $\frac{10}{2L} \sqrt{\frac{T2}{\mu } }$

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put here value

beat frequency = $\frac{10}{2*2} \sqrt{\frac{130}{0.0065}}$ - $\frac{10}{2*2} \sqrt{\frac{120}{0.0065} }$

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