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Tasya [4]
3 years ago
12

A race car starts at rest and speeds up to 40 m/s in a distance of 100 m. Determine the acceleration of the car.

Physics
1 answer:
Finger [1]3 years ago
8 0

Answer:

Acceleration of the car, a=8\ m/s^2

Explanation:

Given that,

Initial speed of the car, u = 0 (at rest)

Final speed of the car, v = 40 m/s

Distance covered, s = 100 m

We need to find the acceleration of the car. It can be calculated using third equation of motion as :

v^2-u^2=2as

a is the acceleration of the car

a=\dfrac{v^2}{2s}

a=\dfrac{(40)^2}{2\times 100}

a=8\ m/s^2

So, the acceleration of the car is 8\ m/s^2. Hence, this is the required solution

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A motorbike is traveling to the left with a speed of 27.0\,\dfrac{\text m}{\text s}27.0 s m ​ 27, point, 0, start fraction, star
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Answer:

\frac{729\frac{m^{2} }{s^{2} } }{83m}≈8.8\frac{m}{s^{2} }

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3 years ago
The two lenses in a microscope are separated by 19.5 cm, and the focal length of the eyepiece lens is 2.75 cm.
Naya [18.7K]

Answer:

Explanation:

a )  If the image of this object is viewed with the eyepiece adjusted for minimum eyestrain (image at the far point of the eye) , the image from object lens must have been formed at the focus of eye lens . So the objective image must have been formed at 19.5 - 2.75 = 16.75 cm from the object lens.

b ) Let the object distance be u

For object lens

v = 16.75 cm , f = .35 cm

1/v - 1/u = 1/f

1/16.75  - 1/u = 1/ .35

.0597 - 1/u = 2.857

1/u = - 2.7973

u = .3575 cm

c ) Angular magnification

= \frac{v_o}{u_o} \times\frac{D}{f_e}

v₀ and u₀ are image and object distance for object lens , D = 25 cm and f_e is focal length of eye lens

= (16.75 / .3575) x( 25 / 2.75)

= 46.85 x 9.09

= 426

7 0
3 years ago
A 500kg elevator is raised to a height of 10 m in 10 seconds. Calculate the power of the motor.
Hunter-Best [27]

Answer:

Correct answer: Third statement  P = 4900 W

Explanation:

Given:

m = 500 kg  the mass of the elevator

h = 10 m  reached height after t = 10 seconds

P = ? power of the motor

The formula for the calculating power of the motor is:

P = W / t

since work is a measure of change in this case of potential energy then it is:

W = ΔEp = Ep - 0 = Ep

In this case we must take g = 9.81 m/s²

Ep = m g h = 500 · 9.81 · 10 = 49,050 W ≈ 49,000 W

Ep ≈ 49,000 W

P = Ep / t = 49,000 / 10 = 4,900 W

P =4,900 W

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3 0
4 years ago
An electron remains suspended between the surface of the Earth (assumed neutral) and a fixed positive point charge, at a distanc
DaniilM [7]

To solve this problem we will apply the concept related to the balance of forces. Here the electrostatic force defined by Coulomb's laws must be proportional to the weight of the electron, defined by Newton's second law, so that such balance exists, both can be described as:

F_e = \frac{kq_1q_2}{d^2}

Here,

q_1 = Charge required

q_2 =Charge of electron

d = Distance

k = Coulomb's constant

F_w = mg

Here,

m = mass

g = Gravitational acceleration

In equilibrium the sum  of Force is zero, then,

\sum F = 0

F_e = F_w

\frac{kq_1q_2}{d^2} = mg

\frac{(8.98755*10^9 )(q_1)(1.6*10^{-19})}{8.8^2} = (9.8)(9.10939*10^{-31} )

q_1 = 4.8075*10^{-19}C

Therefore the charge required for this to happen is 4.8075*10^{-19}C

5 0
3 years ago
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