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3 years ago

14

A car is moving with the velocity of 90km/h. If the car come to rest after 10 seconds. Calculate the final velocity and distance

covered by the car within that time.
plz wirte the full formula

1 answer:

djyliett [7]3 years ago

6 0

Answer:

you can learn from here

https://www.toppr.com/ask/en-bd/question/a-car-is-moving-with-a-velocity-of-10-ms-the-driver-sees-a-wall/

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When you lift a bowling ball with a force of 61.1 N, the ball accelerates upward with an acceleration a. If you lift with a forc

gtnhenbr [62]

Answer:

the weight of the ball is w = 51.94 N ( mass = 5.3 kg)

Explanation:

Following Newton's second law:

net force = mass * acceleration = weight/gravity * acceleration

then denoting 1 and 2 as the first and second lift

F₁ - w= w/g *a₁

F₂ -w = w/g *a₂ = w/g * 2.07a

dividing both equations

(F₂- w)/(F₁ -w)= 2.07

(F₂- w) = 2.07 * (F₁ -w)

1.07*w = 2.07*F₁ - F₂

w = (2.07*F₁ - F₂ )/ 1.07

replacing values

w = (2.07*61.1 N - 70.9 N )/ 1.07 = 51.94 N

then the weight of the ball is w = 51.94 N ( mass = 5.3 kg)

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2 years ago

A 20 kg mass is dropped from a tall rooftop and accelerates at 9.8 m/s2. What is the weight of the dropped object?

patriot [66]

Objects in free fall are weightless.

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2 years ago

Read 2 more answers

Help me rearrange this formula. <br><br>I've been trying but I can't remember how to do it.

NikAS [45]

-- Multiply each side of the formula by 2

-- Then divide each side by t

-- Then subtract V(i) from each side.

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2 years ago

If mars has an orbital period of 1.84 years how far from the sun is it and how

EastWind [94]

<span>141.6 million mi,and idk what u mean by how</span>

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2 years ago

A charged cloud system produces an electric field in the air near earth's surface. a particle of charge -2.0 × 10-9 c is acted o

defon

Part a)

Magnitude of electric field is given by force per unit charge

$E = \frac{F}{q}$

$E = \frac{4.3 * 10^{-6}}{2 * 10^{-9}}$

$E = 2150 N/C$

Part b)

Electrostatic force on the proton is given as

F = qE

$F = 1.6 * 10^{-19} * 2150$

$F = 3.44 * 10^{-16} N$

PART C)

Gravitational force is given by

$F_g = mg$

$F_g = 1.6 * 10^{-27}*9.8$

$F_g = 1.57 * 10^{-26} N$

PART d)

Ratio of electric force to weight

$\frac{F_e}{F_g} = \frac{3.44 * 10^{-16}}{1.57*10^{-26}}$

$\frac{F_e}{F_g} = 2.2 * 10^{10}$

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3 years ago