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2 years ago

A racecar accelerates from rest at 6.5 m/s2 for 4.1 s. How fast will it be going at the end of that time?

Physics

1 answer:

Pie2 years ago

6 0

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

acceleration of the racecar, a = 6.5 m/s²

initial velocity of the car, u = 0

time of motion, t = 4.1 s

The final velocity of the car is given by;

v = u + at

where;

v is the final velocity of the car

suvstitute the givens

v = 0 + (6.5)(4.1)

v = 26.65 m/s.

Therefore, the final velocity of the car is 26.65 m/s.

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A small ball of charge Q and mass m has a velocity v at infinity. It collides head-on with a ball of the same charge and mass wh

Hatshy [7]

Answer:

Explanation:

Kinetic energy of ball in motion = 1/2 m v² . Potential energy = 0

Let the minimum distance between the balls be d on collision.

Electric potential energy at that time= k Q²/d , Here kinetic energy is converted into potential energy . So

1/2 m v² = kQ²/d

d =2 k Q² / mv²,= 18 x 10⁹ x Q²/ m v².

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3 years ago

What happens when a force is applied to an object in stable equilibrium

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Answer:

the object is no longer in equilibrium .

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3 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago

Use the table below to calculate the acceleration of the object. Time (s) 0.0 3.0 6.0 9.0 Velocity (m/s) 0.0 1.2 2.4 3.6

antiseptic1488 [7]

explanation

a=average velocity/average time

average velocity=0.0+1.2+2.4+3.6/4

average velocity=7.2/4

average velocity=1.8 m/s

average time=0.0+3.0+6.0+9.0/4

average time=18/4

average time=4.5 s

a= average velocity/average time

a=1.8/4.5

a=0.4 m/s²

8 0

3 years ago

An object has a mass of 300 g. (a) What is its weight on Earth? (b) What is its mass on the Moon?

Neko [114]

Answer:

The mass of object is 300g.

Its weight on earth is W×0.3kg×9.8m/s2=2.94 N

Its weight on moon is A 300 g would be 48 g on the moon

GOOD LUCK!

7 0

3 years ago