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3 years ago

A car initally at rest accelerates in a straight line at 3 m/s2 what will be its speed after 2 seconds

Physics

1 answer:

Ainat [17]3 years ago

5 0

Answer

In this Question we have given

acceleration, a=3m/s^2

and time, t=2s

we have to find speed, v=?

Apply newton's Ist equation of motion which goes as

$[tex]v =u + at\\ Here v is final velocity\\u - initial velocity\\a-acceleration\\t- time$ [/tex]

In this problem the car starts from rest it means initial velocity, u =0.

Putting values of u, a and t in first equation of motion

$v=0+3*2\\v=6m/s$

Therefore speed of car after 2 second will be 6m/s

Note: In this type of problems. First check the units of all quatities. As all the quantities must be first changed into matching units i.e cgs or s.i. units before applying the equation.

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A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest in 1

attashe74 [19]

The average act on her during the deceleration is 4.47 meters per second.

Explanation:

Given:

youngster mass m = 50.0 kg

She steps off a 1.00 m high platform that is s = 1 meter

She comes to rest in the 10-meter second

To Find:

The average force and momentum

Formulas:

p = m * v

F * Δ t = Δ p

vf^2= vi^2+2as

Solution:

a = 9.8 m/s

vi = 0

vf^2= 0+2(9.8)(1)

vf^2 = 19.6

vf = 4.47 m/s .

Therefore the average force is 4.47 m/s.

5 0

3 years ago

A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.

enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, $KE_{0} = 0.5 mu^{2}$

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, $KE_{f} = 0.5mv^{2}$

a) Using the principle of energy conservation,

$KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s$

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

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3 years ago

A total distance of 4.5km. The overall journey takes 0.62h, whats the average velocity

Westkost [7]

Answer:

7.3km/hr

Explanation:

v=d/t=4.5km/0.62hrs=7.3km/hr

6 0

3 years ago

3.5

11111nata11111 [884]

Answer:

mass of hot water must be 97.80

Explanation:

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3 years ago

An electron emitted from a filament is travelling at 1.5 x 105 m/s when it enters an acceleration of an electron gun in a televi

Crank

Answer:

The acceleration of the electron is 1.457 x 10¹⁵ m/s².

Explanation:

Given;

initial velocity of the emitted electron, u = 1.5 x 10⁵ m/s

distance traveled by the electron, d = 0.01 m

final velocity of the electron, v = 5.4 x 10⁶ m/s

The acceleration of the electron is calculated as;

v² = u² + 2ad

(5.4 x 10⁶)² = (1.5 x 10⁵)² + (2 x 0.01)a

(2 x 0.01)a = (5.4 x 10⁶)² - (1.5 x 10⁵)²

(2 x 0.01)a = 2.91375 x 10¹³

$a = \frac{2.91375 \ \times \ 10^{13}}{2 \ \times \ 0.01} \\\\a = 1.457 \ \times \ 10^{15} \ m/s^2$

Therefore, the acceleration of the electron is 1.457 x 10¹⁵ m/s².

7 0

3 years ago