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Lina20 [59]
2 years ago
10

An object is travelling at a velocity of 5m/s to the right for 5 seconds. If NO forces are applied to the object, what will the

velocity of the object be after 5 seconds.
Physics
1 answer:
creativ13 [48]2 years ago
6 0

Answer:

Below

Explanation:

We can use this formula to find the final velocity of the object :

     vf = vi + at

Now, because no force is applied to the object, the acceleration of this object will be 0 according to Newton's 1st law!

     vf = 5m/s + (0)(5)

     vf = 5 m/s

Hope this helps!

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A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air
Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, v_{o} = 20.0 m/s

Angle made by the ramp, \theta = 22.0^{\circ}

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH

where

H = dsin22^{\circ} = 5sin22^{\circ}

g = 9.8 m/s^{2}

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}

v = \sqrt{400 - 19.6\times 5sin22^{\circ}} = 19.06 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m

(b) now, considering the coefficient of friction bhetween ramp and the ball, \mu = 0.150:

velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH - \mu gd

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5

v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5} = 18.7 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m

6 0
3 years ago
A circuit breaker is rated for a current of 25 a rms at a voltage of 240 v rms. (a) what is the largest value of imax that the b
Mademuasel [1]

The largest value of current that the breaker can carry = Imax=35.4A

Explanation:

Rms value of current= Irms= 25 A

The rms current and the maximum current are related as

Imax= √2 Irms

Imax=√2 (25)

Imax=35.4 A

Thus the maximum current carried by the breaker= 35.4 A

8 0
4 years ago
Why are you able to make someone on a swing have very large swings by only exerting very small forces on them?
Allisa [31]
Because almost all of the force is done by the weight of the person and the mechanism of the swing itself, when you push someone you only give them an increase in velocity, the acceleration comes from the weight at first and then from gravity when the person is coming down, which is why we bend our legs when coming down
3 0
3 years ago
What is the car's average velocity (in m/s) in the interval between t = 1.0 s<br> to t = 1.5 s?
natali 33 [55]

Answer:

1.4 m/s

Explanation:

From the question given above, we obtained the following data:

Initial Displacement (d1) = 0.9 m

Final Displacement (d2) = 1.6 m

Initial time (t1) = 1.5 secs

Final time (t2) = 2 secs

Velocity (v) =..?

The velocity of an object can be defined as the rate of change of the displacement of the object with time. Mathematically, it can be expressed as follow:

Velocity = change of displacement /time

v = Δd / Δt

Thus, with the above formula, we can obtain the velocity of the car as follow:

Initial Displacement (d1) = 0.9 m

Final Displacement (d2) = 1.6 m

Change in displacement (Δd) = d2 – d1 = 1.6 – 0.9

= 0.7 m

Initial time (t1) = 1.5 secs

Final time (t2) = 2 secs

Change in time (Δt) = t2 – t1

= 2 – 1.5

= 0.5 s

Velocity (v) =..?

v = Δd / Δt

v = 0.7/0.5

v = 1.4 m/s

Therefore, the velocity of the car is 1.4 m/s

4 0
3 years ago
What is the difference between real and apparent weightlessness?
MrRissso [65]

Answer:

In space we feel weightlessness because the earth's gravity has less effect on us. The Earth's gravitational attraction at those altitudes is only about 11% less than it is at the Earth's surface. If you had a ladder that could reach as high as the shuttle's orbit, your weight would be 11% less at the top.

Explanation:

Hope this helps:)

4 0
3 years ago
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