All categories

2 years ago

Consider a skateboarder who starts from rest at the top of ramp that is inclined at an angle of 18.0 ∘ to the horizontal.

Physics

1 answer:

kotegsom [21]2 years ago

3 0

Answer:

Explanation:

v= u + at

v is final velocity , u is initial velocity . a is acceleration and t is time

Initial velocity u = 0 . Putting the given values in the equation

v = 0 + g sin 18 x 3.5

= 10.6 m /s

You might be interested in

A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9

Andreyy89

Answer:

$d = 0.0306 m$

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

$U + K = constant$

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

$\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0$

$7.05 + 30.5 = \frac{0.266}{r}$

$r = 7.08 \times 10^{-3} m$

so distance moved by the particle is given as

$d = r_1 - r_2$

$d = 0.0377 - 0.00708$

$d = 0.0306 m$

6 0

2 years ago

A flat uniform circular disk (r= 2.00m,

dusya [7]

Incomplete question.The Complete question is here

A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.

a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.

b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.

Answer:

(a)ω = 1 rad/s

(b)t = 2.41 s

Explanation:

(a) initial angular momentum = final angular momentum

0 = L for disk + L............... for runner

0 = Iω² - mv²r ...................they're opposite in direction

0 = (MR²/2)(ω²) - mv²r ................where is ω is angular speed which is required in part (a) of question

0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)

0=200ω²-200

200=200ω²

ω = 1 rad/s

b.)

lets assume the "starting point" is a point marked on the disk.

The person's angular speed is

v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s

As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.

(angle) + (angle disk turns) = 2π

(1.6 rad/s)(t) + ωt = 2π

t[1.6 rad/s + 1 rad/s] = 2π

t = 2.41 s

6 0

3 years ago

For a top player, a tennis ball may leave the racket on the serve with a speed of 55 m/s (about 120 mi/h). If the ball has a mas

inn [45]

Answer:

Yes is large enough

Explanation:

We need to apply the second Newton's Law to find the solution.

We know that,

$F= ma$

And we know as well that

$a= \frac{v}{t}$

Replacing the aceleration value in the equation force we have,

$F= \frac{mv}{t}$

Substituting our values we have,

$F= \frac{(0.060)(55)}{4*10^{-3}}$

$F=825N$

The weight of the person is then,

$W = mg$

$W = (60)(9.8) = 558N$

<em>We can conclude that force on the ball is large to lift the ball</em>

6 0

2 years ago

Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant

Elena L [17]

Answer:

19 m/s

Explanation:

The complete question requires the final speed to be calculated.

Velocity is the rate and direction at which an object moves. Acceleration is the rate of change of velocity per unit time and can be calculated by the difference in velocity over a given time.

For this question, first the unknown acceleration must be calculated and used to determine the final velocity

Step 1: Calculate the acceleration

$a=\frac{v_{2}-v{1}}{t_{1}}$