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3 years ago

What is the kinetic energy of a golf ball with a mass of 0.046 kg traveling at 15.5 m/s?

Physics

1 answer:

Serjik [45]3 years ago

3 0

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

$k = \frac{1}{2} m {v}^{2} \\$

m is the mass

v is the velocity

From the question we have

$k = \frac{1}{2} \times 0.046 \times {15.5}^{2} \\ = 0.023 \times 240.25 \\ = 5.52575$

We have the final answer as

Hope this helps you

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A positively charged particle of mass 7.2 x 10-8 kg is traveling due east with a speed of 88 m/s and enters a 0.6-T uniform magn

Marianna [84]

Answer:

q = 8.57 10⁻⁵ mC

Explanation:

For this exercise let's use Newton's second law

F = ma

where force is magnetic force

F = q v x B

the bold are vectors, if we write the module of this expression we have

F = qv B sin θ

as the particle moves perpendicular to the field, the angle is θ= 90º

F = q vB

the acceleration of the particle is centripetal

a = v² / r

we substitute

qvB = m v² / r

qBr = m v

q = $\frac{m\ v}{B\ r}$

The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use

v = d / t

the distance is ¼ of the circle,

d = $\frac{1}{4} \ 2\pi r$

d = $\frac{\pi }{2r}$

we substitute

v = $\frac{\pi r}{2t}$

r = $\frac{2 \ t \ v}{\pi }$

let's calculate

r = $\frac{2 \ 2.2 \ 10^{-3} \ 88}{\pi }$ 2 2.2 10-3 88 /πpi

r = 123.25 m

let's substitute the values

q = $\frac{ 7.2 \ 10^{-8} \ 88}{ 0.6 \ 123.25}$ 7.2 10-8 88 / 0.6 123.25

q = 8.57 10⁻⁸ C

Let's reduce to mC

q = 8.57 10⁻⁸ C (10³ mC / 1C)

q = 8.57 10⁻⁵ mC

4 0

3 years ago

The speed of a nerve impulse in the human body is about 100 m/s. If you accidentally stub you toe in the dark, estimate the time

vodomira [7]

Answer:

0.02 s

Explanation:

Take the (+x) direction to be up.

The average velocity v during a time interval Δt is the displacement Δx divided by Δt.

v=Δx/Δt

=x_f-x_i/t_f-t_i (1)

We assume that your height is 1.6m

Solving [1]

Δt=Δx/v

= 0.02 s

4 0

3 years ago

3. A football is kicked with a speed of 35 m/s at an angle of 40°.

jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

$u_y = u sin \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_y = (35)(sin 40)=22.5 m/s$

b) 26.8 m/s

The initial horizontal velocity is given by:

$u_x = u cos \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_x = (35)(cos 40)=26.8 m/s$

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

$v_y = u_y + at$

where

$v_y$ is the vertical velocity at time t

$u_y = 22.5 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, $v_y =0$ ; substituting, we find the time t at which this happens:

$0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s$

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}gt^2$

where

s is the vertical displacement at time t

$u_y = 22.5 m/s$

$g=-9.8 m/s^2$

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

$s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m$

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

$t=2(2.30 s)=4.60 s$

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

$v_x = 26.8 m/s$

Therefore, the horizontal distance travelled during the whole motion is

$d=v_x t = (26.8)(4.60)=123.3 m$

So, the ball lands 123.3 m far from the initial point.

4 0

3 years ago

Explain why the amplitude of the wave did not change when you increased the frequency of the stimulation. how well did the resul

barxatty [35]

The amplitude did not change when the recurrence was expanded on the grounds that the long headstrong time of the heart forestalls adjustment. It is the most extreme removal or separation moved by a point on a vibrating body or wave measured from its balance position. It is equivalent to the one-a large portion of the length of the vibration way.

5 0

3 years ago

An orbiting satellite can become charged by the photoelectric effect when sunlight ejects electrons from its outer surface. Sate

Rufina [12.5K]

Answer:

the longest wavelength of incident sunlight that can eject an electron from the platinum is 233 nm

Explanation:

Given data

Φ = 5.32 eV

to find out

the longest wavelength

solution

we know that

hf = k(maximum) +Ф ...............1

here we consider k(maximum ) will be zero because photon wavelength max when low photon energy

so hf = 0

and hc/ λ = +Ф

so λ = hc/Ф ................2

now put value hc = 1240 ev nm and Φ = 5.32 eV

so hc = 1240 / 5.32

hc = 233 nm

the longest wavelength of incident sunlight that can eject an electron from the platinum is 233 nm

8 0

3 years ago