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andrey2020 [161]
3 years ago
11

A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal

l moves so fast that the string is always taut and perpendicular to the velocity of the ball. As the ball swings from its lowest point to its highest point, what is the work done by gravity and by the tension?
Physics
1 answer:
Brut [27]3 years ago
6 0

Answer:

  • The gravity does a work of - 117.6 Joules.
  • The tension does not do work as the force is perpendicular to the direction of motion at any point in the trajectory.

Explanation:

The work done by the gravity simply is the difference in gravitational potential energy multiplied by -1:

W_g = - \Delta E_p = - (mgh_f  - m g h_i)

where m is the mass of the ball, g is the acceleration due to gravity, h_f is the final height and h_i is the initial height.

So, if the radius is 2.00 m, then the difference of height will be 4 meters:

W_g = - mg (h_f - h_i)

W_g = - 3.00 \ kg \ 9.8 \frac{m}{s^2} \ 4 \m

W_g = - 117.6 Joules

As the tension is perpendicular to the velocity of the ball, the force is always perpendicular to the direction of motion. So, the differential of work will be:

dW = \vec{F}  d\vec{r} = 0

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Need help really fast ! can give brainliest if correct :> thanks <3
Vlada [557]

Answer:

2. Move faster

Explanation:

Because you lighten the weight and pushed at the same speed it is easier to push the 400-grams than the 800-grams.

Have a wonderful day!

8 0
3 years ago
Read 2 more answers
011 10.0 points
Ulleksa [173]

Answer:

2.47 m

Explanation:

Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.

The horizontal velocity of the ball is constant:

v_x = v cos \theta = (25)(cos 35.9^{\circ})=20.3 m/s

and the time taken to cover the horizontal distance d is

t=\frac{d}{v_x}=\frac{52}{20.3}=2.56 s

So this is the time the ball takes to reach the horizontal position of the crossbar.

The vertical position of the ball at time t is given by

y=u_y t - \frac{1}{2}gt^2

where

u_y = v sin \theta =(25)(sin 35.9^{\circ})=14.7 m/s is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:

y=(14.7)(2.56) - \frac{1}{2}(9.8)(2.56)^2=5.52 m

The height of the crossbar is h = 3.05 m, so the ball passes

h' = 5.52- 3.05 = 2.47 m

above the crossbar.

8 0
3 years ago
in a certain experiment, a radio transmitter emits sinusoidal electromagnetic waves of frequency 105.0 mhz in opposite direction
Romashka [77]

As a result, the hollow is 10.90 meters long and the distance between the nodal planes is 1.36 meters.

<h3>Explain electromagnetic waves.</h3>

The oscillations between an electric field and a magnetic field produce waves known as electromagnetic waves, or EM waves.

By definition, we understand that the frequency equals,

f = c/λ

where,

λ = wavelength

c= Speed of light

λ = 2L / n

While the wavelength is equal to,

Where,

L = Length

n = Number of antinodes/nodes

PART A) We know that the first component's wavelength is 110 MHz, so

λ = c/ f

λ = 3*10^8 / 11*10^6

λ = 1.36m

Therefore the distance between the nodal planes is 1.36m

PART B) For this part we need to find the Length through the number of nodes (8) and the wavelength, that is,

λ` = 2l /n

L = 8*2.72/ 2

L = 10.90m

Therefore the length of the cavity is 10.90m.

To know more about electromagnetic waves visit:-

brainly.com/question/3101711

#SPJ4

7 0
1 year ago
A point charge with charge q1 = 3.40 μC is held stationary at the origin. A second point charge with charge q2 = -4.90 μC moves
expeople1 [14]

Answer:

-0.79 J

Explanation:

We are given that

q_1=3.4\mu C=3.4\times 10^{-6} C

1\mu C=10^{-6} C

q_2=-4.9\mu C=-4.9\times 10^{-6} C

x_1=0.125,y_1=0

x_2=0.280,y_2=0.235

We have to find the work done by the electric force on the moving point charge.

r_1=\sqrt{x^2_1+y^2_1}=\sqrt{(0.125)^2+0}=0.125

r_2=\sqrt{(0.280)^2+(0.235)^2}=0.366

Work done,W=kq_1q_2(\frac{1}{r_1}-\frac{1}{r_2})

Where k=9\times 10^9

Using the formula

W=9\times 10^9\times 3.4\times 10^{-6}\times(-4.9\times 10^{-6})(\frac{1}{0.125}-\frac{1}{0.366})

W=-0.79 J

5 0
4 years ago
What does the law of conservation of matter state
r-ruslan [8.4K]

Answer:

Matter can be changed, but not created or destroyed.

Explanation:

Matter can change for through physical and chemical changes, though this happens, matter is conserved. The same amount of matter exists before and after the change.

3 0
2 years ago
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