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Pachacha [2.7K]
3 years ago
5

3- (a) What is the period of rotation of Earth in seconds? (b) What is the angular velocity of Earth? (c) Given that Earth has a

radius of 40,075 km at its equator, what is the linear velocity at Earth’s surface?
Physics
1 answer:
ASHA 777 [7]3 years ago
7 0

Explanation:

a) The Earth makes 1 rotation in 24 hours.  In seconds:

24 hr × (3600 s / hr) = 86400 s

b) 1 rotation is 2π radians.  So the angular velocity is:

2π rad / 86400 s = 7.27×10⁻⁵ rad/s

c) The earth's linear velocity is the angular velocity times the radius:

40075 km × 7.27×10⁻⁵ rad/s = 2.91 km/s

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A ball is thrown straight up into the air and passes a window 3 seconds after being released. It passes the same window on its w
melomori [17]

The  initial velocity of the ball is 55.125 m/s.

<h3>Initial velocity of the ball</h3>

The initial velocity of the ball is calculated as follows;

During upward motion

h = vi - ¹/₂gt²

h = vi - 0.5(9.8)(3²)

h = vi - 44.1 ----------------- (1)

During downward motion

h = vi + ¹/₂gt²

h = 0 + 0.5(9.8)(1.5)²

h = 11.025 ----------- (2)

solve (1) and (2) together, to determine the initial velocity of the ball

11.025 = vi - 44.1

vi = 11.025 + 44.1

vi = 55.125 m/s

Thus, the  initial velocity of the ball is 55.125 m/s.

Learn more about initial velocity here: brainly.com/question/19365526

#SPJ1

8 0
1 year ago
How much energy is stored in a 2.80-cm-diameter, 14.0-cm-long solenoid that has 200 turns of wire and carries a current of 0.800
ozzi

Answer:

The energy stored in the solenoid is 7.078 x 10⁻⁵ J

Explanation:

Given;

diameter of the solenoid, d = 2.80 cm

radius of the solenoid, r = d/2 = 1.4 cm

length of the solenoid, L = 14 cm = 0.14 m

number of turns, N = 200 turns

current in the solenoid, I = 0.8 A

The cross sectional area of the solenoid is given as;

A = \pi r^2\\\\A = \pi (0.014)^2\\\\A = 6.16*10^{-4} \ m^2

The inductance of the solenoid is given by;

L = \frac{\mu_0 N^2A}{l} \\\\L =  \frac{(4\pi*10^{-7})(200^2)(6.16*10^{-4})}{0.14}\\\\L = 2.212*10^{-4} \ H

The energy stored in the solenoid is given by;

E = ¹/₂LI²

E = ¹/₂(2.212 x 10⁻⁴)(0.8)²

E = 7.078 x 10⁻⁵ J

Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J

8 0
3 years ago
A cylinder of mass 14.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed o
aev [14]

Answer:

a) 567J

b) 283.5J

c)850.5J

Explanation:

The expression for the translational kinetic energy is,

E_r = \frac{1}{2} mv^2

Substitute,

14kg for m

9m/s for v

E_r = \frac{1}{2} (14) (9)^2\\= 567J

The translational kinetic energy of the center of mass is 567J

(B)

The expression for the rotational kinetic energy is,

E_R = \frac{1}{2} Iw^2

The expression for the moment of inertia of the cylinder is,

I = \frac{1}{2} mr^2

The expression for angular velocity is,

w = \frac{v}{r}

substitute

1/2mr² for I

and vr for w

in equation for rotational kinetic energy as follows:

E_R = (\frac{1}{2}) (\frac{1}{2} mr^2)(\frac{v}{r} )^2

= \frac{mv^2}{4}

E_R = \frac{14 \times 9^2 }{4} \\\\= 283.5J

The rotational kinetic energy of the center of mass is 283.5J

(c)

The expression for the total energy is,

E = E_r + E_R\\\\

substitute 567J for E(r) and 283.5J for E(R)

E = 567J + 283.5\\= 850.5J

The total energy of the cylinder is 850.5J

6 0
3 years ago
Read 2 more answers
The slope of the following graph represents
guajiro [1.7K]

Answer:

ok jsjajakaka you can come to me when you get home can you please send me the details of the day and I will be there at this time of year is the best way to get a hold of the guy who was the guy who was the guy who was the guy who was the guy who was the guy that was the only thing that was the case but I don't know if I can help you

4 0
3 years ago
Part A (4 pts) Consider light of wavelength λ = 670nm traveling in air. The light is incident at normal incidence upon a thin fi
s344n2d4d5 [400]

Answer:

Explanation:

On both sides of the film , the mediums have lower refractive index.

for interfering pattern from above , for constructive interference of reflected wave from both sides of the film , the condition is

2μt = ( 2n +1 ) λ / 2

μ is refractive index of film ,t is thickness of film λ is wavelength of light

n is order of fringe

for minimum thickness

n = 0

2μt =  λ / 2

t =  λ / 4μ

= 670 / 1.75 x 4

= 95.71 nm .

6 0
3 years ago
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