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madreJ [45]
3 years ago
8

It takes 23 hours 56 minutes and 4 seconds for the earth to make one revolution (mean sidereal day). What is the angular speed o

f the earth? Assume the earth is spherical. Relative to someone on the rotation axis, what is the linear speed of an object on the surface if the radius vector from the center of the earth to the object makes an angle of 48.0� with the axis of rotation. The radius of the earth is 6.37�103 km. What is the acceleration of the object on the surface of the earth in the previous problem? 2. A bicycle is moving at a speed v = 3.80 m/s. If the radius of the front wheel is 0.450 m, how long does it take for that wheel to make a complete revolution? 3. A tire placed on a balancing machine in a service station starts from rest and turns through 19.0 revolutions in 9.53 s before reaching its final angular speed. Calculate its angular acceleration.

Physics
1 answer:
Dominik [7]3 years ago
3 0

Answer:

Explanation:

Check attachment for solution

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cup of hot chocolate has a spoon in it. How does the heat move from the hot chocolate to the metal spoon handle?
serg [7]

The heat moves from the hot chocolate to the handle of the spoon by a process called thermal conduction.

It is the transfer of heat energy from one object to another when they are in contact with eachother.

Hope this answers your question.

5 0
3 years ago
A uniform electric field of magnitude 7.0 ✕ 104 N/C passes through the plane of a square sheet with sides 5.0 m long. Calculate
Vadim26 [7]

Answer:

1.52*10^6 Nm^2/C

Explanation:

Given that:

Electrical field E = 7.0 * 10^{-4}N/C

square side l = 5.0 m

Area A = 5.0 * 5.0

= 25.0 m²

Angle ( θ ) between area vector and E = (90° - 60°)

= 30°

The flux \phi_E can now be determined by using the expression

\phi_E = E*A*Cos \theta

\phi_E = 7.0 * 10^{-4}N/C *25.0m*Cos 30^0

\phi_E = 1515544.457 Nm^2/C

\phi_E = 1.52*10^6 Nm^2/C

5 0
3 years ago
Read 2 more answers
A 12.0 kg rock slides off the edge of a bridge and falls into the water 25.0 meters below. what is the kinetic energy of the roc
Pavel [41]
During the fall, all the initial potential energy of the rock 
U=mgh
has converted into kinetic energy of motion
K= \frac{1}{2}mv^2
where h is the initial height of the rock, m its mass, and v its velocity just before hitting the water. So, for energy conservation, we have
U=K
and so we can find the value of K, the kinetic energy of the rock just before hitting the ground:
K=U=mgh = (12.0 kg)(9.81 m/s^2)(25.0 m)=2943 J
7 0
3 years ago
The illustration in figure below shows a uniform metre rule weighing 30 N pivoted on a wedge placed under the 40 cm mark and car
Nitella [24]

Answer:

W = 30 N

Explanation:

Applying the summation of torques about the wedge for equilibrium, taking the clockwise direction as negative. Since the ruler is balanced horizontally about the wedge. Therefore, the summation of all torques acting about the wedge must be equal to zero.

(70\ N)(40\ cm - 10\ cm)-(30\ N)(50\ cm-40\ cm)-(W)(100\ cm - 40\ cm) = 0\\W(60\ cm) = (70\ N)(30\ cm)-(30\ N)(10\ cm)\\\\W = \frac{1800\ N.cm}{60\ cm}

<u>W = 30 N</u>

6 0
2 years ago
A car traveling in a straight line has a velocity of 5.0 m/s. After an acceleration of 0.75 m/s/s, the cars velocity is 8.0. In
Bogdan [553]
Vs - velocity on beginning
ve - velocity on ending. You've got:
v_s = 5 \frac{m}{s} \\ v_e=8 \frac{m}{s} \\ \hbox{Then:} \\ \Delta v=v_e - v_s = 8 \frac{m}{s} - 5\frac{m}{s}=3 \frac{m}{s} \\ a=0,75 \frac{m}{s^2} \\ \hbox{And from formula:} \\ a=\frac{\Delta v}{\Delta t} \qquad \Rightarrow \qquad  \Delta t= \frac{\Delta v}{a} \\ \hbox{Substitute:} \\ \Delta t=\frac{3\frac{m}{s}}{0,75 \frac{m}{s^2}}= \frac{3}{\frac{3}{4}} s= 3 \cdot \frac{4}{3} s= 4 s
So he needed  4 second. 

3 0
3 years ago
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