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Olegator [25]
3 years ago
13

32. Only a small percentage of the energy

Physics
1 answer:
nordsb [41]3 years ago
5 0
Only a small Percentage of the energy emitted by the sun strikes earth because, since the earth is going in circles round and round, the sun only hits part of the earth and not fully. So for example if the earth is going around and the sun hits Africa then in a couple minutes it will go to the next country and that country will have sunlight. Let me know if you need anything else or if this isn't partially correct. GLAD TO HELP! :)
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A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v
Citrus2011 [14]

Answer:

part (a) v\ =\ 10.42\ at\ 81.17^o towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

  • velocity of the river due to east = v_r\ =\ 1.60\ m/s.
  • velocity of the boat due to the north = v_b\ =\ 10.3\ m/s.

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are 90^o

Let 'v' be the velocity of the boat relative to the shore.

\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.

Let \theta be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore.\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o

part (b)

  • Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river.\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m

7 0
3 years ago
The<br> is the time it takes for a wave to complete one cycle.
hichkok12 [17]

Answer:

Time Period

Explanation:

3 0
3 years ago
Read 2 more answers
What is velocity change divided by time change?
Luda [366]
Acceleration: is the change in velocity divided by the time it takes for the change to occur. . Acceleration is the change in verity divided by the time it takes to make the change, Acceleration has direction.
5 0
4 years ago
What is the maximum acceleration the belt can have without the crate slipping? express your answer using two significant figures
Montano1993 [528]

To prevent the crate from slipping, the maximum force that the belt can exert on the crate must be equal to the static friction force.


Ff = 0.5 * 16 * 9.8 = 78.4 N

a = 4.9 m/s^2


If acceleration of the belt exceeds the value determined in the previous question, what is the acceleration of the crate?


In this situation, the kinetic friction force is causing the crate to decelerate. So the net force on the crate is 78.4 N minus the kinetic friction force.


Ff = 0.28 * 16 * 9.8 = 43.904 N

Net force = 78.4 – 43.904 = 34.496 N

To determine the acceleration, divide by the mass of the crate.

a = 34.496 ÷ 16 = 2.156 m/s^2



8 0
3 years ago
An electron and a proton are fixed at a separation distance of 973 nm. Find the magnitude and the direction of the electric fiel
Vilka [71]

Answer:

The magnitude is: |E|=6084.1\: N/C

The direction of E is in the negative x-direction.

Explanation:

The electric field equation is:

E=k\frac{Q}{r^{2}}

Where:

  • Q is the charge (we can choose the electron or the proton)
  • r is the distance (in our case is at the midpoint 973/2 nm)
  • k is the Coulomb constant (9*10^{9}\: Nm^{2}C^{-2})

Using the electron charge (e = -1.6*10^{-19}\: C)

E=-9*10^{9}\frac{1.6*10^{-19}}{(486.5*10^{-9})^{2}}

The magnitude is:

|E|=6084.1\: N/C

The direction of E is in the negative x-direction.

I hope it helps you!

6 0
3 years ago
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