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Pie
3 years ago
8

How do we know Earth is spinning, not the Sun?

Physics
1 answer:
Tpy6a [65]3 years ago
4 0

Answer:

Because it is said that the earth rotates and revolves around the sun and moon so it is impossible for the earth not to be spinning.

Explanation:

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How will the force of friction affect a wood block being pushed over a table?
olasank [31]

Answer:

D

friction acts in the opposite direction of motion but does not affect the motion of the object

5 0
2 years ago
Calculate the angle θ between the radius-vector of the point and the positive x axis (measured counterclockwise from the positiv
Y_Kistochka [10]

The point obviously is in the 3rs quadrant

So

စ= tan^-1( y/x)-180

စ= -89.7°

6 0
3 years ago
If NEPA charges 5k per kWh, what is the cost of
tiny-mole [99]

Answer:

24k

Explanation:

We multiply by 200V by 24

8 0
2 years ago
(a) Find the magnitude of an earthquake that has an intensity that is 37.25 (that is, the amplitude of the seismograph reading i
MAVERICK [17]

Answer:

The magnitude of an earthquake is 5.6.

Explanation:

The magnitude of an earthquake can be found as follows:

M = log(\frac{I}{S})

Where:

I: is the intensity of the earthquake = 37.25 cm

S: is the intensity of a standard earthquake = 10⁻⁴ cm

Hence, the magnitude is:

M = log(\frac{I}{S}) = log(\frac{37.25}{10^{-4}}) = 5.6

Therefore, the magnitude of an earthquake is 5.6.

I hope it helps you!

6 0
3 years ago
A 4 cm diameter ball is located 40 cm from a point source and 80 cm from a wall. What is the size of the shadow on the wall?
Slav-nsk [51]
<span>11.823 cm There is a slight ambiguity with this question in that I don't know if the measurements are from the surface of the ball, or the center of the ball. I will take this question literally and as such the point light source will be 124 cm from the wall. The key thing to remember is that ball won't be showing an effective diameter of 4 cm to the light source. Instead the shadow line is a tangent to the ball's surface. There is a right triangle where the hypotenuse is the distance from the center of the ball to the light source (42 cm), one leg of the triangle is the radius (2cm). That right triangle will define a chord that will be the effective diameter of the disk casting the shadow. The cosine of the half angle of the chord will be 2/42 = 1/21. The sine of the half angle then becomes sqrt(1-(1/21)^2) = sqrt(440/441) = 2sqrt(110) = 0.99886557. Now multiply that sine by 4 (radius of ball multiplied by 2 since it's the half angle and we want the full side of the chord) and we get an effective diameter of 3.995462279 cm. Now we need to calculate the effective distance that circle is from the wall. It will be slightly larger than 82 cm. The exact value will be 82 + cos(half angle) * radius. So 82 + 1/21 * 2 = 82 + 2/21 = 82.0952381 Now we have the following dimensions with a circle replacing the ball in the original problem. Distance from wall to effective circle = 82.0952381 cm Distance from effective circle to point source = 124 - 82.0952381 = 41.9047619 cm Effective diameter of circle = 3.995462279 cm And because the geometry makes similar triangles, the following ratio applies. 3.995462279/41.9047619 = X/124 Now solve for X 3.995462279/41.9047619 = X/124 124*3.995462279/41.9047619 = X 495.4373226/41.9047619 = X 11.82293611 = X The shadow cast on the wall will be a circle with a diameter of 11.823 cm</span>
7 0
3 years ago
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