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Flauer [41]
3 years ago
7

You are less likely to see a total solar eclipse than a total lunar eclipse because a. the moon’s shadow covers all of Earth dur

ing a solar eclipse. b. new moon phases occur less often than full moon phases. c. only people on the daytime side of Earth can see a solar eclipse. d. the moon’s umbra only covers a small area on Earth’s surface
Physics
2 answers:
Bad White [126]3 years ago
8 0
D. Because the moons shadow during a total lunar eclipse is tinnier than the earth.
frozen [14]3 years ago
4 0

Answer:

You are less likely to see a total solar eclipse than a total lunar eclipse because

d. the moon’s umbra only covers a small area on earth’s surface.

Explanation:

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A window washer who does not want to change his position will want the forces acting on him to be ____________.
natali 33 [55]
My answer is a balanced
6 0
3 years ago
Does parallax affect the precision of a measurement that you make
Paul [167]
Yes, parallax affects the precision of a measurement that you make. It introduces an error in the order of the parallax. It will cause the measurement to be different from the real answer. Hope this answers the question. Have a nice day.
8 0
4 years ago
Read 2 more answers
A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass,
Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2

now we know that

v_0 = 660 m/s

M = 5.98 \times 10^{24} kg

m = 5 \times 10^9 kg

r = 4 \times 10^9 m

R = 6.37 \times 10^6 m

now from above formula

GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2

now we have

2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2

now plug in all data

2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2

v = 7934.2 m/s

5 0
3 years ago
A 90. 0-kg ice hockey player hits a 0. 150-kg puck, giving the puck a velocity of 45. 0 m/s. If both are initially at rest and i
Mice21 [21]

The distance traveled by the hockey player is 0.025 m.

<h3>The principle of conservation of linear momentum;</h3>
  • The principle of conservation of linear momentum states that, the total momentum of an isolated system is always conserved.

The final velocity of the hockey play is calculated by applying the principle of conservation of linear momentum;

m_1v_1 = m_2 v_2\\\\v_1 = \frac{m_2 v_2}{m_1} \\\\v_1 = \frac{0.150 \times 45}{90} \\\\v_1 = 0.075 \ m/s

The time taken for the puck to reach 15 m is calculated as follows;

t = \frac{d}{v} \\\\t = \frac{15\ m}{45 \ m/s} \\\\t = 0.33 \ s

The distance traveled by the hockey player at the calculated time is;

d = vt\\\\d = 0.075 \ m/s \ \times 0.33 \ s\\\\d = 0.025 \ m

Learn more about conservation of linear momentum here: brainly.com/question/7538238

4 0
2 years ago
Merry-go-rounds are a common ride in park playgrounds. The ride is a horizontal disk that rotates about a vertical axis at their
Vera_Pavlovna [14]

Answer:

A = 2.36m/s

B = 3.71m/s²

C = 29.61m/s2

Explanation:

First, we convert the diameter of the ride from ft to m

10ft = 3m

Speed of the rider is the

v = circumference of the circle divided by time of rotation

v = [2π(D/2)]/T

v = [2π(3/2)]/4

v = 3π/4

v = 2.36m/s

Radial acceleration can also be found as a = v²/r

Where v = speed of the rider

r = radius of the ride

a = 2.36²/1.5

a = 3.71m/s²

If the time of revolution is halved, then radial acceleration is

A = 4π²R/T²

A = (4 * π² * 3)/2²

A = 118.44/4

A = 29.61m/s²

7 0
3 years ago
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