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2 years ago

Label the diagram Please

Physics

1 answer:

Nitella [24]2 years ago

8 0

The earth tilt about 23 degrees which leads to the formation of seasons on earth.

<h3>At what degree the earth tilt?</h3>

This is tilted at 23 degrees from the vertical. The northern hemisphere is in the winter season.

This is the path as the earth revolves around the sun. It takes 1 year to complete this path.

This is the path as the earth rotates around its.

So we can conclude that the earth tilt about 23 degrees which leads to the formation of seasons on earth.

Learn more about season here: brainly.com/question/25870256

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Which of the following best describes a noncontact force

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Answer:

a force that is able to act at a distance

Explanation:

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What are some types of landforms on Earth’s surface?<br><br><br><br> PLS ANSWER QUICK 11 POINTS

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3 years ago

Read 2 more answers

A jet plane flying 600 m/s experiences an acceleration of 4g when pulling out of the dive. What is the radius of curvature of th

Gnoma [55]

Answer:

91.84 m/s²

Explanation:

velocity, v = 600 m/s

acceleration, a = 4 g = 4 x 9.8 = 39.2 m/s^2

Let the radius of the loop is r.

he experiences a centripetal force.

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a = v² / r

39.2 x r = 600 x 600

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r = 91.84 m/s²

Thus, the radius of the loop is 91.84 m/s².

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3 years ago

A cyclist rides 6.2 km east, then 9.28 km in a direction 27.27 degrees west of north, then 7.99 km west. A. What is the magnitud

Pani-rosa [81]

Answer:

Explanation:

given,

cyclist ride 6.2 km east and then 9.28 km in the direction of 27.27° west of north and then 7.99 km west.

vertical component = 9.28 cos∅

= 9.28 cos 27.27°

= 8.24 km

horizontal axis component = 9.28 sin ∅

= 9.28 sin 27.27°

= 4.5 km

distance of the final point from the origin

= 7.99 -(6.2-4.5)

= 6.29 km

displacement

$d = \sqrt{6.29^2+8.24^2}$

d = 10.37 km

b) $tan \theta = \dfrac{6.29}{8.24}$

θ = 37.36°

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3 years ago

The tow spring on a car has a spring constant of 3,086 N / m and is initially stretched 18.00 cm by a 100.0 kg college student o

sdas [7]

Answer:

The velocity of the skateboard is 0.774 m/s.

Explanation:

Given that,

The spring constant of the spring, k = 3086 N/m

The spring is stretched 18 cm or 0.18 m

Mass of the student, m = 100 kg

Potential energy of the spring, $P_f=20\ J$

To find,

The velocity of the car.

Solution,

It is a case of conservation of energy. The total energy of the system remains conserved. So,

$P_i=K_f+P_f$

$\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2+20$

$\dfrac{1}{2}\times 3086\times (0.18)^2=\dfrac{1}{2}mv^2+20$

$50-20=\dfrac{1}{2}mv^2$

$30=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{60}{100}}$

v = 0.774 m/s

So, the velocity of the skateboard is 0.774 m/s.

7 0

3 years ago