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3 years ago

How would the seasons be different if Earth were not tilted on its Axis?

Physics

2 answers:

Alja [10]3 years ago

6 0

Answer:

no seasons

Explanation:

if the earth weren't tilted, it would rotate like that as it revolved around the sun, and we wouldn't have seasons—only areas that were colder (near the poles) and warmer (near the Equator).

spin [16.1K]3 years ago

6 0

If the earth was not tilted on its axis there would be no seasons

You might be interested in

Does frequency affect wavelength?

Vinil7 [7]

Answer:

yeah

Explanation:

as wavelength increases frequency decreases and it goes the same for the opposite way

6 0

3 years ago

A 0.86 kg rock is projected from the edge of the top of a building with an initial velocity of 8.65 m/s at an angle 46◦ above th

Georgia [21]

Answer:

Height of the building = 11.4 m

Explanation:

As we know that the stone is projected at an angle 46 degree with speed 8.65 m/s

so the two components of the speed is given as

$v_x = 8.65 cos46$

$v_x = 6 m/s$

vertical component of the speed is given as

$v_y = 8.65 sin46$

$v_y = 6.22 m/s$

now we know that the ball strike at horizontal distance of 13.7 m

so we will have

$x = v_x t$

$13.7 = 6 t$

$t = 2.28 s$

now we know that in vertical direction ball will move under uniform gravity so we can use kinematics

$y = v_y t + \frac{1}{2}at^2$

$y = 6.22(2.28) - \frac{1}{2}(9.81)(2.28^2)$

$y = -11.4 m$

Height of the building = 11.4 m

3 0

4 years ago

Read 2 more answers

You travel in a circle, whose circumference is 8kilometers, at an average speed of kilometer/hour. If you stop at the same point

dimaraw [331]

You skipped over a number in the question, and you didn't tell me what my average speed is. Lucky for you, my average speed has NO EFFECT on the answer to the question.

When you calculate velocity, you only use the straight-line distance between the start-point and the end-point. It doesn't matter what route the thing took to get there, or how much ground it actually covered.

If I travel in a circle and stop at the same point I started from, then the size of the circle doesn't matter, and neither does my speed. The distance between my start-point and my end-point is zero, and my average velocity is zero.

3 0

4 years ago

Sherlock Holmes examines a clue by holding his magnifying glass (with

Alborosie

Answer:

Distance: -30.0 cm; image is virtual, upright, enlarged

Explanation:

We can find the distance of the image using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where:

f = 15.0 cm is the focal length of the lens (positive for a converging lens)

p = 10.0 cm is the distance of the object from the lens

q is the distance of the image from the lens

Solving for q,

$\frac{1}{q}=\frac{1}{f}-\frac{1}{q}=\frac{1}{15.0}-\frac{1}{10.0}=-0.033\\q=\frac{1}{0.033}=-30.0 cm$

The negative sign tells us that the image is virtual (on the same side of the object, and it cannot be projected on a screen).

The magnification can be found as

$M=-\frac{q}{p}=-\frac{-30}{10}=3$

The magnification gives us the ratio of the size of the image to that of the object: since here |M| = 3, this means that the image is 3 times larger than the object.

Also, the fact that the magnification is positive tells us that the image is upright.

8 0

4 years ago

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for the force of static friction:

$f_s = \mu_s N$ --- (1)

where;

$f_s =$ static friction force

$\mu_s =$ coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

$N = ma_{min}$

From equation (1)

$f_s = \mu_s ma_{min}$

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

$F = f_s$