All categories

3 years ago

What physical property of Earth gives rise to the seasons?

Physics

1 answer:

Elza [17]3 years ago

8 0

Answer:

Earth's tilted axis causes the seasons. Throughout the year, different parts of Earth receive the Sun's most direct rays. So, when the North Pole tilts toward the Sun, it's summer in the Northern Hemisphere. And when the South Pole tilts toward the Sun, it's winter in the Northern Hemisphere.

Explanation:

You might be interested in

A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist

Paha777 [63]

Answer:

Part a)

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

$\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

$\frac{1}{2}mv^2 = mgh'$

now we know that

$I = \frac{2}{5}mr^2$

$\omega = \frac{v}{r}$

so we have

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh$

$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh$

$\frac{7}{10}mv^2 = mgh$

$\frac{1}{2}mv^2 = \frac{10}{14}mgh$

so the height on the smooth side is given as

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

6 0

3 years ago

Read 2 more answers

A car travels on a straight, level road. (a) Starting from rest, the car is going 38 ft/s (26 mi/h) at the end of 4.0 s. What is

lbvjy [14]

Answer:

$a)9.5\frac{ft}{s^2}\\ b) 12.66\frac{ft}{s^2}$

Explanation:

A body has acceleration when there is a change in the velocity vector, either in magnitude or direction. In this case we only have a change in magnitude. The average acceleration represents the speed variation that takes place in a given time interval.

$a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{38\frac{ft}{s}-0}{4 s- 0}=9.5\frac{ft}{s^2}\\$

$a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{76\frac{ft}{s}-38\frac{ft}{s}}{7 s- 4s}\\a_{avg}=\frac{38\frac{ft}{s}}{3s}=12.66\frac{ft}{s^2}$

8 0

3 years ago

A quarterback is set up to throw the football to a receiver who is running with a constant velocity v⃗ rv→rv_r_vec directly away

Artist 52 [7]

Answer:

a) V_o,y = 0.5*g*t_c

b) V_o,x = D/t_c - v_r

c) V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

d) Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

Explanation:

Given:

- The velocity of quarterback before the throw = v_r

- The initial distance of receiver = r

- The final distance of receiver = D

- The time taken to catch the throw = t_c

- x(0) = y(0) = 0

Find:

a) Find V_o,y, the vertical component of the velocity of the ball when the quarterback releases it. Express V_o,y in terms of t_c and g.

b) Find V_o,x, the initial horizontal component of velocity of the ball. Express your answer for V_o,x in terms of D, t_c, and v_r.

c) Find the speed V_o with which the quarterback must throw the ball.

Answer in terms of D, t_c, v_r, and g.

d) Assuming that the quarterback throws the ball with speed V_o, find the angle Q above the horizontal at which he should throw it.

Solution:

- The vertical component of velocity V_o,y can be calculated using second kinematics equation of motion:

y = y(0) + V_o,y*t_c - 0.5*g*t_c^2

0 = 0 + V_o,y*t_c - 0.5*g*t_c^2

V_o,y = 0.5*g*t_c

- The horizontal component of velocity V_o,x witch which velocity is thrown can be calculated using second kinematics equation of motion:

- We know that V_i, x = V_o,x + v_r. Hence,

x = x(0) + V_i,x*t_c

D = 0 + V_i,x*t_c

V_o,x + v_r = D/t_c

V_o,x = D/t_c - v_r

- The speed with which the ball was thrown can be evaluated by finding the resultant of V_o,x and V_o,y components of velocity as follows:

V_o = sqrt ( V_o,x^2 + V_o,y^2)

V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

- The angle with which it should be thrown can be evaluated by trigonometric relation:

tan(Q) = ( V_o,y / V_o,x )

tan(Q) = ( (0.5*g*t_c)/ (D/t_c - v_r) )

Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

6 0

3 years ago

Larry drops a 5kg ball off of a building. The ball hits the ground 4.7s later. How tall is the building?

musickatia [10]

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 4.7 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (4.7 s) + ½ (9.8 m/s²) (4.7 s)²

Δy ≈ 110 m

8 0

3 years ago

A meter stick is held vertically above your hand, with the lower end between your thumb and first finger. On seeing the meter st

inessss [21]

Answer:

$t=0.193s$

Explanation:

What is said is that the meter fell d=18.3cm=0.183m under the action of gravity. We can use the formula for accelerated motion:

$d=v_0t+\frac{at^2}{2}$

Since it departed from rest it will mean that:

$d=\frac{at^2}{2}$

So our time will be:

$t=\sqrt{\frac{2d}{a}}$

Which for our values is:

$t=\sqrt{\frac{2(0.183m)}{(9.81m/s^2)}}=0.193s$

7 0

3 years ago