All categories

2 years ago

The earth's surface absorbs ------of the sun's energy that reaches it.

Physics

2 answers:

RideAnS [48]2 years ago

7 0

Answer:

Explanation:

three-quarters

n200080 [17]2 years ago

6 0

Answer:

D): The Earth`s surface absorbs half the sun`s energy when it reaches it.

Explanation:

When solar energy is absorbed partially from the Earth`s surface it is absorbed as heat.

(Wrote this off my notes, your welcome)

You might be interested in

Practice Exercises Name: : Billy-Joe stands on the Talahatchee Bridge kicking stones into the water below a) If Billy-Joe kicks

babymother [125]

Answer:

a) The answer is 11,7m

b) The time it takes to fall will be shorter

Explanation:

We will use the next semi-parabolic movement equations

$H=Hi+Viy*t+1/2*g*t$

$X=Xi+Vx*t$

Where g(gravity acceleration)=9,81m/s^2

Also Xi, Hi and Viy are zero, as the stones Billy-Jones is kicking stay still before he moves them, so we take that point as the reference point

The first we must do is to find how much time the stones take to fall, this way:

$t=(5.40m)/(3.50m/s)$

Then t=1,54s

After that we need to replace t to find H, this way

$H=(1/2)*(9,81m/s^2)*(1,54s)^2$

Then H=11,7m

b) The stones will fall faster as the stones will be kicked harder, it will cause the stones move faster, it means, more horizontal velocity. In order to see it better we could assume the actual velocity is two times more than it is, so it will give us half of the time, this way:

$t=(5,40m)/(2*3,50m/s)$

Then, t=0,77

4 0

3 years ago

A 2300 kg sailboat is moving west at 5.5 m/s when an eastward wind

pogonyaev

The boat is initially at equilibrium since it seems to start off at a constant speed of 5.5 m/s. If the wind applies a force of 950 N, then it is applying an acceleration a of

950 N = (2300 kg) a

a = (950 N) / (2300 kg)

a ≈ 0.413 m/s²

Take east to be positive and west to be negative, so that the boat has an initial velocity of -5.5 m/s. Then after 11.5 s, the boat will attain a velocity of

v = -5.5 m/s + a (11.5 s)

v = -0.75 m/s

which means the wind slows the boat down to a velocity of 0.75 m/s westward.

5 0

3 years ago

What is the audible range of human hearing

Phantasy [73]

Answer:

on an average, 20Hz to 20kHz

8 0

3 years ago

What are the major weather fronts? How do they differ from each other?

dmitriy555 [2]

Answer:

they differ each other bc it is a lot of with earth quakes and everything.

Explanation:

3 0

3 years ago

A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati

Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

$\tau = I * \alpha (1)$

Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
τ = F*r (2)
For a solid uniform disk, the rotational inertia regarding an axle passing through its center is just I = m*r²/2 (3).
Replacing (2) and (3) in (1), we can solve for α, as follows:

$\alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)$

Since the angular acceleration is constant, we can use the following kinematic equation:

$\omega_{f}^{2} - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)$

Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

$0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)$

Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

$\omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)$

Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

$v = \omega * r (8)$

where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of the disk, r becomes the radius of the disk, 0.200 m.
Replacing this value and (7) in (8), we get:

$v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)$

There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

$a_{t} = \alpha * r (9)$

where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
Replacing this value and (4), in (9), we get:

$a_{t} = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)$

Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

$a_{c} = \omega^{2} * r (11)$

Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
Replacing this value and (7) in (11) we get:

$a_{c} = \omega^{2} * r = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)$

The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

$a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)$

6 0

2 years ago