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3 years ago

Where is the sun's energy most concentrated

Physics

1 answer:

Maru [420]3 years ago

5 0

B. At the equator

Explanation:

The energy coming from the Sun hits the Earth's surface at different angles, depending on the latitude of the place. The more perpendicular the ray of lights hit the surface, the more the energy transmitted to the Earth's surface, the warmer the location.

The angle at which the ray of lights hit the Earth is related to the latitude: in particular, the ray of lights arrive perpendicular at the equator ( $0^{\circ}$ ), they arrive at larger angle in the United States (which is located at intermediate latitudes) and they arrive at the largest angles at the poles. For this reason, the sun's most energy is concentrated at the equator.

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Are we really real or are we all an allusion?Do you believe in past lives?How do you think you died in past life?

OLEGan [10]

Answer:

I don't even know

Explanation:

Shower thoughts

8 0

3 years ago

Read 2 more answers

What would happen if you use a thicker wire around the iron nail of an electromagnet? (thats the whole question)

puteri [66]

Answer:

When we have a current I, we will have a magnetic field perpendicular to this current.

Then if we have a wire in a "spring" form. then we will have a magnetic field along the center of this "spring".

Now suppose we put an iron object in the middle (where the magnetic field is) then we will magnetize the iron object.

Of course, the intensity of the magnetic field is proportional to the current, given by:

B = (μ*I)/(2*π*r)

Where:

μ is a constant, I is the current and r is the distance between to the current.

Now remember that for a resistor:

R = ρ*L/A

R is the resistance, ρ is the resistivity, which depends on the material of the wire, L is the length of the wire, and A is the cross-section of the wire.

If we increase the area of the wire (if we use a thicker wire).

And the relation between resistance and current is:

I = V/R

Where V is the voltaje.

Now, if we use a thicker wire, then the cross-section area of the wire increases.

Notice in the resistance equation, that the cross-section area is on the denominator, then if we increase the area A, the resistance decreases.

And the resistance is on the denominator of the current equation, then if we decrease R, the current increases.

If the current increases, the magnetic field increases, which means that we will have a stronger electromagnet.

3 0

3 years ago

A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The

nordsb [41]

Answer:

$t=6.96s$

Explanation:

From this exercise, our knowable variables are hight and initial velocity

$v_{oy}=96ft/s$

$y_{o}=112ft$

To find how much time does the ball strike the ground, we need to know that the final position of the ball is y=0ft

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

$0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}$

Solving for t using quadratic formula

$t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

$a=-\frac{1}{2} (32.2)\\b=96\\c=112$

$t=-0.999s$ or $t=6.96s$

Since time can't be negative the answer is t=6.96s

7 0

3 years ago

Calculate the tension on the rope. A. 57.89 N B. 32.73 N C. 69.84 N D. 12.55 N

Ratling [72]

Hi there!

$\large\boxed{\text{B. 32.73N}}$

To calculate the tension, we must calculate the acceleration of the system.

Begin with a summation of forces:

∑F = -M₁gsinФ + T - T + M₂g

Simplify and solve for acceleration: (Tensions cancel out)

$a = \frac{-M_1gsin\theta + T - T + M_2g}{M_1+M_2}$

Plug in values. Let g = 10 m/s²

$a = \frac{-3(10)(sin30)+8(10)}{3+8} = 5.91 m/s^{2}$

Now, to find tension, let's sum up the forces acting on ONE block. For simplicity, we can look at the hanging block:

∑F = -T + W

ma = -T + W

Rearrange to solve for T:

T = W - ma

We know the acceleration, so plug in the values:

T = (8)(10) - (8)(5.91) = 32.73 N

3 0

3 years ago

A solid cylinder with a mass of 2.46 kg and a radius of 0.049 m starts from rest at a height of 4.80 m and rolls down a 24.3 ◦ s

irina1246 [14]

Answer:

$v_{f}\approx 2.097\,\frac{m}{s}$

Explanation:

Let assume that the solid cylinder rolls down a frictionless incline. The translational speed can be found by using the Principle of Energy Conservation and the Work-Energy Theorem:

$m_{cyl}\cdot g\cdot y_{o} = \frac{1}{2}\cdot m_{cyl} \left( 1+ \frac{1}{R}\right)\cdot v_{f}^{2}$

$g\cdot y_{o} = \frac{1}{2}\cdot\left( 1+ \frac{1}{R}\right)\cdot v_{f}^{2}$

The translational speed is:

$v_{f} = \sqrt{\frac{2\cdot g\cdot y_{o}}{\left(1 + \frac{1}{R} \right)}}$

$v_{f} = \sqrt{\frac{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (4.80\,m)}{\left(1 + \frac{1}{0.049\,m} \right)} }$

$v_{f}\approx 2.097\,\frac{m}{s}$

7 0

3 years ago