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1 year ago

At which of the following angles will the sunlight received at a location on Earth spread out over the largest area?

1 answer:

3 0

The angle at which the sunlight received at a location on Earth spread out over the largest area is 10°. Last option is correct.

<h3>What is sunlight?</h3>

The light coming from the Sun reaching the Earth's surface is called as Sunlight.

When the sun is overhead, the intensity is high because sun's rays are perpendicular to the earth's surface, so the energy spreads over a small area and the heat is too high in that region.

When, the angle is smaller, the sunlight will spread out over a larger area.

Thus, at 10° the sunlight received at a location on Earth spread out over the largest area. Last option is correct.

Learn more about Sunlight.

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Mercury has one of the lowest specific heats. This fact added to its liquid state at most atmospheric temperatures make it effec

UNO [17]

The specific heat of mercury is 149.4 J/(kgK)

Explanation:

When a substance is supplied with an amount of energy Q, its temperature increases according to the equation:

$\Delta T=\frac{Q}{mC_s}$

where

$\Delta T$ is the increase in temperature

m is the mass of the sample

$C_s$ is its specific heat capacity

For the sample of mercury in this problem we have

Q = 275 J

m = 0.450 kg

$\Delta T = 4.09 K$

Therefore, by re-arranging the equation we find the mercury's specific heat:

$C_s = \frac{Q}{m\Delta T}=\frac{275}{(0.450)(4.09)}=149.4 J/(kgK)$

Learn more about specific heat capacity:

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5 0

3 years ago

A circuit is built based on the circuit diagram shown. What is the current in the 50 Ω resistor

In-s [12.5K]

Answer:

1.2 A

Explanation:

From the diagram attached, The three resistors are parallel because the each ends of the resistors are connected together. Since they are in parallel, the voltage across each resistor is the same. The voltage source connected in parallel to the resistors is 60 V. Therefore the voltage across the 50 Ω resistor is 60 V. Using ohm law:

Voltage (V) = Current (I) × Resistance (R)

V = IR

I = V/R

I = 60 V/ 50 Ω

I = 1.2 A

The current in the 50 Ω resistor is 1.2 A

5 0

2 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

VMariaS [17]

At point C because it is at the lowest position.

7 0

2 years ago

The height of a typical playground slide is about 1.8 m and it rises at an angle of 30 ∘ above the horizontal.

Salsk061 [2.6K]

Answer:

$5.94\ \text{m/s}$

$1.7$

$0.577$

Explanation:

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

$\theta$ = Angle of slope = $30^{\circ}$

v = Velocity of child at the bottom of the slide

$\mu_k$ = Coefficient of kinetic friction

$\mu_s$ = Coefficient of static friction

h = Height of slope = 1.8 m

The energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.8}\\\Rightarrow v=5.94\ \text{m/s}$

The speed of the child at the bottom of the slide is $5.94\ \text{m/s}$

Length of the slide is given by

$l=h\sin\theta\\\Rightarrow l=1.8\sin30^{\circ}\\\Rightarrow l=0.9\ \text{m}$

$v=\dfrac{1}{2}\times5.94\\\Rightarrow v=2.97\ \text{m/s}$

The force energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2+\mu_kmg\cos\theta l\\\Rightarrow \mu_k=\dfrac{gh-\dfrac{1}{2}v^2}{gl\cos\theta}\\\Rightarrow \mu_k=\dfrac{9.81\times 1.8-\dfrac{1}{2}\times 2.97^2}{9.81\times 0.9\cos30^{\circ}}\\\Rightarrow \mu_k=1.73$

The coefficient of kinetic friction is $1.7$ .

For static friction

$\mu_s\geq\tan30^{\circ}\\\Rightarrow \mu_s\geq0.577$

So, the minimum possible value for the coefficient of static friction is $0.577$ .

8 0

2 years ago

A wave is a _____ that transmits energy

galben [10]

Answer is disturbance

3 0

3 years ago

Read 2 more answers