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Fiesta28 [93]
2 years ago
7

Near the poles, more energy is reflected back into space than is absorbed. near the poles, more energy is reflected back into sp

ace than is absorbed.
a. True
b. False
Physics
1 answer:
VladimirAG [237]2 years ago
7 0
Correct answer: a) True.

In fact, the soil at the poles consists mainly of snow. Snow has an albedo of 90%: albedo is the fraction of light (and so, of energy) that is reflected back to the space. This means that at the poles, about 90% of the light is reflected back to the space, while only 10% is absorbed.
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When you are ice skating, to get started, you push your stake backwards on the ice and, as a result, begin to move forward. whic
LUCKY_DIMON [66]

C. Newtons third law of motion

Because eventually, the frictional forces will slow you to a halt. Newton's Third Law of Motion For every action there is an equal and opposite reaction. When they push off against the ice, or "stroke" with their skates, they are applying a force down and back against the ground.

Hope this helps!

8 0
2 years ago
A circuit consists of a series combination of 6.50 −kΩ and 4.50 −kΩ resistors connected across a 50.0-V battery having negligibl
Vlada [557]

Answer:

Part A: 16.1 V

Part B: 20.5 V

Part C: 21.5%

Explanation:

The voltmeter is in parallel with the 4.5-kΩ resistor and the combination is in series with the 6.5-kΩ resistor. The equivalent resistance of the parallel combination is given as

\dfrac{1}{R_E}=\dfrac{1}{4.50}+\dfrac{1}{10.0}

R_E=\dfrac{4.50\times10.0}{4.50+10.0} = 3.10

Part A

The voltmeter reading is the potential difference across the parallel combination. This is found by using the voltage-divider rule.

V_1 = \dfrac{3.10}{3.10+6.50}\times50.0 = \dfrac{3.10}{9.60}\times50.0 = 16.1 \text{ V}

Part B

Without the voltmeter, the potential difference across the 4.5-kΩ resistor is found using the same rule as above:

V_2 = \dfrac{4.50}{4.50+6.50}\times50.0 = \dfrac{4.50}{11.0}\times50.0 = 20.5 \text{ V}

Part C

The error in % is given by

\dfrac{20.5-16.1}{20.5}\times100\% = \dfrac{4.4}{20.5}\times100\% = 21.5\%

4 0
3 years ago
Read 2 more answers
A silver wire with resistivity 1.59 × 10-8 Ω-m carries a current density of 4.0 A/mm2, What is the magnitude of the electric fie
Nat2105 [25]

Answer:

Electric field, E = 0.064 V/m

Explanation:

It is given that,

Resistivity of silver wire, \rho=1.59\times 10^{-8}\ \Omega-m

Current density of the wire, J=4\ A/mm^2=4\times 10^6\ A/m^2

We need to find the magnitude of the electric field inside the wire. The relationship between electric field and the current density is given by :

E=J\times \rho

E=4\times 10^6\times 1.59\times 10^{-8}

E = 0.0636 V/m

or

E = 0.064 V/m

So, the magnitude of electric field inside the wire is 0.064 V/m. Hence, this is the required solution.

6 0
2 years ago
if you have a kinetic energy of 1470 J, and you are 60kg mass and 0 m above the ground, what is you velocity?
laiz [17]

Answer:

The 39.

Explanation:

8 0
2 years ago
Read 2 more answers
A stone is thrown with an initial speed of 12 m/s at an angle of 30o above the horizontal from the top edge of a cliff. If it ta
kherson [118]

Answer:

d=58m

Explanation:

From the question we are told that:

Initial Speed U=12m/s

Time T=5.6s

Angle \theta=30

Generally the  Newton's equation for motion is mathematically given by

d=d'+ut+\frac{at^2}{2}

d=12cos30*5.6

d=58m

8 0
3 years ago
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