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docker41 [41]
3 years ago
6

What did ancient astronomers think areas of the moon called mares might be?

Physics
1 answer:
Bumek [7]3 years ago
4 0

The ancient astronomers think areas of the moon called mares might be Seas.

Option D  

<u>Explanation</u>:

The surface area of Earth's moon is dark, large, and is basaltic plains which are formed by ancient volcanic eruptions. They were dubbed as Maria, "ancient astronomers" who misunderstood them as actual seas. They are less reflective than highlands. Due to their iron-rich composition, they tend to appear dark from the naked eye. The Maria cover approximately about 16% of surface mostly on side that is visible from Earth. The few Maria on side that is too far are much smaller and residing mostly in very large craters. The ancient astronomers mistook the surface area as look like actual seas.

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A 22.0 kg child slides down a slide that makes a 37.0° angle with the horizontal. (a) What is the magnitude of the normal force
patriot [66]

Answer:

(a) 172.185 N

(b) 53^{\circ}

Solution:

As per the question:

Mass of the child, m = 22.0 kg

Angle, \theta = 37.0^{\circ}

Now,

(a) The magnitude of the normal force exerted by the slide on the child:

F_{N} = mgcos\theta

F_{N} = 22\times 9.8cos37^{\circ} = 172.185\ N

Now,

(b) The angle from the horizontal at which the force is directed is:

90^{\circ} - 37^{\circ} = 53^{\circ}

6 0
3 years ago
What type of wave is the highest on the electromagnetic spectrum?
san4es73 [151]
Gamma waves are on the far right
8 0
3 years ago
An electric heating element has a resistance of 16 Ω and is connected to a voltage of 120 V. How much current will flow in this
sergeinik [125]

This is weird.

All three 'choices' are true.

Line um up. (a) shows how to solve the problem. (b) does it. and (c) is the answer.

7 0
3 years ago
All of the following processes are the ones that lead directly to the formation of sedimentary rock EXCEPT-
emmasim [6.3K]

Answer:

melting of rock deep underground.

Explanation:

The melting of rocks deep underground does not produce sedimentary rocks. Most igneous rocks are produced by this process.

When molten rocks underground called magma is solidified in the subsurface, it results into the formation of igneous bodies.

  • Sedimentary rocks forms by the accumulation of sediments.
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4 0
2 years ago
An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,
insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}

Where

t_{hf} = Temperature at the inside of the pipe = 300 °C

t_{f} = Temperature at the outside of the pipe = 20 °C

r₁ =internal  radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

k_A = 15 W/m·K

k_B = 0.038 W/m·K

h_{hf} = 75 W/m²·K

h_{cf} = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

Q = \frac{t_A-t_B}{R_T} Where R_T for the air film on the pipe outer surface is given by

R_T= \frac{1}{\alpha A}

where A =area of the outside of the pipe

= \frac{1}{10*2\pi*0.07*1 } = 0.227 K/W

Therefore

131.32 W = \frac{t_A-20}{0.227} which gives

t_A = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

8 0
3 years ago
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