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2 years ago

Kon'nichiwa~please help me with this question!!

Physics

1 answer:

andrezito [222]2 years ago

4 0

Let's check the relationship

$\\ \rm\hookrightarrow g=\dfrac{GM}{r^2}$

$\\ \rm\hookrightarrow g\propto G$

Raindrops will fall faster . .
Also walking on ground would become more difficult as g increases.

Option C is wrong by now .Let's check D once

$\\ \rm\hookrightarrow T\propto \dfrac{1}{\sqrt{g}}$

So time period of simple pendulum would decrease.

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PLEASE HELP !!

Paraphin [41]

Distance from the reference point

7 0

3 years ago

Comparing objects in a related group can reveal patterns among them. These patterns in turn can help us learn more about those o

rosijanka [135]

Answer:

C. All planets orbit the Sun in the same direction.

E. The orbits of the planets are evenly distributed in distance from the Sun.

F. All planets orbit the Sun in a roughly flat plane.

Explanation:

As we can see that due to big bang theory of planet formation we can say all planets revolve around the sun in circular orbit with almost in same plane

Size of the planets are unevenly distributed and they all revolve around the sun in same directions

so here correct options are

C. All planets orbit the Sun in the same direction.

E. The orbits of the planets are evenly distributed in distance from the Sun.

F. All planets orbit the Sun in a roughly flat plane.

5 0

3 years ago

What kind of physics is used to describe the interaction of matter and energy from the beginning of the cosmic singularity to th

lisov135 [29]

Cosmology physics is used to describe the interaction of matter and energy from the beginning of the cosmic singularity to the Planck time.

Astronomy's field of cosmology examines the beginnings and development of the universe, from the Big Bang through the present and into the future. The current cosmology model for the development of the universe is the Big Bang theory, which relies on the validity of general relativity. There is a singularity there as well.

According to the Big Bang theory, the cosmos was once compressed into an infinitely small point 13.77 billion years ago. Prior to $10^{-43}$ seconds after the Big Bang, during the Planck Era, gravity, nuclear strong, nuclear weak, and electromagnetic forces are thought to have been merged into a single "super" force.

Learn more about big bang theory here;

brainly.com/question/18297161

#SPJ4

3 0

2 years ago

Bob is pulling a 30kg filing cabinet with a force of 200N , but the filing cabinet refuses to move. The coefficient of static fr

vredina [299]

Answer:

$2.0\cdot 10^2 N$

Explanation:

The cabinet does not move: this means that the net force acting on it is zero.

Along the horizontal direction, we have two forces:

- The push exerted by Bob, F = 200 N, forward

- The frictional force, $F_f$ , which acts in the opposite direction (backward)

Since the net force must be zero, we have:

$F-F_f = 0$

So solving the equation we can find the magnitude of the friction force:

$F_f = F = 200 N=2.0 \cdot 10^2 N$

7 0

4 years ago

The heat loss from a boiler is to be held at a maximum of 900Btu/h ft2 of wall area. What thickness of asbestos (k= 0.10 Btu/h f

zmey [24]

Answer:

a. 0.122 ft b. -70 Btu/h ft² c. 633.33 °F

Explanation:

a. Since the rate of heat loss dQ/dt = kAΔT/d where k = thermal conductivity, A = area, ΔT = temperature gradient and d = thickness of insulation.

Now [dQ/dt]/A = kΔT/d

Given that [dQ/dt]/A = rate of heat loss per unit area = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), ΔT = T₂ - T₁ = 500 °F - 1600 °F = -1100 °F. We need to find the thickness of asbestos, d. So,

d = kΔT/[dQ/dt]/A

d = 0.10 Btu/h ft ℉ × -1100 °F/-900Btu/h ft²

d = 0.122 ft

b. If the 3 in thick Kaolin is added to the outside of the asbestos, and the outside temperature of the asbestos is 250℉, the heat loss due to the Kaolin is thus

[dQ/dt]/A = k'ΔT'/d'

k' = 0.07 Btu/h ft ℉(for Kaolin), ΔT' = T₂ - T₁ = 250 °F - 500 °F = -250 °F and d' = 3 in = 3/12 ft = 0.25 ft

[dQ/dt]/A = 0.07 Btu/h ft ℉ × -250 °F/0.25 ft

[dQ/dt]/A = -70 Btu/h ft²

c. To find the temperature at the interface, the total heat flux equals the individual heat loss from the asbestos and kaolin. So

[dQ/dt]/A = k(T₂ - T₁)/d + k'(T₃ - T₂)/d' where [dQ/dt]/A = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), k' = 0.07 Btu/h ft ℉(for Kaolin), T₁ = 1600 °F, T₂ = unknown and T₃ = 250℉.

Substituting these values into the equation, we have

-900Btu/h ft² = 0.10 Btu/h ft ℉(T₂ - 1600 °F)/0.122 ft + 0.07 Btu/h ft ℉(250℉ - T₂)/0.25 ft

-900Btu/h ft² = 0.82 Btu/h ft ℉(T₂ - 1600 °F) + 0.28Btu/h ft ℉(250℉ - T₂)

-900 °F = 0.82(T₂ - 1600 °F) + 0.28(250℉ - T₂)

-900 °F = 0.82T₂ - 1312°F + 70 °F - 0.28T₂

collecting like terms, we have

-900 °F + 1312°F - 70 °F = 0.82T₂ - 0.28T₂

342 °F = 0.54T₂

Dividing both sides by 0.54, we have

T₂ = 342 °F/0.54

T₂ = 633.33 °F

8 0

3 years ago

Kon'nichiwa~please help me with this question!!​

Kon'nichiwa~please help me with this question!!