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MissTica
3 years ago
10

What process is a method of heat transfer but does NOT contribute significantly to heating the surface or atmosphere of the eart

h?
A) conduction
B) convection
C) insulation
D) radiation
Physics
1 answer:
klasskru [66]3 years ago
7 0

Answer:

conduction

Explanation:

Conduction is a method of heat transfer, where two objects exchange heat by touching, but does not contribute to moving heat through earth's atmosphere, because the atmosphere is so sparse it does not transmit heat through conduction effectively. Insulation is not a method of heat transfer.

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What does it mean when we say two objects are in equilibrium
cupoosta [38]

An object is considered to be in a condition of equilibrium when it is balanced with regard to all external forces.

Equilibrium:

An object is considered to be in equilibrium if both its angular acceleration and the acceleration of its center of mass are equal to zero. In layman's terms: The item must either be at rest or moving at a constant speed if it is not accelerating because F = ma (force = mass x acceleration). Even in motion, a body can be in equilibrium. This kind of equilibrium is referred to as a dynamic equilibrium.

A weight suspended by a spring or a brick laying on a flat surface is an example. The equilibrium is unstable if the force with the smallest deviation tends to increase the displacement. As an example, imagine a ball bearing on the edge of a razor blade.

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2 years ago
Which expression describes the third vector in the diagram, which leads from the tip of B to the tip of Ā?
podryga [215]

Answer:

axb

Explanation:

because the singular order

6 0
3 years ago
According to kinetic molecular theory, which statement best describes collisions between gas particles?
Mkey [24]
A they are elastic.
being in random motion means they're not predictable. 
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3 years ago
During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.
lord [1]

Answer:

(a) F_{sm} = 4.327\times 10^{20}\ N

(b) F_{em} = 1.983\times 10^{20}\ N

(c) F_{se} = 3.521\times 10^{20}\ N

Solution:

As per the question:

Mass of Earth, M_{e} = 5.972\times 10^{24}\ kg

Mass of Moon, M_{m} = 7.34\times 10^{22}\ kg

Mass of Sun, M_{s} = 1.989\times 10^{30}\ kg

Distance between the earth and the moon, R_{em} = 3.84\times 10^{8}\ m

Distance between the earth and the sun, R_{es} = 1.5\times 10^{11}\ m

Distance between the sun and the moon, R_{sm} =  1.5\times 10^{11}\ m

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

F_{G} = \frac{Gmm'_{2}}{r^{2}}                             (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}

F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}

F_{sm} = 4.327\times 10^{20}\ N

(b) The force exerted by the Earth on the Moon is given by eqn (1):

F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}

F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}

F_{em} = 1.983\times 10^{20}\ N

(c) The force exerted by the Sun on the Earth is given by eqn (1):

F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}

F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}

F_{se} = 3.521\times 10^{20}\ N

7 0
3 years ago
A car accelerates from rest at 3m/s2 along a straight road how far has the car traveled after 4s?
murzikaleks [220]

Answer:

24 m

Explanation:

x = 1/2at^{2} + v_{0}t + x_{0}\\a = 3 m/s^{2}\\v_{0} = 0

7 0
3 years ago
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