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Aleks [24]
3 years ago
8

If two objects are the same size but one object is 3 times hotter than the other object, the hotter object emits If two objects

are the same size but one object is 3 times hotter than the other object, the hotter object emits 3 times more energy per time per area. 9 times more energy per time per area. 12 times more energy per time per area 81 times more energy per time per area. none of the above
Physics
1 answer:
Illusion [34]3 years ago
8 0

Answer:

81 times the first body

Explanation:

Let the temperature of the first object is T.

The temperature of the second object is 3T.

According to the Stefan's law, the energy radiated per unit area per unit time is given by

E \alpha T^{4}

So, for the first body

E_{1} \alpha T^{4}      .... (1)

for the second body

E_{2} \alpha (3T)^{4}    ..... (2)

By equation (1) and (2),

E2 = 81 times E1

So, the second body gives 81 times more energy per unit area per unit time

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Waves can travel but the medium does not "travel" with it, even though they may move up and down.
Sonja [21]

Answer:

While all waves can transmit energy through a medium, certain waves can also transmit energy through empty space. A medium is a material through which waves can travel. It can be a solid, liquid, or gas. When waves travel through a medium, the particles of the medium are not carried along with the wave.

Explanation:

hope it helps

4 0
3 years ago
Read 2 more answers
Assume that when you stretch your torso vertically as much as you can, your center of mass is 1.0 m above the floor. The maximum
Elenna [48]

1) 0.77 m

2) 0.23 m

Explanation:

1)

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

\frac{1}{2}mv^2 = mgh

where

m is the mass of the man

v is the speed after jumping

g=9.8 m/s^2 is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s

In the acceleration phase, we know that the initial velocity is

u=0

And the force exerted on the floor is 2.3 times the gravitational force, so

F=2.3 mg

This means the net force on you is

F_{net} = F-mg=2.3mg-mg=1.3 mg

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g

Now we can use the  following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

v^2-u^2=2as

where s is the quantity we want to find. Solving for s,

s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m

2)

At the beginning, we are told that the height of the center of mass above the floor is

h = 1.0 m

During the acceleration phase and the crouch, the height of the center of mass of the body decreases by

\Delta h = -0.77 m

This means that the lowest point reached by the center of mass above the floor during the crouch is

h'=h+\Delta h = 1.0 - 0.77 = 0.23 m

This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

3 0
3 years ago
Use the velocity vs time graph to analyze the motion of the object.
arsen [322]

Explanation:

the object has constant velocity for 2 seconds and it get a constant accelration (2ms-2)

7 0
2 years ago
A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud
IceJOKER [234]

Answer: 1.51 km

Explanation:

<u>Coulomb's Law:</u> The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or,   \vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

<u>Given:</u>

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

<u>So,</u>

\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514  \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}

Therefore, the separation between the two charges (r) = 1.51 km

3 0
2 years ago
44.2 cm + 0.123 cm = cm
trapecia [35]
The answer is 44.323 cm
3 0
3 years ago
Read 2 more answers
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