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3 years ago

If you cook a marshmallow a few feet above a camp fire, it will cook mostly

Physics

1 answer:

nasty-shy [4]3 years ago

6 0

Answer:

A) radiation.

Explanation:

Hello,

In this case, heat transfer from a body with a high temperature to a body with a lower temperature may occur when the bodies are not in direct physical contact with each other or when they are separated in space and it is is called heat radiation. Solids, liquids, or gaseous substances are able to emit energy via a process of electromagnetic radiation because of vibrational and rotational movement of their molecules and atoms.

Hence, since the camp fire is not in contact with the marshmallow and it still cooks it, we can infer that the mechanism of heat transfer is A) radiation. Moreover, both conduction and convection occur when the bodies are in contact.

Regards.

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Which diagram best represents the electric field around a negatively charged conducting sphere? (See pic)

dalvyx [7]

The answer is D !!!!!!!

3 0

3 years ago

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, $r=4\ AU=1.496\times 10^{11}\ m$

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

$T^2=\dfrac{4\pi^2}{GM}r^3$

M is the mass of the sun

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3$

$T^2=\sqrt{9.96\times 10^{14}}\ s$

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0

3 years ago

Can someone please help me with this physics question? I'm desperate!

Lelu [443]

Answer:

a) 2·√10 seconds

b) Linda should be approximately 30.6 meters

c) Jenny's speed at the 100-m mark is approximately 6.325 m/s

Explanation:

The speed with which Linda is running = 8.6 m/s

The point Jenny starts = The 80-m mark

The acceleration of Jenny = 1.0 m/s²

a) The time it takes Jenny to run from the 80-m mark to the 100-m mark, <em>t</em>, is given as follows

Δs = u·t + (1/2)·a·t²

Δs = Distance = 100-m - 80-m = 20-m

u = The initial velocity of Jenny = 0

a = Jenny's acceleration = 1.0 m/s²

∴ 20 = 0×t + (1/2) × 1 × t² = t²/2

20 = t²/2

t = √(20 × 2) = 2·√10

The time it takes Jenny to run from the 80-m mark to the 100-m mark = 2·√10 seconds

b) The distance Linda runs in t = 2·√10 seconds, d = v × t

Given that Linda's velocity, v = 8.6 m/s, we have;

d = 8.0 × 2·√10 = 16·√10

The distance Linda runs in t = 2·√10 seconds = 16·√10 meters ≈ 50.6 meters

Therefore, Linda should be approximately (50.6 - 20) meters = 30.6 meters behind Jenny when Jenny starts running

c) Jenny's speed at the 100 m mark is given as follows;

v = u + a·t

t = 2·√10 seconds, a = 1.0 m/s², u = 0

∴ v = 0×t + 1.0×2·√10 = 2·√10 ≈ 6.325

Jenny's speed at the 100-m mark ≈ 6.325 m/s

3 0

2 years ago

In a tug-of-war game on one campus, 15 students pull on a rope at both ends in an effort to displace the central knot to one sid

mojhsa [17]

Answer:

Net pull = 110 N to the left

Explanation:

Group the different pulls according to the direction (right or left)

2 pull 196 N each to the right

4 pull 98 N each to the left

5 pull 62 N each to the left

3 pull 150 N each to the right

1 pull 250 N to the left

Since positive direction is to the right, the pulls to the left will have a minus (-)

$Net Force= 2(196)+4(-98)+5(-62)+3(150)+1(-250) \\Net Force = -110$

The resulting force is negative, meaning the direction is to the left

6 0

3 years ago

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Agata [3.3K]

Answer:

i think its C hope this is rhight have a good day

Explanation:

6 0

3 years ago

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