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olga nikolaevna [1]
3 years ago
15

PLEASE HELPPPPPPPPPPPPP

Physics
2 answers:
sertanlavr [38]3 years ago
8 0

Answer: sliding

Explanation:

A snow skier slowing to a stop after skiing down a mountain is an example of sliding friction

mario62 [17]3 years ago
3 0

Answer:

static friction

Explanation:

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Question 3 of 10
Stels [109]

Answer:

30N

Explanation:

6 0
3 years ago
Which list correctly describes the usual order of planets inward toward the sun?
Sati [7]

Option B

Neptune, Uranus, Saturn, Jupiter, Mars, Earth, Venus, Mercury correctly describes the usual order of planets inward toward the sun

<u>Explanation:</u>

Our solar system continues much considerably than the eight planets that revolve around the Sun. The position of the planets in the solar system, commencing inward to the sun is the accompanying: Neptune, Uranus, Saturn, Jupiter, Mars, Earth, Venus, Mercury.

Most next to the Sun, simply rocky material could resist the heat. For this logic, the first four planets: Mercury, Venus, Earth, and Mars are terrestrial planets.  The four large outer worlds — Jupiter, Saturn, Uranus, and Neptune: because of their enormous size corresponding to the terrestrial planets. They're also frequently composed of gases like hydrogen, helium, and ammonia preferably than of rocky surfaces.

7 0
3 years ago
A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge
Alexxx [7]

Explanation:

It is given that, a long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire.

The charge per unit length of the wire is \lambda and the net charge per unit length is 2 \lambda.

We know that there exist zero electric field inside the metal cylinder.

(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let \lambda_i\ and\ \lambda_o are the charge per unit length on the inner and outer surfaces of the cylinder.

For inner surface,

\phi=\dfrac{q_{enclosed}}{\epsilon_o}

E.A=\dfrac{q_{enclosed}}{\epsilon_o}

0=\dfrac{\lambda_i+\lambda}{\epsilon_o}

\lambda_i=-\lambda  

For outer surface,

\lambda_i+\lambda_o=2\lambda

-\lambda+\lambda_o=2\lambda

\lambda_o=3\lambda

(b) Let E is the electric field outside the cylinder, a distance r from the axis. It is given by :

E_o=\dfrac{\lambda_o}{2\pi \epsilon_o r}

E_o=\dfrac{3\lambda}{2\pi \epsilon_o r}

Hence, this is the required solution.

6 0
3 years ago
Positive feedback interactions in earth’s systems are always a result of human action. T/F
VLD [36.1K]

Positive feedback interactions in earth’s systems are always a result of human action is a FALSE statement.

<u>Explanation:</u>

Earth is a unstable equilibrium which tends to move out of equilibrium, but several other factors try to bring it back in equilibrium again. Earth has a different actions going on both on its surface and also inside it.

Human can alter, or can modify a very small part of the events that occur on earth’s surface. But they don’t have any control on what’s going inside earth’s core. So positive feedback interactions are not only the results of human interaction, but also different other factors.  

4 0
4 years ago
You plug a string of 100 lights in series into a 120 V power outlet, and each light has a resistance of 3.00 _. If each light ha
UNO [17]
V = IR

By completing the equation, i found that the total power equation is : 4.8, 
Which means that it's not exceed the power rating.

So i believe the answer would be :  The string will remain lit

hope this helps
6 0
4 years ago
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