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ANTONII [103]
2 years ago
13

What are the basic si units for the speed of light?.

Physics
1 answer:
DanielleElmas [232]2 years ago
7 0
C=meters/second or C=m/s
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Earl is using his hands to hold a metal pan 10 centimeters above a hot burner. How can this scenario be changed to demonstrate c
Olenka [21]

Answer: touch the pan to the burner

Explanation:

There are three modes of heat transfer:

conduction, convection and radiation.  

For conduction, the heat transfers from a hot object to a cold object when the two are in contact.

For convection there is bulk motion of fluid occurs which transfers the heat.

For heat transfer by radiation, medium is not required.

Thus, to demonstrate conduction between pan and burner, the pan must touch the burner.

8 0
3 years ago
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A camper stands in a valley between two parallel cliff walls. He claps his hands and notices that the echo from the nearby wall
Stells [14]
5,658 ft is your answer I believe
6 0
2 years ago
Does salt or fresh water make a difference in floating of ships? plz answer ASAP will give brainliest to the best answer
ch4aika [34]

Answer:

The density of salt water is more than that of density of fresh water, so so salt and fresh water makes a difference in floating of ships.

5 0
3 years ago
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A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m bef
bekas [8.4K]

Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

μ_k = 0.1773

3 0
3 years ago
Determine W (or fuel energy) required to launch a satellite of mass m at rest from a launching pad placed at the surface earth,
jeka57 [31]

Answer:

5. 9GmM/(10R)

Explanation:

m is the mass of the satellite

M is the mass of the earth

W is the energy required to launch the satellite

Energy at earth surface = Potential energy (PE) + W

W = Energy at earth surface - Potential energy (PE)

But PE = -\frac{GMm}{R}

Therefore: W = Energy at earth surface - \frac{GMm}{R}

Energy at earth surface (E) at an altitude of 5R = -\frac{GMm}{5r} +\frac{1}{2}mV^2

But V=\sqrt{\frac{GM}{5R} }

Therefore: E=-\frac{GMm}{5R}+\frac{1}{2}m(\sqrt{\frac{GM}{5R} } )^2=  -\frac{GMm}{5R}+\frac{GMm}{10R}  = -\frac{GMm}{10R}

W = E - PE

W=-\frac{GMm}{10R}-(-\frac{GMm}{R})=-\frac{GMm}{10R}+\frac{GMm}{R}=\frac{9GMm}{10R} \\W=\frac{9GMm}{10R}

7 0
3 years ago
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