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Ksivusya [100]
2 years ago
15

A sled sliding on a flat,icy surface with a constant velocity is best described by

Physics
1 answer:
AysviL [449]2 years ago
5 0
The first one is actually Newton's First Law of Motion. 

Newton's First Law of Motion is commonly stated as "an object in motion will remain in motion unless acted upon by an outside force." This sled is moving across the ice at the same constant speed and velocity meaning it will won't stop unless an outside force acts upon it. An example being someone coming and grabbing the sled or maybe the sled slips off of the ice and finds itself stuck in a pile of snow. Similarly, if the sled was not moving, it would not move unless acted upon again by an outside force. 

You believed it to be Newton's Third Law, but for future reference, his third law is more about every action having an equal opposite reaction. This wouldn't explain what is currently happening with the sled. A helpful tip into figuring out where to start in figuring out the answer I got is in the example you gave it literally states "with a constant velocity" which also ends up being in the description of Newton's First Law. Though, don't rely on that because the answer could have had a made up definition.

Unfortunately, I do not know the answer to your other question. I apologize. Hope this helped.
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A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a
aleksley [76]

Answer:

 T₂ =602  °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

Gauge pressure at condition 1 given = 100 KPa

The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\

T₂ = 875 K

T₂ =875- 273 °C

T₂ =602  °C

5 0
2 years ago
(a) Two point charges totaling 8.00 μC exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the
Anastaziya [24]
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7 0
2 years ago
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The colored lines in the figure represent paths taken by different people walking around in a city. Assume that each city block
jarptica [38.1K]

Answer: 592.37m

Explanation:

Person D is the blue line.

The total displacement is equal to the difference between the final position and the initial position, if the initial position is (0,0) we have that he first goes down two blocks, then right 6 blocks. then up 4 blocks, then left 1 block.

Now i will considerate that the positive x-axis is to the right and the positive y-axis is upwards.

Then the new position will be, if B is a block:

P =(6*B - 1*B, -2*B + 4*B) = (5*B, 2*B)

And we know that B = 110m

P = (550m, 220m)

Now, then the displacement will be equal to the magnitude of our vector, (because the difference between P and the initial position is equal to P, as the initial position is (0,0))  this is:

P = √(550^2 + 220^2) = 592.37m

4 0
2 years ago
When compared to others, how is a greater velocity represented on a motion map?
kakasveta [241]
Motion map has the points spaced farther apart (because the car would go a further distance in each second), and the velocity vectors (arrows) are longer, because the car is moving faster. So 'with longer vectors' is the correct answer
8 0
2 years ago
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A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet
Gnom [1K]

Answer:

d. 37 °C

Explanation:

m_{m} = mass of lump of metal = 250 g

c_{m} = specific heat of lump of metal  = 0.25 cal/g°C

T_{mi} = Initial temperature of lump of metal = 70 °C

m_{w} = mass of water = 75 g

c_{w} = specific heat of water = 1 cal/g°C

T_{wi} = Initial temperature of water = 20 °C

m_{c} = mass of calorimeter  = 500 g

c_{c} = specific heat of calorimeter = 0.10 cal/g°C

T_{ci} = Initial temperature of calorimeter = 20 °C

T_{f} = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) +  m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C

6 0
2 years ago
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