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mr_godi [17]
2 years ago
14

How does time relate to motion

Physics
1 answer:
docker41 [41]2 years ago
5 0
As we move, time goes up. Think of it on a graph; as time increases on the x axis, motion can either stay the same, increase, or decrease.
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A meteorologist plans to release a weather balloon from ground level, to be used for high-altitude atmospheric measurements. The
Slav-nsk [51]

Answer:

563.86 N

Explanation:

We know the buoyant force F = weight of air displaced by the balloon.

F = ρgV where ρ = density of air = 1.29 kg/m³, g = acceleration due to gravity = 9.8 m/s² and V = volume of balloon = 4πr/3 (since it is a sphere) where r = radius of balloon = 2.20 m

So, F = ρgV = ρg4πr³/3

substituting the values of the variables into the equation, we have

F =  1.29 kg/m³ × 9.8 m/s² × 4π × (2.20 m)³/3

= 1691.58 N/3

= 563.86 N

8 0
3 years ago
a 60-kg skier starts at the top of a 10 meter slope. At the bottom, she is traveling 10 m/s. How much energy does she loose to f
Alex
She would lose <span> 2,880 Joules (J) of energy </span>
7 0
4 years ago
Read 2 more answers
Please help , will surely mark As brainlest
Oliga [24]
The north arrow of the compasses will all point to the south since opposite poles attract. I have put an example picture below.

For the horse shoe magnet the magnetic field line will also go from north to south. I have put a picture of this below



8 0
3 years ago
0.5000 kg of water at 35.00 degrees Celsius is cooled, with the removal of 6.300 E4 J of heat. What is the final temperature of
kogti [31]

Answer:

=30.22°C

Explanation:

The enthalpy change made the water to cool.

Enthalpy change = MC∅ where M is the mass of the water, C is the specific heat capacity of water and ∅ the temperature change.

ΔH=MC∅

∅=ΔH/MC

=(6.3×10⁴J)/(0.5kg×4186J/(kg°C))

=30.1

Final temperature = 35.00°C-30.1°C

=4.9°C

5 0
3 years ago
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An object 82 cm high forms a virtual image 4.1 cm high located 4.6 cm behind a mirror. Find the object distance.
ioda

Answer:

The object distance is 92 cm.  

Explanation:

let v be the image distance and h be the height of the image, let u the be the object distance and H be the height of the object.

then, the magification of the mirror is given by:

m = -v/u and m = h/H

so, -v/u = h/H

         u = -v×H/h

            = -(-4.6)×(82)/(4.1)

            = 92 cm

Therefore, the object distance is 92 cm.

8 0
3 years ago
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