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just olya [345]
3 years ago
12

The two main states of mechanical energy are ___________ and potential energy.

Physics
2 answers:
shutvik [7]3 years ago
6 0
<span>Potential energy and Kinetic energy</span>
alexgriva [62]3 years ago
4 0
Kinetic.

Potential Energy is the state of energy where something has a high energy that could potentially occur, and kinetic energy is the state of energy where something is occuring.
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Tony walks at an average speed of 70 m/min from his home to
aksik [14]

Answer:

The answer is 30 min

Explanation:

t = s/v

v = 70 m/min, s = 2100 m

t = 2100/70 = 30 min.

Hope it helps you! \(^ᴥ^)/

6 0
2 years ago
Name 4 element of weather ​
andriy [413]

Answer:

They are temperature, atmospheric pressure, wind, humidity, precipitation, and cloudiness. Together, these components describe the weather at any given time.

7 0
2 years ago
A mass M tied to a light string is moving in a vertical circle. The tension in the string at the top is TT and TB at the bottom
natita [175]

Answer:

Explanation:

Tension provides centripetal force in the circular motion . In circular motion work done by force = torque x angle

torque is zero as , centripetal force passes through axis of rotation that is center.

So work done by centripetal force  = 0

So work done by tension on M = 0

5 0
3 years ago
Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.8 kg and contains
OLEGan [10]

Answer:

C = 771.35 J/kg°C

Explanation:

Here, e consider the conservation of energy equation. The conservation of energy principle states that:

Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container

Since,

Heat Given or Absorbed by a material = m C ΔT

Therefore,

m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃

where,

m₁ = Mass of Metal Piece = 2.3 kg

C = Specific Heat of Metal = ?

ΔT₁ = Change in temperature of metal piece = 165°C - 18°C = 147°C

m₂ = Mass of Metal Container = 3.8 kg

ΔT₂ = Change in temperature of metal piece = 18°C - 15°C = 3°C

m₃ = Mass of Water = 20 kg

C₃ = Specific Heat of Water = 4200 J/kg°C

ΔT₃ = Change in temperature of water = 18°C - 15°C = 3°C

Therefore,

(2.3 kg)(C)(147°C) = (3.8 kg)(C)(3°C) + (20 kg)(4186 J/kg°C)(3°C)

C[(2.3 kg)(147°C) - (3.8 kg)(3°C)] = 252000 J

C = 252000 J/326.7 kg°C

<u>C = 771.35 J/kg°C</u>

5 0
3 years ago
A packet is dropped from a stationary helicopter, hovering at a height 'h' from the ground level, reaches the ground in 12s. Cal
Ksju [112]
Use kinematic equations to solve:

1) yf = yo + vo*t + 1/2at²

yf = final height
yo = initial height
vo = initial velocity
a = acceleration
t = time

yf - yo = vo*t + 1/2at²

yf - yo = h

vo = 0

Thus,

h = 1/2at²

h = 1/2(9.8)(12)² = 705.6 m

2) vf = vo + at

vo = 0

Thus,

vf = at

vf = (9.8)(12) = 117.6 m/s
4 0
3 years ago
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