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marin [14]
3 years ago
5

A 100 kg bag of sand has a weight on 100 N. When dropped its acceleration is what?

Physics
1 answer:
Anettt [7]3 years ago
4 0

Answer:

The acceleration of the bag of sand is 1\ m/s^2.

Explanation:

We have,

Mass of a bag of sand is 100 kg

Weight of the bag of sand is 100 N

It is required to find the acceleration of the bag when it is dropped. The weight of an object is given by :

F =ma

When it is dropped, a = g

g=\dfrac{F}{m}\\\\g=\dfrac{100}{100}\\\\g=1\ m/s^2

So, the acceleration of the bag of sand is 1\ m/s^2.

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Rotational speed refers to the number of complete _____ in a given amount of time.
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Answer:

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2 years ago
The potential difference between two points, A and B, in an electric field is 2.00 volts. The energy required to move a charge o
Pavlova-9 [17]

Answer:

W_A_B=-1.6\times 10^{-18} J

Explanation:

Let A and B be two points located in a uniform electric field, A being a distance d from B in the direction of the field. The work that an external force must do to bring a unit positive charge q from the reference point to the point considered against the electric force at constant speed, mathematically is expressed by:

V_B_A=\frac{W_A_B}{q}

Therefore, isolating W_A_B and replacing the data provided:

W_A_B=V_B_A *q=-2*(8\times 10^{-19}) =-1.6\times 10^{-18}J

6 0
3 years ago
Calculate the amount of heat needed to raise 1.0 kg of ice at -20 degrees Celsius to steam at 120 degree Celsius
CaHeK987 [17]

Answer:

801.1 kJ

Explanation:

The ice increases in temperature from -20 °C to 0 °C and then melts at 0 °C.

The heat required to raise the ice to 0 °C is Q₁ = mc₁Δθ₁ where m =  mass of ice = 1 kg, c₁ = specific heat capacity of ice = 2108 J/kg°C and Δθ₁ = temperature change. Q₁ = 1 kg × 2108 J/kg°C × (0 - (-20))°C = 2108 J/kg°C × 20  °C = 4216 J

The latent heat required to melt the ice is Q₂ = mL₁ where L₁ = specific latent heat of fusion of ice = 336000 J/kg. Q₁ = 1 kg × 336000 J/kg = 336000 J

The heat required to raise the water to 100 °C is Q₃ = mc₂Δθ₂ where m =  mass of ice = 1 kg, c₂ = specific heat capacity of water = 4187 J/kg°C and Δθ₂ = temperature change. Q₃ = 1 kg × 4187 J/kg°C × (100 - 0)°C = 4187 J/kg°C × 100  °C = 418700 J

The latent heat required to convert the water to steam is Q₄ = mL₂ where L = specific latent heat of vapourisation of water = 2260 J/kg. Q₄ = 1 kg × 2260 J/kg = 2260 J

The heat required to raise the steam to 120 °C is Q₅ = mc₃Δθ₃ where m =  mass of ice = 1 kg, c₃ = specific heat capacity of steam = 1996 J/kg°C and Δθ₃ = temperature change. Q₃ = 1 kg × 1996 J/kg°C × (120 - 100)°C = 1996 J/kg°C × 20  °C = 39920 J

The total amount of heat Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅ = 4216 J + 336000 J

+ 418700 J + 2260 J + 39920 J = 801096 J ≅ 801.1 kJ

4 0
3 years ago
What is the force on a 52 coulomb charge in a field of 5 N/C
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Https://courses.physics.illinois.edu/phys102/sp2013/lectures/lecture2.pdf



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