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lys-0071 [83]
3 years ago
11

jim(mass=100kg) rollerblades on a smooth horizontal floor at a constant speed of 2.0 m/s for a distance of 5m in 5 seconds. What

is the power in this process? (assume g is 10 m/s^2)
Physics
1 answer:
azamat3 years ago
7 0
Since Jim's speed is constant and he is moving in a straight line, he is not accelerating, and we know the net force on him is zero. There is no Force anywhere doing any work. So no power is being added to him or dissipated by him.
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Answer:

Oxygen-Symbol-O Potassium Symbol-K

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A proton, initially traveling in the +x-direction with a speed of 5.05×10^5 m/s , enters a uniform electric field directed verti
Korvikt [17]

Answer:

The strength of the electric field is 1.35\times10^{4}\ N/C.

Explanation:

Given that,

Speed v= 5.05\times10^{5}\ m/s

Time t= 3.90\times10^{-7}\ s

Angle = 45°

We need to calculate the acceleration

Using equation of motion

v = u+at

5.05\times10^{5}=0+a\times3.90\times10^{-7}

a =\dfrac{5.05\times10^{5}}{3.90\times10^{-7}}

a=1.29\times10^{12}\ m/s^2

We need to calculate the strength of the electric field

Using relation of newton's second law and electric force

F= ma=qE

ma = qE

E=\dfrac{ma}{q}

Put the value into the formula

E=\dfrac{1.67\times10^{-27}\times1.29\times10^{12}}{1.6\times10^{-19}}

E=1.35\times10^{4}\ N/C

Hence, The strength of the electric field is 1.35\times10^{4}\ N/C.

3 0
3 years ago
When steam condenses 1. All of these occur. 2. None of these occur. 3. molecules move closer together. 4. it changes from the ga
True [87]

2. None of these occur

5 0
3 years ago
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Early psychologists were predominately all of the following except __________. A. white B. male C. European D. wealthy
yulyashka [42]

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Explanation: on the e2020 test its right

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4 years ago
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A 26.4 g silver ring (cp = 234 J/kg·°C) is heated to a temperature of 66.2°C and then placed in a calorimeter containing 4.94 ✕
Slav-nsk [51]

Answer:

The final temperature of the mixture = 64.834 °C.

Explanation:

Heat lost by the silver ring = heat gained by the water + heat transferred to the surrounding.

c₁m₁(t₁-t₃) = c₂m₂(t₃-t₂) + Q..............Equation 1

Where c₁ = specific heat capacity of the silver copper, m₁ = mass of the silver copper, t₁ = initial temperature of the silver copper, t₃ = final temperature of the mixture. c₂ = specific heat capacity of water, t₂ = initial temperature of water, m₂ = mass of water, Q = energy transferred to the surrounding.

making t₃ the subject of the equation,

t₃ = [c₁m₁t₁+c₂m₂t₂-Q]/(c₁m₁+c₂m₂)........................ Equation 2

Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92×10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

t₃ = [(234×26.4×66.2)+(4200×0.0492×24)-0.136]/[(234×26.4)+(4200×0.0492)]

t₃ = (408957.12+4959.36-0.136)/(6177.6+206.64)

t₃ = (413916.48-0.136)/6384.24

t₃ = 413916.34/6384.24

Thus the final temperature of the mixture = 64.834 °C.

6 0
3 years ago
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