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FromTheMoon [43]
3 years ago
9

while working out a man performed 2525j of work in 19seconds . what was his power A:132.9w. B:241.5w C 47.975w. D100.5w

Physics
2 answers:
Olenka [21]3 years ago
7 0
A: 132.9w because 2525\19 is how much energy transferred per second which is also known as the power
tamaranim1 [39]3 years ago
6 0

Answer:

Power of the man is 132.9 W.

Explanation:

Given that,

Work performed by the man is 2525 J

Time for which the work is performed is 19 second

To find :

Power of the man

Solve :

Let P is the power of the man. The total work done divided by time taken is called the power of an object. Its formula is given by :

P=\dfrac{W}{t}

P=\dfrac{2525\ J}{19\ s}

P=132.89\ W

or

P=132.9\ W

Therefore, the power of the man is 132.9 W.

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Problem 3: Thermal expansionThe steel rod has the length 2 m and cross-section area 200 cm2at the room temperature 20◦C. Weapply
zalisa [80]

Answer:

a) 2.00024 m

b) 0.036%

c) 436.67°C

Explanation:

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Initial cross sectional Area = A₀ = 200 cm² = 0.02 m²

We can obtain initial volume = V₀ = A₀L₀ = 0.02 × 2 = 0.04 m³

Initial Temperature = T₀ = 20°C

Coefficient of linear expansivity = α = (2 × 10⁻⁶) (°C)⁻¹

a) New length of the rod after heating to 80°C

Linear expansion is given as

ΔL = L₀ × α ×ΔT

ΔL = 2 × 2 × 10⁻⁶ × (80 - 20) = 0.00024 m = 0.24 mm

New length = old length + expansion = 2 + 0.00024 = 2.00024 m

b) The percentage of the volume change of the rod.

Volume expansion is given by

ΔV = V₀ × (3α) × ΔT

Volume expansivity ≈ 3 × (linear expansivity)

ΔV = 0.04 × (3×2×10⁻⁶) × (80 - 20) = 0.0000144 m³

Percentage change in volume = 100% × (ΔV/V₀) = 100% × (0.0000144/0.04) = 0.036%

c) The maximal temperature we can allow if the volume should not increase by more than half percent.

For a half percent increase in volume, the corresponding change in volume needs to be first calculated.

Percentage change in volume = 100% × (ΔV/V₀)

0.5 = 100% × (ΔV/0.04)

(ΔV/0.04) = 0.005

ΔV = 0.0002 m³

Then we now investigate the corresponding temperature that causes this.

ΔV = V₀ × (3α) × ΔT

0.0002 = 0.04 × (3×2×10⁻⁶) × ΔT

ΔT = (0.0002)/(0.04 × 3 × 2 × 10⁻⁶) = 416.67°C

Maximal temperature = T₀ + ΔT = 20 + 416.67 = 436.67°C

4 0
3 years ago
A student of weight 652 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude
miskamm [114]

Answer:

Explanation:

Given

Weight of person W=652\ N

At highest point Magnitude of the normal force F=585\ N

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Negative sign shows force is in upward direction

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F_n=652+67

F_n=719\ N  

8 0
3 years ago
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