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2 years ago

13.

Physics

1 answer:

lisov135 [29]2 years ago

5 0

The net torque on the seesaw is 294 Nm.

<h3>What is torque?</h3>

Torque is the force that tends to rotate the body to which it was applied.

To calculate the net torque, we use the formula below

Formula:

τ = mgd-m'gd'........... Equation 1

Where:

τ = Net torque

m = Jenny's mass
m' = Tom's mass
d = Jenny's distance from the pivot
d' = Tom's distance from the pivot
g = acceleration due to gravity

From the question,

Given:

m = 40 kg
m' = 30 kg
d = 1.5 m
d' = 1 m
g = 9.8 m/s²

Substitute these values into equation 1

τ = (40×1.5×9.8)-(30×1×9.8)
τ = 588-294
τ = 294 Nm

Hence, The net torque on the seesaw is 294 Nm.

Learn more about torque here: brainly.com/question/14839816

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An object at 27°C has its temperature increased to 37°C.

Firlakuza [10]

Answer:

b. 14

Explanation:

$T_{i}$ = Initial temperature = 27 °C = 27 + 273 = 300 K

$T_{f}$ = Final temperature = 37 °C = 37 + 273 = 310 K

$P_{i}$ = Initial Power radiated by the object

$P_{f}$ = Final Power radiated by the object

We know that the power radiated is directly proportional to fourth power of the temperature. hence

$\frac{P_{f}}{P_{i}} = \frac{T_{f}^{4} }{T_{i}^{4} }\\\frac{P_{f}}{P_{i}} = \frac{(310)^{4} }{(300)^{4} }\\\frac{P_{f}}{P_{i}} = 1.14\\P_{f} = (1.14) P_{i}$

Percentage increase in power is given as

$\frac{(P_{f} - P_{i})\times100}{P_{i}} \\\frac{((1.14) P_{i} - P_{i})\times100}{P_{i}} \\14$

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3 years ago

Two cars are heading towards each other but are 12 km apart. one car is going 70 km/hr, and the other is going 50 km/hr. how muc

vova2212 [387]

<span>12-50t=70t, t= 0.1h = 6 minutes.</span>

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3 years ago

A thin stream of water flows smoothly from a faucet and falls straight down. at one point the water is flowing at a speed of v1

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3 years ago

A) the unstretched length of each elastic rope is 24m. The rope obeys hookes law. The vertical distance between P and Q is 35m.

solong [7]

Explanation:

a) The rope obeys Hooke's law, so:

F = k Δx

The elastic energy in the rope is:

EE = ½ k Δx²

Or, in terms of F:

EE = ½ F Δx

Use trigonometry to find the stretched length.

cos 20° = 35 / x

x = 37.25

So the displacement is:

Δx = 37.25 − 24

Δx = 13.25

The elastic energy per rope is:

EE = ½ (3.7×10⁴ N) (13.25 m)

EE = 245,000 J

There's two ropes, so the total energy is:

2EE = 490,000 J

Rounded to one significant figure, the elastic energy is 5×10⁵ J.

b) The elastic energy in the ropes is converted to gravitational energy.

EE = PE = mgh

5×10⁵ J = (1.2×10³ kg) (9.8 m/s²) h

h = 42 m

Rounded to one significant figure, the height is 40 m. So the claim is not justified.

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