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3 years ago

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5. One electrical appliance operates with a voltage of 120 V, while another operates with 240 V. Based on this information alone

, is it correct to say that the second appliance uses more power than the first? Give your reasoning.

1 answer:

serious [3.7K]3 years ago

4 0

Answer:

Yes it is true saying that 2nd applience use more power than first one

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What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force $F_{1}=18\ N$

Force $F_{2}=26\ N$

Force $F_{3}=14\ N$

We need to calculate the torque due to force F₁

Using formula of torque

$\tau_{1}=-F_{1}d_{1}$

$\tau_{1}=-F_{1}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{1}=-18\times\dfrac{0.2}{2}$

$\tau_{1}=-1.8\ N-m$

We need to calculate the torque due to force F₂

Using formula of torque

$\tau_{2}=F_{2}d_{2}$

$\tau_{2}=F_{2}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{2}=26\times\dfrac{0.2}{2}$

$\tau_{2}=2.6\ N-m$

We need to calculate the torque due to force F₃

Using formula of torque

$\tau_{3}=F_{3}d_{3}$

$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{3}=0.1(14\sin45+14\cos45)$

$\tau_{3}=1.92\ N-m$

We need to calculate the net torque on the square plate

$\tau=\tau_{1}+\tau_{2}+\tau_{3}$

$\tau=-1.8+2.6+1.92$

$\tau=2.72\ N-m$

Hence, The net torque on the square plate is 2.72 N-m.

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A person on a road trip drives a car at different constant speeds over several legs of the trip. She drives for 10.0 min at 50.0

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<h2>The average speed for the entire trip is 47.5 m/s .</h2>

It is given that for different time span car have different speed and also the person spend $40\ min=\dfrac{40}{60}\ hrs =0.67\ hrs\ .$ in eating lunch and buying gas.

We know , average speed is total distance covered by total time taken .

Therefore , average speed , $v=\dfrac{total\ distance }{total\ times}$

$v=\dfrac{\dfrac{10}{60}\times 50+\dfrac{19}{60}\times 100+\dfrac{60}{60}\times 55}{\dfrac{10}{60}+\dfrac{10}{60}+\dfrac{60}{60}+ \dfrac{40}{60}}\\\\\\v=47.5\ m/s$

Hence, this is the required solution.

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