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3 years ago

14. A group of students tested different materials in an electric circuit. The table shows the results of their experiment.

Physics

1 answer:

kvasek [131]3 years ago

8 0

Answer: There is no pic homie

Explanation:I am on the same question, so I can’t help :(

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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.

3 0

4 years ago

True or False? Sometimes, substances can be added to a reaction to cause the reaction to start or speed up. These are not consid

Sphinxa [80]

The correct anwser is A true

8 0

4 years ago

Read 2 more answers

Jane described two substances which participate in photosynthesis. Substance A: Combines with water Substance B: Gaseous product

saul85 [17]

<span>I am pretty sure the correct answer is C.
It makes the most sense.
Please correct me if I'm wrong.</span>

5 0

3 years ago

Read 2 more answers

A ball is dropped from rest and falls to the floor. The initial gravitational potential energy of the ball-Earth-floor system is

Cloud [144]

Answer:

The mechanical energy of the ball-Earth-floor system the instant the ball left the floor is 7 Joules.

Explanation:

It is given that,

Initial gravitational potential energy of the ball-Earth-floor system is 10 J.

The ball then bounces back up to a height where the gravitational potential energy is 7 J.

Let U is the mechanical energy of the ball-Earth-floor system the instant the ball left the floor. Due to the conservation of energy, the mechanical energy is equal to difference between initial gravitational potential energy and the after bouncing back up to a height.

Initial mechanical energy is 10 + 0 = 10 J

Mechanical energy just before the collision is 0 + 10 = 10 J

Final mechanical energy, 7 + 0 = 7 J

Hence, the mechanical energy of the ball-Earth-floor system the instant the ball left the floor is 7 Joules.

4 0

3 years ago

A second class lever needs less effort to lift a heavy load.why give reason

Marrrta [24]

Answer:second class lever needs less force to lift a load

Explanation: because the load is between the fulcrum/pivot point and the force. The fulcrum is far away from the force

7 0

3 years ago