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3 years ago

13.

Physics

1 answer:

lisov135 [29]3 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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A long, hollow wire with inner radius a and outer radius b carries a uniform current density J. What is the magnetic field as a

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Answer:

The magnetic field in the region a < r < b is $B=\frac{u_{0}I(r^{2}-a^{2}) }{2\pi r(b^{2}-a^{2}) }$

Explanation:

If we have the a < r < b. The formula of current is:

$J=\frac{I_{total} }{A}$

Where:

A = area enclosed by the loop.

Itotal = total current in loop.

$J=\frac{I}{\pi b^{2}-\pi a^{2} }$

$I_{enclosed} =JA_{enclosed}$

$I_{enclosed} =\frac{I(\pi r^{2}- \pi a^{2})}{\pi b^{2}-\pi a^{2} }$

If we have the Ampere`s law:

$\int\limits^a_b {B} \, ds =u_{0} I_{enclosed} \\2B\pi r=u_{0} (\frac{I(\pi r^{2}-\pi ^{2} }{\pi ^{2}-\pi a^{2} } )\\B=\frac{u_{0}I(r^{2}-a^{2}) }{2\pi r(b^{2}-a^{2}) }$

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3 years ago

Question 2 (ID=81813)

zaharov [31]

Answer:

The answer is B

Explanation:

5/2=2.5

2.5x2=5

Hope this helps ik its kinda confusing lol

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3 years ago

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A cruise ship makes its way from one island to another. The ship is in motion compared with which reference point?

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A reference point would be something not on the ship which could be used to calculate distance traveled.

Answer: C.) A lighthouse on a nearby Island

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3 years ago

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Calculate the speed of a proton after it accelerates from rest through a potential difference of 350 V.

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The speed of a proton after it accelerates from rest through a potential difference of 350 V is $25.86 \times 10^4 ~m/s$ .

Initial velocity of the proton $u = 0$

Given potential difference $\Delta V = 350V$

let's assume that the speed of the proton is $v$ ,

Since the proton is accelerating through a potential difference, proton's potential energy will change with time. The potential energy of a particle of charge $q$ when accelerated with a potential difference $\Delta V$ is,

$U = q \Delta V$

Due to Work-Energy Theorem and Conservation of Energy - <em>If there is no non-conservative force acting on a particle then loss in Potential energy P.E must be equal to gain in Kinetic Energy K.E</em> i.e

$\Delta K = \Delta V$

If the initial and final velocity of the proton is $u$ and $v$ respectively then,

change in Kinetic Energy $\implies \Delta K = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - 0$

change in Potential Energy $\implies \Delta U = q\Delta V$

from conservation of energy,

$v= \sqrt{\frac{2q\Delta V}{m}}$

so, $v = \sqrt{\frac{2\times 350 \times 1.6\times 10^{-19}}{1.67 \times 10^{-27}}$

$= 25.86 \times 10^4 ~m/s$

To read more about the conservation of energy, please go to brainly.com/question/14668053

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2 years ago

Learning Goal: To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circ

andreev551 [17]

Answer:

v = 15.8 m/s

Explanation:

Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement. So the variation of mechanical energy is equal to the work of the fictional force

$W_{fr}$ = ΔEm = $Em_{f}$ -Em₀

Let's write the mechanical energy at each point

Initial

Em₀ = Ke = ½ k x²

Final

$Em_{f}$ = K + U = ½ m v² + mg y

Let's use Hooke's law to find compression

F = - k x

x = -F / k

x = 4400/1100

x = - 4 m

Let's write the energy equation

fr d = ½ m v² + mgy - ½ k x²

Let's clear the speed

v² = (fr d + ½ kx² - mg y) 2 / m

v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50) 2/60.0

v² = (160 + 8800 - 1470) / 30

v = √ (229.66)

v = 15.8 m/s

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