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Aliun [14]
3 years ago
6

which changes will decrease the electric force between two positively charged objects? select two options

Physics
1 answer:
aliina [53]3 years ago
4 0
Is they mad that you not with them same goes that ain’t be in my dm
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Determine the amount of work done by the applied force when a 87 N force is applied to move a 15 kg object a horizontal
elena-s [515]

Answer:

391.5 J

Explanation:

The amount of work done can be calculated using the formula:

  • W = F║d
  • where the force is parallel to the displacement

Looking at the formula, we can see that the mass of the object does not affect the work done on it.

Substitute the force applied and the displacement of the object into the equation.

  • W = (87 N)(4.5 m)
  • W = 391.5 J  

The amount of work done on the object is 391.5 J in order to move it 4.5 meters with an applied force of 87 Newtons.

5 0
2 years ago
Read 2 more answers
An engine flywheel initially rotates counterclockwise at 6.77 rotations/s. Then, during 23.9 s, its rotation rate changes to 3.5
olganol [36]

Answer:

-2.70 rad/s²

Explanation:

Given that

ω1 = initial angular velocity of the flywheel, which is 6.77 rev/s

If we convert it to rad/s, we have

(6.77 x 2π) rad/s = 13.54π rad/s

ω2 = final angular velocity of the flywheel = -3.51 rev/s,

On converting to rad/s also, we have

(-3.51 x 2π) rad/s = 7.02π rad/s

α = average angular acceleration of the flywheel = ?

Δt = elapsed time = 23.9 s

Now, using the formula, α = (ω2 - ω1)/Δt. On substituting, we have

α = (-7.02π rad/s - 13.54π rad/s)/23.9 s

α = -20.56π rad/s / 23.9 s

α = -64.59 rad/s / 23.9 s

α = -2.70 rad/s²

Therefore, the average angular acceleration of the flywheel is -2.70 rad/s²

7 0
3 years ago
A thin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. It had a wall thickness
Mekhanik [1.2K]

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

\sigma_h = \frac{Pd}{2t}

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

\sigma_l = \frac{Pd}{4t}

The letters have the same meaning as before.

Then he hoop stress would be,

\sigma_h = \frac{Pd}{2t}

\sigma_h = \frac{75 \times 8}{2\times 0.25}

\sigma_h = 1200psi

And the longitudinal stress would be

\sigma_l = \frac{Pd}{4t}

\sigma_l = \frac{75\times 8}{4\times 0.25}

\sigma_l = 600Psi

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

\tau_{max} = \frac{\sigma_h}{2}

\tau_{max} = \frac{1200}{2}

\tau_{max} = 600Psi

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

6 0
3 years ago
Which electrical device makes it possible to transmit electrical energy efficiently from a power plant to users?
Debora [2.8K]
Solar pannels are the energy sorce that lets you transfer energy from a plant source to the use or the consumer 
6 0
3 years ago
Read 2 more answers
You push on a box with 100 N of force, causing it to accelerate at 5 m/s?.
Xelga [282]
<h3><u>Given</u><u>:</u><u>-</u></h3>

Force,F = 100 N

Acceleration,a = 5 m/s²

<h3><u>To</u><u> </u><u>be</u><u> </u><u>calculated:-</u><u> </u></h3>

Calculate the mass of the box .

<h3><u>Formula</u><u> </u><u>used:-</u><u> </u></h3>

\bf \: Force = Mass  \times  Acceleration

<h3><u>Solution:-</u><u> </u></h3>

\sf \: Force = Mass × Acceleration

★ Substituting the values in the above formula,we get:

\sf \implies \: 100 = Mass \times 5

\sf \implies \: Mass =   \cancel\dfrac{100}{5}

\sf \implies \: Mass = 20 \: kg

4 0
3 years ago
Read 2 more answers
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