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Mashcka [7]
3 years ago
10

If we decrease the amount of force applied to an object, and all other factors remain the same, the amount of work completed wil

l
Physics
1 answer:
Nat2105 [25]3 years ago
4 0
A ) decrease.
B ) increase.
C ) increase, then decrease.
D ) not change.

The answer is A) decrease

Take pushing a box, for example-- You  push your hardest then give out, still trying to push the box. You are doing less work than what you have started with!

( Mind marking me for branliest? ; ) )
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yes This is correct Answer

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The metal cover that seals the top of the cylinder
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Answer:a lid

Explanation bob pulled the lid off the jar of pickles

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A neutron consists of one "up" quark of charge +2e/3 and two "down" quarks each having charge -e/3. If we assume that the down q
Anon25 [30]

Answer:

The magnitude of the electrostatic force is 120.85 N

Explanation:

We can use Coulomb's law to find the electrostatic force between the down quarks.

In scalar form, Coulomb's law states that for charges q_1 and q_2 separated by a distance d, the magnitude of the electrostatic force F between them is:

F = k \frac{|q_1q_2|}{d^2}

where k is Coulomb's constant.

Taking the values:

d = 4.6 \ 10^{-15} m

q_1 = q_2 = - \frac{e}{3} = - \frac{1.6 \ 10^{-19} \ C}{3}

and knowing the value of the Coulomb's constant:

k = 8.99 \ 10 ^{9} \frac{N m^2}{C^2}

Taking all this in consideration:

F = 8.99 \ 10 ^{9} \frac{N m^2}{C^2} \frac{ (- \frac{1.6 \ 10^{-19} \ C}{3} ) ^2}{(4.6 \ 10^{-15} m)^2}

F = 120.85  \ N

8 0
3 years ago
3 kg of wet clothes are hung on the middle of a clothesline with posts 6 ft apart. The clothesline sags down by 3 feet. What is
miskamm [114]

Answer:

Tension in string equals 14.715 Newtons

Explanation:

The situation is represented in the figure attached below:

For equilibrium of the clothes along y- axis we have

\sum F_{v}=0\\\\\Rightarrow 2Tcos(\theta )=W\\\\\therefore T=\frac{W}{2cos(\theta )}

Applying values we get

\sum F_{v}=0\\\\\Rightarrow 2Tcos(\theta )=W\\\\\therefore T=\frac{3\times 9.81}{2\times 1}=14.715N(\because cos(\theta )=\frac{3}{3}=1)

4 0
3 years ago
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An object exhibits SHM with an angular frequency w = 4.0 s-1 and is released from its maximum displacement of A = 0.50 m at t =
vivado [14]

Explanation:

It is given that,

Angular frequency, \omega=4\ s^{-1}

Maximum displacement, A = 0.5 m at t = 0 s

We need to find the time at which it reaches its maximum speed. Firstly, we will find the maximum velocity of the object that is exhibiting SHM.

v_{max}=A\times \omega

v_{max}=0.5\times 4

v_{max}=2\ m/s............(1)

Acceleration of the object, a=\omega^2A

a=4^2\times 0.5

a=8\ m/s^2...............(2)

Using first equation of motion we can calculate the time taken to reach maximum speed.

v=u+at

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t = 0.25 s

So, the object will take 0.25 seconds to reach its maximum speed. Hence, this is the required solution.

4 0
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