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Tresset [83]
3 years ago
12

You move 25N object 5 meters. how much work did you do

Physics
1 answer:
Katen [24]3 years ago
8 0

Work = Force(N) * Distance(m)

         = 25 * 5

Work = 125 Joules

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Euglena are _______.<br><br> A heterotrophs<br><br> B autotrophs
zysi [14]
The Euglena is unique in that it is both heterotrophic (must consume food) and autotrophic (can make its own food).
6 0
4 years ago
The ball will oscillate along the z axis between z=dz=d and z=−dz=−d in simple harmonic motion. What will be the angular frequen
Eddi Din [679]

Answer:

\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }

Explanation:

Additional information:

<em>The ball has charge </em>-q_0<em>, and the ring has  positive charge </em>+Q<em> distributed uniformly along its circumference. </em>

The electric field at distance z along the z-axis due to the charged ring is

E_z= \dfrac{kQz}{(z^2+a^2)^{3/2}}.

Therefore, the force on the ball with charge -q_0 is

F=-q_oE_z

F=- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}

and according to Newton's second law

F=ma=m\dfrac{d^2z}{dz^2}

substituting F we get:

- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=m\dfrac{d^2z}{dz^2}

rearranging we get:

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=0

Now we use the approximation that

z^2+a^2\approx a^2 <em>(we use this approximation instead of the original </em>d^2+a^2\approx a^2<em> since </em>z<em>, our assumption still holds )</em>

and get

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(a^2)^{3/2}}=0

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{a^{3}}=0

Now the last equation looks like a Simple Harmonic Equation

m\dfrac{d^2z}{dz^2}+kz=0

where

\omega=\sqrt{ \dfrac{k}{m} }

is the frequency of oscillation. Applying this to our equation we get:

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Q}{a^{3}}z=0\\\\m=m\\\\k= \dfrac{kq_0Q}{a^{3}}

\boxed{\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }}

5 0
4 years ago
at 1 p.m. a car traveling at a constant velocity of 78 km per hour towards the West it's 34 km to the west of our school how far
Fittoniya [83]

83 km/h * 2.5 hours (3:30 - 1:00) = 207.5 km  

207.5 km - 15 km = 192.5 km

3 0
3 years ago
A stone is thrown vertically upward with an initial speed of 24.7 m/s. Neglect air resistance. The speed of the stone when it is
fredd [130]

Answer:

The speed of the stone when it is 4.66 m higher is 236.057 m/s.

Explanation:

Given the initial velocity and vertical distance, we can use the fourth kinematic equation (v^{2} =v_{o}^{2}+2ay) to find v final, or the v to the left of the equal sign. We know v_{o} (initial velocity) is 24.7 m/s, y (change in vertical distance) is 4.66 m, and a is another way to write g (acceleration due to gravity), or 9.8 m/s^{2}.

From here you could plug in the values and solve for v final, but to make the solving process simpler, we can simplify the given equation, <em>then </em>plug in the known values.

To isolate v final, we can take the square root of v^{2} and do the same to the right side of the equation. Therefore, we can find v final with: v_{o} \sqrt{2ay}, where v initial is outside of the square root because it squared...

If we plug in the known values to the simplified equation, we get: v=24.7m/s*\sqrt{2(9.8m/s)(4.66 m)}

The final answer is 236.057 m/s.

4 0
2 years ago
Which change would result in a stronger electromagnet?
Sonbull [250]
I'd say B.) Increasing the voltage of the battery.
6 0
3 years ago
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