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

Euglena are _______. A heterotrophs 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:

The ball has charge $-q_0$ , and the ring has positive charge $+Q$ distributed uniformly along its circumference.

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

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

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, then 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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