Which is a group pragram acoholics recover in

Which of the following represents an image that is located in front of a lens?

Lostsunrise [7]

-di represents an image in front of a lens

5 0

3 years ago

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In order for a wave to occur, there must be a ______________ in the medium.

natta225 [31]

I believe the answer is “amplitude”.

3 0

4 years ago

A pitcher hurls a 0.25kg ball around a vertical circular path of radius 0.6 m, applying a tangential force of 30N, before releas

Sever21 [200]

Thank you for posting your question here at brainly. Below is the solution:

Ke up top = 1/2*.25 *225
gain of Pot energy = .25*9.81*1.2 
work input = (1/4)(2 pi *.6)*30 

so 
sum of those 3 energies = 
(1/2)(.25)v^2

3 0

4 years ago

At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r

Oksana_A [137]

Answer:

For proton:

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

For electron:

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s

Explanation:

For a proton we have:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

$W = \Delta Vq$ (1)

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C

V: is the potential

Also, the work is equal to:

$W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}$ (2)

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

$v_{a}$ : is the velocity in the point a

$v_{b}$ : is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

$\Delta Vq = \frac{1}{2}mv_{a}^{2}$

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}$

$v_{a} = 2.9 \cdot 10^{4} m/s$

For an electron we have:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

$\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}$

Since $v_{a}$ = 0 (starts from rest) and $m_{e}$ = 9.1x10⁻³¹ kg (electron's mass), we have:

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}$

$v_{b} = 1.2 \cdot 10^{6} m/s$

I hope it helps you!

6 0

4 years ago

What would increase the force of gravity between two objects?

yulyashka [42]

By Newton's Law of Universal Gravitation.

F = GMm/r²

Where F is Force of Gravitation, M = Mass of first object, m = mass of second object, r = distance of separation

From the formula, you can see that if the masses, M and m, increased, the value of F would definitely increase as well.

And if r increased the value of F would be reduced because you would be dividing by a bigger number, but when the value of r is decreased the value of F would be increased, because you would then be dividing by something smaller. Note the r is at the denominator of the formula.

So F would increase if there was increase in Masses and decrease in distance.

So the answer is C. a and b.

3 0

3 years ago