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2 years ago

Hello pick all the right ones

Physics

1 answer:

hammer [34]2 years ago

6 0

Answer:

ok where's the problem so i can help u

Explanation:

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A positive charge is moved from point A to point B along an equipotential surface. How much work is performed or required in mov

Nitella [24]

Answer:

No work is performed or required in moving the positive charge from point A to point B.

Explanation:

Lets take

Q= Positive charge which move from point A to point B along

Voltage difference,ΔV =V₁ - V₂

The work done

W = Q . ΔV

Given that charge is moved from point A to point B along an equipotential surface.It means that voltage difference is zero.

ΔV = 0

W = Q . ΔV

W = Q x 0

W= 0 J

So work is zero.

5 0

2 years ago

If the pitch of the sound coming out of a speaker increases, which statement is true about the sound wave?

S_A_V [24]

Answer:

frequency and amplitude increases

3 0

1 year ago

Calculating ph how is ph related to the concentration of hydronium ions. True or False

Dmitry_Shevchenko [17]

I think this is TRUE. Ph is calculating the acidic acids in water. And you need to know the concentration of hydronium ions. Hope this helped !

7 0

2 years ago

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie

leva [86]

To solve the problem it is necessary to apply conservation of the moment and conservation of energy.

By conservation of the moment we know that

$MV=mv$

Where

M=Heavier mass

V = Velocity of heavier mass

m = lighter mass

v = velocity of lighter mass

That equation in function of the velocity of heavier mass is

$V = \frac{mv}{M}$

Also we have that $m/M = 1/7 times$

On the other hand we have from law of conservation of energy that

$W_f = KE$

Where,

W_f = Work made by friction

KE = Kinetic Force

Applying this equation in heavier object.

$F_f*S = \frac{1}{2}MV^2$

$\mu M*g*S = \frac{1}{2}MV^2$

$\mu g*S = \frac{1}{2}( \frac{mv}{M})^2$

$\mu = \frac{1}{2} (\frac{1}{7}v)^2$

$\mu = \frac{1}{98}v^2$

$\mu = \frac{1}{g(98)(5.1)}v^2$

Here we can apply the law of conservation of energy for light mass, then

$\mu mgs = \frac{1}{2} mv^2$

Replacing the value of $\mu$

$\frac{1}{g(98)(5.1)}v^2 mgs = \frac{1}{2}mv^2$

Deleting constants,

$s= \frac{(98*5.1)}{2}$

$s = 249.9m$

7 0

3 years ago

Which two actions best demonstrate taking responsibility?

kow [346]

Answer:

Leading your people into the right direction, and always knowing what is best for you and your people.

Explanation:

6 0

3 years ago