Can somebody answer this for me asap!

Which is best supported by the data in the chart?

bija089 [108]

Answer:

Option C: Current X has a lower potential difference than Current Y.

Explanation:

The chart above only shows the potential difference of difference current.

A careful observation of the chart shows that Current X has a lower potential difference than Current Y.

7 0

3 years ago

A particle traveling in a circular path of radius 300 m has an instantaneous velocity of 30 m/s and its velocity is increasing a

hodyreva [135]

Answer:

5 m/s2

Explanation:

The total acceleration of the circular motion is made of 2 components: centripetal acceleration and linear acceleration of 4 m/s2. They are perpendicular to each other.

The centripetal acceleration is the ratio of instant velocity squared and the radius of the circle

$a_c = \frac{v^2}{r} = \frac{30^2}{300} = \frac{900}{300} = 3 m/s^2$

So the magnitude of the total acceleration is

$a = \sqrt{a_c^2 + a_l^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 m/s^2$

4 0

3 years ago

Nuclear _________ reactions convert protons into helium; thus, becoming the source of all energy radiated by the sun. A) fission

Bogdan [553]

Nuclear fusion reactions convert protons into helium; thus, becoming the source of all energy radiated by the sun.

Answer: Option D

<u>Explanation:</u>

Nuclear fusion reaction is one among the two nuclear reactions in which the atoms nucleus interact with each other to produce the products. In nuclear fusion, two smaller atoms react together to form a new atom with bigger size.

So large amount of energy is required to start the nuclear fusion reaction. The fusion reactions mostly occurs in stars. The illumination in Stars even the Sun is due to nuclear fusion reaction occurring with the atoms present in them.

Mostly the energy radiated by Sun is due to the protons-protons chain reaction. In this chain reaction, the protons get converted into helium due to nuclear fusion reaction, thus becoming the source of all energy radiated by the Sun.

3 0

3 years ago

A man starts walking north at 3 ft/s from a point P. Five minutes later a woman starts walking south at 4 ft/s from a point 500

mrs_skeptik [129]

Answer:

ds/dt = 6.98 ft/s

Explanation:

Given:

- The speed of man due north Vm = 3 ft/s

- The speed of woman due south Vw = 4 ft/s

- Woman starts walking 5 mins later than man

Find:

At what rate are the people moving apart 15 min after the woman starts walking?

Solution:

- The total time for which the man is walking due north from P, is Tm:

Tm = 5 + 15 = 20 mins

- The total distance traveled by man in Tm mins is:

Dm = Tm*Vm

Dm = 20*60*3

Dm = 3,600 ft

- The total time for which the woman is walking due south from 500 ft due east from P, is Tw:

Tw = 15 = 15 mins

- The total distance traveled by man in Tw mins is:

Dw = Tw*Vw

Dw = 15*60*4

Dw = 3,600 ft

- The displacement between man and woman at any instance is (s) which can be related by pythagoras theorem as follows:

s^2 = (dm + dw)^2 + 500^2

Where, dm : Distance travelled by man at any time Tm

dw : Distance travelled by woman at any time Tw

- Differentiate s with respect to t:

2s*ds/dt = 2*(dm + dw)*(Vm + Vw)

s*ds/dt = (dm + dw)*(Vm + Vw)

ds/dt = [ (dm + dw)*(Vm + Vw) ] / s

- Evaluate the rate of separation of man and woman ds/dt by evaluating at instance Tm = 20 mins and Tw = 15 mins. We have:

ds/dt = [ (Dm + Dw)*(Vm + Vw) ] / sqrt ( (Dm + Dw)^2 + 500^2 )

- Plug in the values:

ds/dt = [ (3600 + 3600)*(3 + 4) ] / [sqrt ( (3600 + 3600)^2 + 500^2 )]

ds/dt = 6.98 ft/s

7 0

3 years ago

A bucket of mass M (when empty) initially at rest and containing a mass of water is being pulled up a well by a rope exerting a

Naily [24]

Answer:

$V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT$

Explanation:

Given that

Constant rate of leak =R

Mass at time T ,m=RT

At any time t

The mass = Rt

So the total mass in downward direction=(M+Rt)

Now force equation

(M+Rt) a =P- (M+Rt) g

$a=\dfrac{P}{M+Rt}-g$

We know that

$a=\dfrac{dV}{dt}$

$\dfrac{dV}{dt}=\dfrac{P}{M+Rt}-g$

$\int_{0}^{V}V=\int_0^T \left(\dfrac{P}{M+Rt}-g\right)dt$

$V=\dfrac{P}{R}\ ln\dfrac{M+RT}{M}-gT$

$V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT$

This is the velocity of bucket at the instance when it become empty.

6 0

3 years ago