2. A hanging wind-chime on a calm day would have kinetic or potential energy?

A river has a steady speed of 0.566 m/s. A student swims upstream a distance of 1.57 km and returns (still swimming) to the star

MArishka [77]

Answer:

3503.72 seconds

819.96 seconds

Explanation:

$V_r$ =Velocity of river = 0.566 m/s

$V_s$ =Velocity of student = 1.17 m/s

Distance to travel = 1.57 km = 1570 m

So,

Time = Distance / Speed

$\frac{1570}{V_s-V_r}+\frac{1570}{V_s+V_r}=t\\\Rightarrow t=\frac{1570}{1.17-0.566}+\frac{1570}{1.17+0.566}\\\Rightarrow t=3503.72\ s$

Time taken by the student to complete the trip is 3503.72 seconds

In still water

$\frac{1570}{V_s}+\frac{1570}{V_s}=t\\\Rightarrow t=\frac{1570}{1.17}+\frac{1570}{1.17}\\\Rightarrow t=2683.76\ s$

The difference in time between moving water and still water is 3503.72-2683.76 = 819.96 seconds

5 0

3 years ago

The battery charger for an mp3 player contains a step-down transformer with a turns ratio of 1:38, so that the voltage of 120 v

matrenka [14]

Transformer contains two coils: primary and secondary. They allow change of voltage to lower or higher value. In first case we have step-down and in second case we have step-up transformer.
Formula used for transformer is: $\frac{N_{1} }{N_{2}} = \frac{V_{1}}{V_{2}}$

Where:N1 = number of turns on primary coilN2 = number of turns on secondary coilV1 = voltage on primary coilV2 = voltage on secondary coil
In a step-down transformer primary coil has more turns than secondary coil. So the ratio 1:38 means that for each turn on secondary coil we have 38 turns on primary coil.
We can solve the equation for V2: $V_{2} = \frac{ V_{1}* N_{2} }{ N_{1} } \\ V_{2} = \frac{ 120* x }{ 38x} } \\ V_{2} = 3.16V$

Secondary coil provides voltage of 3.16V.

7 0

3 years ago

Which of the advantages to social media as a new media could also be viewed as a disadvantage

Bad White [126]

Fast Communication, can be used for good and can be used for bad hackers.

4 0

4 years ago

Read 2 more answers

A car is moving north on a freeway. If a bug is flying south on the freeway, is the momentum of the bug positive or negative?

faltersainse [42]

Negative

Because the car is moving up and the bug is moving down. but it also depends on the weather so choice between one of those two I think is Negative but I may be wrong.

3 0

2 years ago

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r

aivan3 [116]

Answer:

Explanation:

Given that,

Initial angular velocity is 0

ωo=0rad/s

It has angular velocity of 11rev/sec

ωi=11rev/sec

1rev=2πrad

Then, wi=11rev/sec ×2πrad

wi=22πrad/sec

And after 30 revolution

θ=30revolution

θ=30×2πrad

θ=60πrad

Final angular velocity is

ωf=18rev/sec

ωf=18×2πrad/sec

ωf=36πrad/sec

a. Angular acceleration(α)

Then, angular acceleration is given as

wf²=wi²+2αθ

(36π)²=(22π)²+2α×60π

(36π)²-(22π)²=120πα

Then, 120πα = 8014.119

α=8014.119/120π

α=21.26 rad/s²

Let. convert to revolution /sec²

α=21.26/2π

α=3.38rev/sec

b. Time Taken to complete 30revolution

θ=60πrad

∆θ= ½(wf+wi)•t

60π=½(36π+22π)t

60π×2=58πt

Then, t=120π/58π

t=2.07seconds

c. Time to reach 11rev/sec

wf=wo+αt

22π=0+21.26t

22π=21.26t

Then, t=22π/21.26

t=3.251seconds

d. Number of revolution to get to 11rev/s

∆θ= ½(wf+wo)•t

∆θ= ½(0+11)•3.251

∆θ= ½(11)•3.251

∆θ= 17.88rev.

5 0

3 years ago

Read 2 more answers