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-BARSIC- [3]
2 years ago
5

A stone is dropped from the

Physics
1 answer:
ICE Princess25 [194]2 years ago
8 0
  • Height=h=500m
  • Acceleration=g=10m/s^2
  • Initial velocity=u=0
  • Speed of sound=c=340m/s
  • TIME TAKEN BY STONE TO HIT WATER=t
  • Time taken by sound to hear back=T

Now

\\ \sf\longmapsto h=ut+\dfrac{1}{2}gt^2

\\ \sf\longmapsto h=0t+\dfrac{1}{2}10t^2

\\ \sf\longmapsto 500=5t^2

\\ \sf\longmapsto t^2=100

\\ \sf\longmapsto t=10s

Now

\\ \sf\longmapsto h=cT

\\ \sf\longmapsto T=\dfrac{h}{c}

\\ \sf\longmapsto T=\dfrac{500}{340}

\\ \sf\longmapsto T=1.47\approx 1.5s

Total time:-

\\ \sf\longmapsto T_{net}=t+T=10+1.5=11.5s

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The main advantage of an electromagnet over a permanent magnet is that the magnetic field can be quickly changed by controlling the amount of electric current in the winding. However, unlike a permanent magnet that needs no power, an electromagnet requires a continuous supply of current to maintain the magnetic field.
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3 years ago
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For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro
viktelen [127]

Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

where,

N_A=Avogadro's Number

n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3

Now,

the drift velocity, V_d

V_d=\frac{J}{ne}

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

4 0
3 years ago
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It takes 180 kj of work to accelerate a car from 21.0 m/s to 27.0 m/s. what is the car's mass?
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A 330.-ohm resistor is connected to a 5.00-volt battery. What is the current through the resistor?
Gelneren [198K]

I = 0.0152 A = 15.2 mA

Explanation:

Using Ohm's law,

V = IR or

I = V/R

= (5.00 V)/(330. ohms)

= 0.0152 A

= 15.2 mA

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2 years ago
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A gas has an initial volume of 2.5 L at a temperature of 275 K and a pressure of 2.1 atm. The pressure of the gas increases to 2
olchik [2.2K]

Answer:

2.10L

Explanation:

Given data

V1= 2.5L

T1= 275K

P1= 2.1atm

P2= 2.7 atm

T2= 298K

V2= ???

Let us apply the gas equation

P1V1/T1= P2V2/T2

substitute into the expression we have

2.1*2.5/275= 2.7*V2/298

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Cross multiply

275*2.7V2= 298*5.25

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V2= 1564.5/742.5

V2= 2.10L

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7 0
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