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The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 1025 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 55 volts?

Answers

Answer 1

Answer:

Explanation:

Given data:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Distance d between the plates = 1 mm = 1×10⁻³m

Voltage of the battery is emf = 100 V

Resistance = 1025 ohm

Solution:

In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by

$V = V_(0)(1-e^{(-t)/(RC) } )\n(V)/(V_(0) ) = 1-e(^{(-t)/(RC) }) \ne^{(-t)/(RC) } = 1- (V)/(V_(0) )$

Taking natural log on both sides,

$e^{(-t)/(RC) } = 1- (V)/(V_(0) ) \n(-t)/(RC) = ln(1-(V)/(V_(0) ) )\nt = -RCln(1 - (V)/(V_(0) ))$

$t = -RC ln (1-(V)/(V_(0) ))$ (1)

Now we can calculate the capacitance by using the area of the plates.

C = ε₀A/d

= $((8.85*10^(-12))) (20*10^(-4)) )/(1*10^(-3) )$

= 18×10⁻¹²F

Now we can get the time when the voltage drop from 100 to 55 V by putting the values of C, V₀, V and R in the equation (1)

$t = -RC ln (1-(V)/(V_(0) ))$

= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)

= 15×10⁻⁹s

= 15 ns

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A golf pro swings a golf club, striking a golf ball that has a mass of 55.0 g. The club is in contact with the ball for only 0.00340 s. After the collision, the ball leaves the club at a speed of 46.0 m/s. What is the magnitude of the average force (in N) exerted on the ball by the club?
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A horizontal cylindrical tank 8.00 ft in diameter is half full of oil (60.0 Ib/ft3). Find the force on one end

Answers

Answer:

Assuming h as the height of the cylindrical tank

$F=480\pi h \,g\,\, (lb)/(ft)$

Explanation:

Assuming that the height is $h$ we can find the volume of the cylindrical tank, then:

$V=\pi*r^2*h$

The diameter is 8.00 ft then $r=4.00 ft$ the total volume of the tank is:

$V=\pi (4.00 ft)^2 h=16\pi h\,\, ft^2$

But the tank is half full of oil, then we need half of the volume. For that reason the volume of oil is:

$V_(oil)=(16\pi h)/(2)ft^2=8\pi h \,\,ft^2$

We know the density of the oil $\rho=60.0\,lb/ft^3$ , with this we can fing the mass of oil that we have because:

$\rho=(m)/(V)$ then $m=\rho V$

Then the mass of oil that we have is:

$m=(60.0(lb)/(ft^3))(8\pi h\,\,ft^2)$

$m=480\pi h (lb)/(ft)$

Note that with the value of h we have the mass in correct units.

Finally to find the force we now that $F=mg$ then we just need to multiply the mass by the gravity.

$F=480\pi h \,g\,\, (lb)/(ft)$

The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about 100 cm . a) What is their initial "launch" speed off the ground?b)How long are they in the air?

Answers

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\n\Rightarrow -u^2=2as-v^2\n\Rightarrow u=√(v^2-2as)\n\Rightarrow u=√(0^2-2* -9.81* 1)\n\Rightarrow u=4.45\ m/s$

a) The vertical speed when the player leaves the ground is 4.45 m/s

$v=u+at\n\Rightarrow t=(v-u)/(a)\n\Rightarrow t=(0-4.45)/(-9.81)\n\Rightarrow t=0.45\ s$

Time taken to reach the maximum height is 0.45 seconds

$s=ut+(1)/(2)at^2\n\Rightarrow 1=0t+(1)/(2)* 9.81* t^2\n\Rightarrow t=\sqrt{(1* 2)/(9.81)}\n\Rightarrow t=0.45\ s$

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

A car with tires of radius 0.25 m come to a stop from 28.78 m/s (100 km/hr) in 50.0 m without any slipping of tires. Find: (a) the angular acceleration of the wheels; (b) number of revolutions made while coming to rest.

Answers

Answer:

The answer is below

Explanation:

a) Using the formula:

$\omega^2=\omega_o^2+2\alpha \theta\n\n\omega=final\ angular\ velocity,\omega_o=initial\ anglular\ velocity,\alpha= angular\ acceleration,\n\theta=angular\ distance\n\nGiven\ that:\n\ninitial\ velocity(u)=28.78m/s,distance(s)=50\ m,radius(r)=0.25\ m,\nfinal/ velocity(v)=0(stop)\n\n\omega=v/r=(28.78m/s)/(0.25m) =115.12\ rad/s,\omega_o=0,\theta=s/r=(50\ m)/(0.25\ m)=200\ rad\n \n\omega^2=\omega_o^2+2\alpha \theta\n\n115.12^2=0^2+2\alpha(200)\n\n2\alpha(200)=13252.6144\n\n\alpha=33.13\ rad/s^2$

$\theta=200\ rad=200\ rad*(1\ rev)/(2\pi\ rad)=31.83\ rev$

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is = 2.93 × 109 W/m2. What is the rms value of (a) the electric field and (b) the magnetic field in the electromagnetic wave emitted by the laser?

Answers

A. The rms value of electric field be "1.05 × 10⁶ N/C".

B. The rms value of magnetic field will be "3.5 × 10⁻³ T".

Magnetic and Electric field

According to the question,

Intensity of the wave, S = 2.93 × 10⁹ W/m²

Free space permittivity, $\epsilon_0$ = 8.86 × 10⁻¹²

Speed of light, c = 3 × 10⁸

A. We know that,

The rms value of electric field,

→ $E_(rms)$ = $\sqrt{(S)/(\epsilon_0 c) }$

By substituting the values,

= $\sqrt{(2.93* 10^9)/((8.85* 10^(-12))(3* 10^8)) }$

= 1.05 × 10⁶ N/C

and,

B. We know that,

The rms value of magnetic field,

→ $B_(rms)$ = $(E_(rms))/(c)$

By substituting the values,

= $(1.05* 10^6)/(3* 10^8)$

= 3.5 × 10⁻³ T

Thus the above response is appropriate.

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To solve this problem, it is necessary to apply the concepts related to the electric field according to the intensity of the wave, the permittivity constant in free space and the speed of light.

As well as the expression of the rms of the magnetic field as a function of the electric field and the speed of light.

PART A) The expression for the rms of electric field is

$E_(rms) = \sqrt{(S)/(\epsilon_0 c)}$

Where,

S= Intensity of the wave

$\epsilon_0$ = Permitivitty at free space

c = Light speed

Replacing we have that,

$E_(rms) = \sqrt{((2.93*10^9))/((8.85*10^(-12))(3*10^8))}$

$E_(rms) = 1.05*10^6N/C$

The RMS value of electric field is $1.05*10^6N/C$

PART B) The expression for the RMS of magnetic field is,

$B_(rms) = (E_(rms))/(c)\nB_(rms) = (1.05*10^6)/(3*10^8)\nB_(rms) =3.5*10^(-3)T$

The RMS of the magnetic field is $3.5*10^(-3)T$

A 12.0 kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.50 m, is whirled in a vertical circle with a constant angular speed of 120 rev/min. The cross-sectional area of thewire is 0.014 cm2. Calculate the elongation of the wire when the mass is a) at its lowest point of the path and b) at the highest point of its path

Answers

Answer:

a) the elongation of the wire when the mass is at its lowest point on the path = 0.5 cm

b) the elongation of the wire when the mass is at its highest point on the path = 0.42 cm

Explanation:

Given that;

the angular speed $\omega = 120 \ rev/min$

Then converting it to rad/s ; we have:

= $(120 \ rev/min )((2 \ \pi \ rad )/(1 \ rev) ) ((1 \ min )/(60 \ s) )$

= 12.57 rad/s

The cross-sectional area of the wire A = 0.014 cm²

A = (0.014 cm²) ( $(10^(-4) \ m^2)/(1 \ cm^3)$ )

A = $0.014*10^(-4) \ m^2$

mass (m) = 12.0 kg

R = 0.5 m

g = 9.8 m/s²

To calculate for the mass when its at the lowest point of the path; we use the Newton's second law of motion; which is expressed as:

$T - mg = ma_(rad)$

where;

$a_rad = ( radical \ acceleration ) = R \omega^2$

Now; we can rewrite our equation as;

$T -mg = m R \omega ^2$

$T = mR \omega^2 + mg$

$T = m( R \omega^2 + g)$

Replacing our given values ; we have:

$T = 12.0( 0.5(12.57)^2 + 9.8)$

$T = 12.0( 0.5(158.0049) + 9.8)$

$T = 12.0( 79.00245 + 9.8)$

$T = 12.0( 88.80245)$

T = 1065.6294 N

T ≅ 1066 N

Determining the elongation $\delta l$ in the wire by using the equation

Y = $(Tl)/(AY)$

Making $\delta l$ the subject of the formula; we have

$\delta l = (Tl)/(AY)$

where ;

l = length of the wire

T =Tension in the wire

A = cross - sectional area

Y = young's modulus

Then;

$\delta l = ((1066 N)(0.5m))/((0.014*10^(-4)m^2)(7.0*10^(10)Pa))$

$\delta l$ = $( 0.00544 m ) *((10 ^2 cm)/(1m) )$

$\delta l$ = 0.5 cm

Thus, the elongation of the wire when the mass is at its lowest point on the path = 0.5 cm

Using Newton's second law of motion also for the mass at its highest point of the path; we have:

$T +mg = m R \omega ^2$

$T = mR \omega^2- mg$

$T = m( R \omega^2 - g)$

Replacing our given values ; we have:

$T = 12.0( 0.5(12.57)^2 - 9.8)$

$T = 12.0( 0.5(158.0049)-9.8)$

$T = 12.0( 79.00245 - 9.8)$

$T = 12.0( 69.20245)$

T = 830.4294 N

T = 830 N

Determining the elongation $\delta l$ in the wire by using the equation

Y = $(Tl)/(AY)$

Making $\delta l$ the subject of the formula; we have

$\delta l = (Tl)/(AY)$

where ;

l = length of the wire

T =Tension in the wire

A = cross - sectional area

Y = young's modulus

Then;

$\delta l = ((830 N)(0.5m))/((0.014*10^(-4)m^2)(7.0*10^(10)Pa))$

$\delta l$ = $( 0.00424 m ) *((10 ^2 cm)/(1m) )$

$\delta l$ = 0.42 cm

Thus, the elongation of the wire when the mass is at its highest point on the path = 0.42 cm

Burns produced by steam at 100°C are much more severe than those produced by the same mass of 100°C water. Calculate the quantity of heat in (Cal or kcal) that must be removed from 6.1 g of 100°C steam to condense it and lower its temperature to 46°C. Specific heat of water = 1.00 kcal/(kg · °C); heat of vaporization = 539 kcal/kg; specific heat of human flesh = 0.83 kcal/(kg · °C).

Answers

Final answer:

To calculate the quantity of heat that must be removed from 6.1 g of 100°C steam, we need to consider both the change in temperature and the phase change from steam to liquid. The specific heat of water is used to calculate the heat required to lower the temperature, while the heat of vaporization is used to calculate the heat required to condense the steam. Adding these two heat values together gives us the total amount of heat that must be removed from the steam, which is approximately 3.61164 kcal.

Explanation:

When steam at 100°C condenses and its temperature is lowered to 46°C, heat must be removed from the steam. To calculate the amount of heat, we can use the specific heat of steam and the latent heat of vaporization. First, we calculate the heat required to lower the temperature of the steam from 100°C to 46°C using the specific heat of water. We then calculate the heat required to condense the steam using the latent heat of vaporization. Finally, we add these two heat values together to obtain the total amount of heat that must be removed from the steam.

Given:

Mass of steam = 6.1 g
Temperature change = 100°C - 46°C = 54°C
Specific heat of water = 1.00 kcal/(kg · °C)
Heat of vaporization = 539 kcal/kg

Calculations:

Heat required to lower the temperature of the steam:
Q1 = mass × specific heat × temperature change
= 6.1 g × (1.00 kcal/(kg · °C) ÷ 1000 g) × 54°C
Heat required to condense the steam:
Q2 = mass × heat of vaporization
= 6.1 g × (539 kcal/kg ÷ 1000 g)
Total heat required:
Q = Q1 + Q2

Calculation:

Q1 = 0.32874 kcal
Q2 = 3.2829 kcal
Q = Q1 + Q2 = 0.32874 kcal + 3.2829 kcal = 3.61164 kcal

Therefore, the quantity of heat that must be removed from 6.1 g of 100°C steam to condense it and lower its temperature to 46°C is approximately 3.61164 kcal.

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Final answer:

To condense and cool 6.1 g of 100°C steam to 46°C, 3.2879 kcal must be removed for condensation, and 0.3304 kcal for cooling, for a total of 3.6183 kcal.

Explanation:

Calculating the Quantity of Heat for Condensation and Cooling

To calculate the quantity of heat that must be removed from 6.1 g of 100°C steam to condense it and lower its temperature to 46°C, we need to consider two processes: condensation and cooling. For condensation, we use the heat of vaporization, and for cooling, we use the specific heat of water.

Calculate the heat released during condensation of steam into water at 100°C:
Heat = mass × heat of vaporization
Heat (in kcal) = (6.1 g) × (539 kcal/kg) × (1 kg / 1000 g)
Heat = 3.2879 kcal
Calculate the heat released when the water cools from 100°C to 46°C:
Heat = mass × specific heat × change in temperature
Heat (in kcal) = (6.1 g) × (1.00 kcal/kg°C) × (1 kg / 1000 g) × (100°C - 46°C)
Heat = 0.3304 kcal

Total heat removed is the sum of the heat from both steps: 3.2879 kcal + 0.3304 kcal = 3.6183 kcal.

Learn more about Heat Transfer here:

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