Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is 0.070m2, and the magnitude of the fluid velocity is 3.50m/s.

part A

What is the fluid speed at point in the pipe where the cross-sectional area is 0.105m2?

part b

What is the fluid speed at point in the pipe where the cross-sectional area is 0.047m2?

part c

Calculate the volume of water discharged from the open end of the pipe in 1.00hour.

Answer:

Velocity of a proton, [tex]v=1.7\times 10^6\ m/s[/tex]    

Explanation:

It is given that,

Potential difference, [tex]V=15\ kV=15\times 10^3\ V[/tex]

Let v is the velocity of a proton that has been accelerated by a potential difference of 15 kV.

Using the conservation of energy as :

[tex]\dfrac{1}{2}mv^2=qV[/tex]

q is the charge of proton

m is the mass of proton

[tex]v=\sqrt{\dfrac{2qV}{m}}[/tex]

[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 15\times 10^3\ V}{1.67\times 10^{-27}\ kg}[/tex]

[tex]v=1695361.75\ m/s[/tex]

[tex]v=1.69\times 10^6\ m/s[/tex]

or

[tex]v=1.7\times 10^6\ m/s[/tex]

So, the velocity of a proton is [tex]1.7\times 10^6\ m/s[/tex]. Hence, this is the required solution.

Answer:

66.2 sec

Explanation:

C₁ = 1.0 F

C₂ = 1.0 F

ΔV = Potential difference across the capacitor = 6.0 V

C = parallel combination of capacitors

Parallel combination of capacitors is given as

C = C₁ + C₂

C = 1.0 + 1.0

C = 2.0 F

R = resistance = 33 Ω

Time constant is given as

T = RC

T = 33 x 2

T = 66 sec

V₀ = initial potential difference across the combination = 6.0 Volts

V = final potential difference = 2.2 volts

Using the equation

[tex]V = V_{o} e^{\frac{-t}{T}}[/tex]

[tex]2.2 = 6 e^{\frac{-t}{66}}[/tex]

t = 66.2 sec

Answer:

Explanation:

When a body starts sliding on an inclined plane, the acceleration of body is due to its impotent 9f weight. The component of weight is mg Sin theta along the plane.

Thus, the acceleration is g Sin theta

= 9.8 × Sin 30 = 4.9 m/s^2

Answer:

Explained

Explanation:

The most dense planet in  our solar system is Earth. Earth is most dense planet because

The least dense planet of the solar system is Saturn because Saturn is mostly made of gases and its size is smaller than the Jupiter. Jupiter has more gravity hence its density is higher to Saturn. Moreover, Uranus and Neptune are ice giants. Although they are also made of gases but due their distance from sun most of these gases have solidified. Hence making them more dense than Saturn and Jupiter.

Saturn is the least dense planet, while Earth is the densest planet in our solar system.

The least dense planet in our solar system is Saturn, while the densest planet is Earth. Saturn is a gas giant composed mostly of hydrogen and helium, which have lower densities compared to the rocks and metals found on rocky planets like Earth. Earth, on the other hand, has a higher density due to its solid composition and a core made of iron and nickel.

Answer:

(a). The magnitude of the gravitational force is [tex]7\times10^{-7}\ N[/tex]

(b). The magnitude of the gravitational force is [tex]1.35\times10^{-6}\ N[/tex]

Explanation:

Given that,

Mass of baby = 4.20 kg

Mass of father = 100 kg

Distance = 0.200 m

We need to calculate the gravitational force

Using gravitational formula

[tex]F = \dfrac{G\times m_{b}m_{f}}{r^2}[/tex]

Put the value in to the formula

[tex]F=\dfrac{6.67\times10^{-11}\times4.20\times100}{0.200^2}[/tex]

[tex]F=7\times10^{-7}\ N[/tex]

(b). We need to calculate the gravitational force

Using gravitational formula

[tex]F = \dfrac{G\times m_{b}m_{f}}{r^2}[/tex]

Put the value in to the formula

[tex]F=\dfrac{6.67\times10^{-11}\times4.20\times1.9\times10^{27}}{(6.29\times10^{11})^2}[/tex]

[tex]F=1.35\times10^{-6}\ N[/tex]

Hence, (a). The magnitude of the gravitational force is [tex]7\times10^{-7}\ N[/tex]

(b). The magnitude of the gravitational force is [tex]1.35\times10^{-6}\ N[/tex]

The magnitude of the gravitational force is 7×10⁻⁷ N and the magnitude of the gravitational force due to Jupiter is 1.35×10⁻6 N

Gravitational potential energy is the energy which a body posses because of its position.

The gravitational potential energy of a body is given as,

[tex]G=\dfrac{Fr^2}{Mm}[/tex]

Here, (m) is the mass of the body, (F) is the gravitational force and (r) is the height of the body.

The mass of baby is 4.20 kg and mass of father is 100 kg. The distance between them is 0.200 m. Put the values in the above formula as,

[tex]6.67\times10^{-11}=\dfrac{F(0.200)^2}{100(4.20)}\\F=7\times10^{-7}\rm \; N[/tex]

Put the values in the above formula again as,

[tex]6.67\times10^{-11}=\dfrac{F(6.29\times10^{11})^2}{(1.9\times10^{27})(4.20)}\\F=1.35\times10^{-6}\rm \; N[/tex]

Thus, the magnitude of the gravitational force is 7×10⁻⁷ N and the magnitude of the gravitational force due to Jupiter is 1.35×10⁻6 N.

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Answer:

COP = 4.25

Explanation:

It is given that,

Cooling temperature, [tex]T_c=-26^{\circ}C=273-26=247\ K[/tex]

Heating temperature, [tex]T_h=50^{\circ}C=323\ K[/tex]

We need to find the coefficient of performance. It is given by :

[tex]COP=\dfrac{T_h}{T_h-T_c}[/tex]

[tex]COP=\dfrac{323}{323-247}[/tex]

COP = 4.25

So, the best coefficient of performance of a refrigerator is 4.45 Hence, this is the required solution.

Answer:

(a) 110 m/s

(b) 42 m/s

Explanation:

mass, m = 400 kg, F = 3200 N, V1 = 2 m/s,

acceleration, a = Force / mass = 3200 / 400 = 8 m/s^2

(a) Use first equation of motion

v = V1 + a t

v = 2 + 8 x 1 = 10 m/s

(b) Again using first equation of motion

v = V1 + a t

v = 2 + 8 x 5 = 42 m/s

Thus, the velocity of plane after 1 second is 10 m/s and after 5 second the velocity is 42 m/s.

The 400 kg mine car moves a distance of 6 m after one second and 110 m after five seconds given an initial velocity of 2 m/s and a force of 3200 N.

The problem describes a physics scenario where a 400 kg mine car is pulled up an incline by a cable and motor. The force on the cable is given as F = 3200N and the initial velocity, V1, is given as 2 m/s. We can calculate the distance the car moves on the plane at different times using the physics equations of motion.

Let's use the equation of motion: s = ut + 1/2 at², where 's' is the distance moved, 'u' is the initial velocity, 'a' is the acceleration, and 't' is the time.

Given that the net force is equal to mass times acceleration (F = ma), we can calculate the acceleration, 'a', as F/m. So, a = 3200N/400kg = 8 m/s².

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Answer:

2.14×10⁸ m/s

Explanation:

L=Required Length=0.7 m

L₀=Initial length=1 m

c=speed of light=3×10⁸ m/s

[tex]From\ Lorentz\ Contraction\ relation\\L=L_0\sqrt{1-\frac {v^2}{c^2}}\\\Rightarrow 0.7=1\sqrt {1-\frac {v^2}{c^2}}\\\Rightarrow 0.49=1-\frac {v^2}{c^2}\\\Rightarrow 0.49-1=-\frac {v^2}{c^2}\\\Rightarrow -0.51=-\frac {v^2}{c^2}\\\Rightarrow 0.51=\frac {v^2}{c^2}\\\Rightarrow v^2=0.51\times c^2\\\Rightarrow v=\sqrt{0.51} \times c\\\Rightarrow v=0.71\times 3\times 10^8\\\Rightarrow v=2.14\times 10^8\ m/s\\\therefore velocity\ of\ stick\ should\ be\ 2.14\times 10^8\ m/s[/tex]

Answer:

0.86 m/s

Explanation:

A = cross-sectional area of the duct = 900 cm² = 900 x 10⁻⁴ m²

v = speed of air in the duct

t = time period of circulation = 40 min = 40 x 60 sec = 2400 sec

V = Volume of the air in the room = volume of room = 7 x 11 x 2.4 = 184.8 m³

Volume of air in the room is given as

V = A v t

inserting the values

184.8 = (900 x 10⁻⁴) (2400) v

v = 0.86 m/s

Answer:

Explanation:

(a) It is called conductors.

The conductors are the materials which can allow the current to pass through it.

(b) The example of conductor is copper, iron, etc.

The best conductor of electricity is silver.

(c) The example of semiconductor is silicon, germanium, etc.

The semiconductors are the materials which are insulators at normal temperature, but if the temperature increases, the conductivity of semiconductor increases.

(d) An example of insulator is wood.

Insulators are the materials which do not allow the current to pass through them.

Answer:

420 N

Explanation:

m = 0.42 Kg, u = 17 m/s, v = 0 m/s, t = 0.017 s

By first law of Newtons' laws of motion, the rate of change in momentum is force, F = m (v - u) / t

F = 0.42 x ( 0 - 17) / 0.017

F = - 420 N

Negative sign shows hat the force is resistive that means the ball finally comes to rest.

Answer:

[tex]\tau = 7.63 Nm[/tex]

Explanation:

As we know that moment of force is given as

[tex]\tau = \vec r \times \vec F[/tex]

now we have

[tex]\vec r = 1.2 m[/tex]

[tex]\vec F = 14 N[/tex]

now from above formula we have

[tex]\tau = r F sin\theta[/tex]

here we know that

[tex]\theta = 27 degree[/tex]

so we have

[tex]\tau = (1.2)(14) sin27[/tex]

[tex]\tau = 7.63 Nm[/tex]

The flux of the electric field through a Gaussian surface depends on the charge enclosed by the surface and the area of the surface. In this scenario, Gaussian surfaces A and B enclose the same positive charge +Q, but the area of surface A is three times larger than that of surface B. Therefore, the flux through surface A is three times larger than the flux through surface B.

The flux of the electric field through a Gaussian surface depends on the charge enclosed by the surface and the area of the surface. In this scenario, Gaussian surfaces A and B enclose the same positive charge +Q, but the area of surface A is three times larger than that of surface B. Since the flux of electric field is proportional to the area of the surface, the flux through surface A is three times larger than the flux through surface B. Therefore, the correct answer is E) three times larger than the flux of electric field through Gaussian surface B.

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Final answer:

The correct answer is D) equal to the flux of the electric field through Gaussian surface B. The electric flux depends on the charge enclosed by the Gaussian surface, not on its size or shape, according to Gauss's law.

Explanation:

The question asks whether Gaussian surfaces A and B, enclosing the same positive charge +Q, with surface A having three times the area of surface B, have different electric fluxes. According to Gauss's law, the electric flux (Φ) through a closed surface is proportional to the charge enclosed (Φ = Q/ε0). Since both surfaces enclose the same charge, the flux through each surface must be the same, regardless of their respective areas.

Thus, the correct answer is D) equal to the flux of the electric field through Gaussian surface B because the flux depends only on the amount of enclosed charge, not on the size or shape of the Gaussian surface.

Answer:

a)

14 rad/s

b)

11.2 m/s

Explanation:

a)

d = diameter of tire = 0.80 m

r = radius of tire = (0.5) d = (0.5) (0.80) = 0.40 m

v = speed of bicycle = 5.6 m/s

w = angular speed of the tire

Speed of cycle is given as

v = r w

5.6 = (0.40) w

w = 14 rad/s

b)

v' = speed of blue dot

Speed blue of dot is given as

v' = v + rw

v' = 5.6 + (0.40) (14)

v' = 11.2 m/s

The angular speed of the bicycle tire is 14 rad/s. The speed of the blue dot when it is 0.80 m above the road is the same as the bicycle's speed which is 5.6 m/s.

This is a problem which involves the calculation  of the angular speed and linear speed of an object in circular motion, here the circular motion being the rotation of a bicycle tire.

Given the linear speed of the bicycle and the radius of the tire, we would use the equation that links linear speed v and angular speed a, expressed as v = ra. We can find angular speed by rearranging this formula as a = v/r. Hence, we can calculate the angular speed of the tire as 5.6 m/s divided by the radius of the tire, which is half of the diameter, so 0.4m, which equals 14 rad/s.

Next, when a point on the tire (the blue dot) is 0.8 m above the ground it is at the top of tire's circular path, so its speed is equivalent to the linear speed of the bicycle, 5.6 m/s. This is because at this point the dot is not in contact with the ground hence it isn't stationary relative to the ground unlike the point at the bottom of the tire.

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Answer:

Explanation:

Joule is SI unit of work

Work = force x distance

Work = mass x acceleration x distance

Unit of mass is kg

Unit of acceleration is m/s^2

Unit of distance is m

So, unit of work = kg x m x m /s^2

So, Joule = kg m^2 / s^2

Answer:

Number of electrons, [tex]n=5.62\times 10^{21}[/tex]

Explanation:

It is given that,

Resistance, R = 4 ohms

Current, I = 3 A

Time, t = 5 min = 300 s

We need to find the number of electrons pass through the resistor during this time interval. Let the number of electron is n.

i.e. q = n e ...............(1)

And current, [tex]I=\dfrac{q}{t}[/tex]

[tex]I\times t=n\times e[/tex]

[tex]n=\dfrac{It}{e}[/tex]

e is the charge of an electron

[tex]n=\dfrac{3\ A\times 300\ s}{1.6\times 10^{-19}}[/tex]

[tex]n=5.62\times 10^{21}[/tex]

So, the number of electrons pass through the resistor is [tex]5.62\times 10^{21}[/tex]. Hence, this is the required solution.

Answer:

10.95 minute

Explanation:

V = 120 V, I = 2 A, m = 0.489 kg, c = 4186 J/kgC, T1 = 23 C , T2 = 100 C

Let time be the t.

Heat energy is equal to electrical energy

m x c (T2 - T1) = V x I x t

0.489 x 4186 x (100 - 23) = 120 x 2 x t

t = 656.73 second

t = 10.95 minute

Answer:

(a) θ = 55.85 degree

(b) 7.89 km

Explanation:

Using vector notations

A = 1.88 km south = 1.88 (- j) km = - 1.88 j km

B = 9.05 km 47 degree north of east

B = 9.05 ( Cos 47 i + Sin 47 j) km

B = (6.17 i + 6.62 j) km

Net displacement is

D = A + B

D = - 1.88 i + 6.17 i + 6.62 j = 4.29 i + 6.62 j

(a) Angle made with positive X axis

tanθ = 6.62 / 4.29 = 1.474

θ = 55.85 degree

(b) distance = [tex]Distance = \sqrt{(4.29)^{2} + (6.62)^{2}}[/tex]

distance = 7.89 km

To find the car's final location, add the displacements in the north and east directions. The car's final location is 6.4 km north and 6.0 km east relative to the origin.

To find the car's final location, we need to add the displacements in both the north and east directions.

First, let's calculate the north displacement by using the distance formula:

North Displacement = 9.05 km * sin(47°) = 6.4 km

Next, let's calculate the east displacement by using the distance formula:

East Displacement = 9.05 km * cos(47°) = 6.0 km

Therefore, the car's final location relative to the origin is 6.4 km north and 6.0 km east.

Answer:

118.06 days

Explanation:

d = distance of the center of moon from surface of planet = 2.315 x 10⁵ km = 2.315 x 10⁸ m

R = radius of the planet = 4.15 x 10³ km = 4.15 x 10⁵ m

r = center to center distance between the planet and moon = R + d

M = mass of the planet = 7.15 x 10²² kg

T = Time period of revolution around the planet

Using Kepler's third law

[tex]T^{2}=\frac{4\pi ^{2}r^{3}}{GM}[/tex]

[tex]T^{2}=\frac{4\pi ^{2}(R + d)^{3}}{GM}[/tex]

[tex]T^{2}=\frac{4(3.14)^{2}((4.15\times 10^{5}) + (2.315\times 10^{8}))^{3}}{(6.67\times 10^{-11})(7.15\times 10^{22})}[/tex]

T = 1.02 x 10⁷ sec

we know that , 1 day = 24 h = 24 x 3600 sec = 86400 sec

T = [tex](1.02 \times 10^{7} sec)\frac{1 day}{86400 sec}[/tex]

T = 118.06 days

Answer:

The voltage of the generator is 7.27 kV.

Explanation:

Given that,

Output of generator = 440 V

Current = 20 A

Resistance = 0.60 Ω

Power loss =0.010%

We need to calculate the total power of the generator

Using formula of power

[tex]P=VI[/tex]

Where, V = voltage

I = current

Put the value into the formula

[tex]P=440\times20[/tex]

[tex]P=8800\ W[/tex]

Th power lost on the transmission lines

[tex]P_{L}=0.010\% P[/tex]

[tex]P_{L} = 0.010\%\times8800[/tex]

[tex]P_{L}=0.88\ W[/tex]

The current passing through the transmission line

[tex]I'=\sqrt{\dfrac{P_{L}}{R}}[/tex]

[tex]I'=\sqrt{\dfrac{0.88}{0.60}}[/tex]

[tex]I'=1.211\ A[/tex]

We need to calculate the voltage of the generator

Using formula of voltage

[tex]V_{g}=\dfrac{P}{I'}[/tex]

Put the value into the formula

[tex]V_{g}=\dfrac{8800}{1.211}[/tex]

[tex]V_{g}=7.27\times10^{3}\ V[/tex]

[tex]V_{g}=7.27\ kV[/tex]

Hence, The voltage of the generator is 7.27 kV.

The voltage of the generator output needs to be stepped up with a transformer to limit power losses in transmission.

To limit power losses in transmission, the voltage of the generator output needs to be stepped up with a transformer. The formula to calculate power loss is P_loss = I^2 x R. Given that the power loss should not exceed 0.010% of the original power, we can calculate the maximum allowable power loss. By rearranging the formula, we can find the voltage required for the generator output with a transformer.

First, we calculate the original power using P = V x I. Then, we calculate the maximum allowable power loss as a percentage of the original power. Next, we substitute the given values into the formula and solve for the new voltage using the rearranged formula.

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Answer:

Part a)

x = 3.95 cm

Part b)

x = - 11.9 cm

Explanation:

Part a)

Since both charges are of same sign

so the position at which net force is zero between two charges is given as

[tex]\frac{kq_1}{r_1^2} = \frac{kq_2}{(15.8 - r)^2}[/tex]

here we know that

[tex]q_1 = Q[/tex]

[tex]q_2 = 9Q[/tex]

[tex]\frac{Q}{r^2} = \frac{9Q}{(15.8 - r)^2}[/tex]

square root both sides

[tex](15.8 - r) = 3r[/tex]

[tex]r = 3.95 cm[/tex]

Part b)

Since both charges are of opposite sign

so the position at which net force is zero will lie on the other side of smaller charges is given as

[tex]\frac{kq_1}{r_1^2} = \frac{kq_2}{(19.6 + r)^2}[/tex]

here we know that

[tex]q_1 = Q[/tex]

[tex]q_2 = -7Q[/tex]

[tex]\frac{Q}{r^2} = \frac{7Q}{(19.6 + r)^2}[/tex]

square root both sides

[tex](19.6 + r) = 2.64r[/tex]

[tex]r = 11.9 cm[/tex]

so on x axis it will be at x = - 11.9 cm

A. C = 75μF and B. V = 10V.

We have to use the equation k = C/C₀ and k = V₀/V which both are the dielectric constant.

A. The capacitance after the slab is inserted.

With C₀ = 15μF and k = 5.0. Clear k for the equation k = C/C₀:

C = k*C₀

C = (5.0)(15x10⁻⁶F) = 0.000075F

C = 75μF

B. The voltage across the capacitor's plates after the slab is inserted.

With V₀ = 50V and k = 5.0. Clear V from the equation k = V₀/V:

V = V₀/k

V = 50V/5.0

V = 10V

Dielectric constant of the capacitor is the ratio between capacitance of capacitor before and after slab inserted.

The capacitance of capacitor is the ratio of the electric charge stored inside the capacitance to the potential difference.

The capacitance of the capacitor is 15-uF and it is connected to  50-V battery. The value of the dielectric constant is 5.0.

The dielectric constant of the slab is the ratio of capacitance of the capacitor after the slab and the capacitance of the before the slab inserted.

As the value of the dielectric constant is 5.0. Thus the capacitance of the capacitor after the slab is inserted given as,

[tex]0.5=\dfrac{C}{15\times10^{-6}}\\C=75\rm \mu F[/tex]

Thus, the capacitance of the capacitor after the slab is inserted is 75-uF.

The dielectric constant of the slab is the ratio of voltage across the capacitor's of the capacitor before the slab inserted and the voltage across the capacitor after the slab inserted.

As the value of the dielectric constant is 5.0. Thus the voltage across the capacitor's plates after the slab is inserted can be given as,

[tex]V=\dfrac{50}{5.0}\\C=10\rm V[/tex]

Thus, the voltage across the capacitor's plates after the slab is inserted is 10 volts.

Dielectric constant of the capacitor is the ratio between capacitance of capacitor before and after slab inserted.

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Answer:

3.958 × 10²⁶ watt

Explanation:

Given:

Distance between earth and sun, [tex]R_e[/tex] = 150 ×10⁸ m

Total solar irradiance at earth orbit  = 1400 watt/m²

Now,

Area irradiated ([tex]A_e[/tex]) will be = [tex]4\pi R_e^2[/tex]

⇒ [tex]A_e[/tex] =  [tex]4\pi \times (150\times 10^{8})^2[/tex]

⇒ [tex]A_e[/tex] =  [tex]2.827\times 10^{23}m^2[/tex]

Therefore, the flux = Total solar irradiance at earth orbit × [tex]A_e[/tex]

the flux =   [tex]1400watt/m^2\times 2.827\times 10^{23}m^2[/tex]

⇒the flux = 3.958 × 10²⁶ watt

Answer:

a)

51764.7 ohm

b)

28.5 A

Explanation:

a)

t = time taken to charge to 90.0% of its final value = 16.2 s

T = time constant

C = capacitance = 136 x 10⁻⁶ F

R = resistance of the resistor

Q₀ = Maximum charge stored

Q = Charge after time "t" = 0.90 Q₀

Using the equation

[tex]Q = Q_{o}(1 - e^{\frac{-t}{T}})[/tex]

[tex]0.90 Q_{o} = Q_{o}(1 - e^{\frac{-16.2}{T}})[/tex]

[tex]0.90 = (1 - e^{\frac{-16.2}{T}})[/tex]

T = 7.04 s

Time constant is given as

T = RC

7.04 = R (136 x 10⁻⁶)

R = 51764.7 ohm

b)

V = Potential difference of voltage source = 265 volts

q = Amount of charge discharged = 0.90 Q₀ = 0.90 (CV) = (0.90) (136 x 10⁻⁶) (265) = 0.032436 C

t = time taken to discharge = 1.14 x 10⁻³ s

Current is given as

[tex]i = \frac{q}{t}[/tex]

i = 28.5 A

For the RC charging circuit of a camera flash unit has the value of resistance and current as,

RC circuit is the circuit in which the voltage or current passes through the resistor and capacitor consist by the circuit.

The formula for the RC circuit can be given as,

[tex]Q=Q_o\left(1-e^{\dfrac{-1}{RC}}\right )[/tex]

Here, [tex](Q_o)[/tex] is the initial charge and (R) is the resistance.

The  capacitor charges to 90.0% of its final value. Thus, the charge for the first case can be given as,

[tex]Q=0.90 Q_o[/tex]

The capacitance of 136 μF. Put the values in the above formula as,

[tex]0.90Q_o=Q_o\left(1-e^{\dfrac{-1}{R\times136\times10^{-6}}}\right )\\R=51765\rm ohm[/tex]

Thus, the value of the resistance R is 51765 ohms.

The voltage source of the RC circuit is 265 V  and the capacitor discharges 90.0% of its full charge in 1.14 ms. Thus, the amount of charge discharged is,

[tex]Q=0.9\times136\times10^{-6}\times265\\Q=3.2436\times10^{-2}\rm C[/tex]

The value of current is the ratio of discharge per second time (1.14 ms.). Therefore,

[tex]I=\dfrac{3.2436\times10^{-2}}{1.14\times10^{-3}}\\I=28.5\rm A[/tex]

Thus, for the RC charging circuit of a camera flash unit has the value of resistance and current as,

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Answer:

Magnetic force, F =  0.18 N

Explanation:

It is given that,

Current flowing in the wire, I = 0.2 A

Length of the wire, L = 3 m

Magnetic field, B = 0.3 T

It is placed perpendicular to the magnetic field. We need to find the magnitude of force on the wire. It is given by :

[tex]F=ILB\ sin\theta[/tex]

[tex]F=0.2\ A\times 3\ m\times 0.3\ T\ sin(90)[/tex]

F = 0.18 N

So, the magnitude of force on the wire is 0.18 N. Hence, this is the required solution.

Answer:

Impulse, J = 2.1 kg-m/s

Explanation:

Given that,

Mass of baseball, m = 0.14 kg

It was approaching him at 50.0 m/s and, as a result, the ball leaves the bat at 35.0 m/s in the direction of the pitcher. We need to find the magnitude of Impulse delivered to the baseball.

The change in momentum is equal to the Impulse imparted to the ball i.e.

[tex]J=m(v-u)[/tex]

[tex]J=0.14(-35-50)[/tex]

J = -2.1 kg-m/s

So, the Impulse delivered to the baseball is 2.1 kg-m/s

Answer:

a)  The elevator was  31.64 m high up when the bolt came loose.

b)  Speed of the bolt when it hits the bottom of the shaft = 25.41 m/s

Explanation:

a) Considering motion of bolt:-

Initial velocity, u =  5 m/s

Acceleration , a = -9.81 m/s²

Time = 3.1 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= 5 x 3.1 - 0.5 x 9.81 x 3.1²

    s = 0 x t + 0.5 x 9.81 x t²

    s = -31.64 m

The elevator was  31.64 m high up when the bolt came loose.

b) We have equation of motion v = u + at

  Initial velocity, u =  5 m/s

 Acceleration , a = -9.81 m/s²

 Time = 3.1 s  

Substituting

  v = u + at

  v  = 5 - 9.81 x 3.1 = -25.41 m/s

Speed of the bolt when it hits the bottom of the shaft = 25.41 m/s

Answer:

(iii) 6

Explanation:

Part a)

Since it is given that both ends are fixed and it is vibrating in 5th harmonic

So here it will have 5 number of loops in it

so we can draw it in following way

each loop will have 1 antinode and two nodes which means number of nodes is one more than number of anti-nodes

So there are 5 loops which means it will have 5 antinodes

and hence there will be 6 nodes in it

so correct answer will be

(iii) 6

Answer:

a) Rotational Inertia = Decreases

Because the distance from the axis is decreasing

b) Angular momentum = Remains the same

because there is no external torque

c) Angular speed = increases

because here rotational inertia decreases due to which angular speed will increase

Explanation:

Here the Beetle is initially moving along the rim of the disc

So here during the motion of beetle there is no external force on the system of beetle and the disc.

So here we can also say that there is no torque acting on the system

so angular momentum of the disc + beetle system will remain conserved

so here we have

[tex]I_1\omega_1 = I_2 \omega_2[/tex]

here as the beetle crawls towards the centre of the disk then

a) Rotational Inertia = Decreases

Because the distance from the axis is decreasing

b) Angular momentum = Remains the same

because there is no external torque

c) Angular speed = increases

because here rotational inertia decreases due to which angular speed will increase

Answer:  [tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^{2+}(aq)+2H_2O(l)[/tex]

Explanation:

Reduction is a process where electrons are gained and acidic solution means presence of [tex]H^+[/tex] ions.

Reduction of [tex]MnO_2[/tex] to [tex]Mn^{2+}[/tex]

Mn is in +4 oxidation state in [tex]MnO_2[/tex] which goes to +2 state in [tex]Mn^{2+}[/tex] by gain of 2 electrons.

[tex]MnO_2(s)\rightarrow Mn^{2+}(aq)[/tex]

In order to balance oxygen atoms:

[tex]MnO_2(s)\rightarrow Mn^{2+}(aq)+2H_2O[/tex]

In order to balance hydrogen atoms:

[tex]MnO_2(s)+4H^+(aq)\rightarrow Mn^{2+}(aq)+2H_2O(l)[/tex]

In order to balance charges:

[tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^{2+}(aq)+2H_2O(l[/tex]

Thus the net balanced half reaction for the reduction of solid manganese dioxide to manganese ion in acidic aqueous solution is:

[tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^{2+}(aq)+2H_2O(l)[/tex]

Final answer:

To balance the reduction of solid manganese dioxide (MnO2) to manganese ion (Mn2+) in acidic aqueous solution, the balanced half-reaction is MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l).

Explanation:

To balance the reduction of solid manganese dioxide (MnO2) to manganese ion (Mn2+) in acidic aqueous solution, we can follow these steps:

1. Write the half-reaction for reduction, adjusting the physical states of the reactants and products:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

2. Balance the number of Mn and O atoms on each side of the equation:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

3. Balance the H atoms by adding H+ ions:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

4. Balance the charges by adding electrons:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

Therefore, the balanced half-reaction for the reduction of solid manganese dioxide to manganese ion in acidic aqueous solution is:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)