A real object with height of 3.20 cm is placed to the left of a converging lens whose focal length is 90cm. The image is on the right of the lens and 4.50cm tall and inverted. Where is the object? Where is the image? Is the image real or virtual?

The area of a square is given by:

A = s²

A is the square's area

s is the length of one of the square's sides

Let us take the derivative of both sides of the equation with respect to time t in order to determine a formula for finding the rate of change of the square's area over time:

d[A]/dt = d[s²]/dt

The chain rule says to take the derivative of s² with respect to s then multiply the result by ds/dt

dA/dt = 2s(ds/dt)

A) Given values:

s = 14m

ds/dt = 3m/s

Plug in these values and solve for dA/dt:

dA/dt = 2(14)(3)

dA/dt = 84m²/s

B) Given values:

s = 25m

ds/dt = 3m/s

Plug in these values and solve for dA/dt:

dA/dt = 2(25)(3)

dA/dt = 150m²/s

When the side of the square is 14 m, the rate at which the area is changing is 84 m²/s.

When the side of the square is 25 m, the rate at which the area is changing is 150 m²/s.

The given parameters;

The area of the square is calculated as;

A = L²

The change in the area is calculated as;

[tex]\frac{dA}{dt} = 2l\frac{dl}{dt}[/tex]

When the side of the square is 14 m, the rate at which the area is changing is calculated as;

[tex]\frac{dA}{dt} = 2l \frac{dl}{dt} \\\\\frac{dA}{dt} = 2 \times l \times \frac{dl}{dt}\\\\\frac{dA}{dt} = 2 \times 14 \times 3\\\\\frac{dA}{dt} = 84 \ m^2/s[/tex]

When the side of the square is 25 m, the rate at which the area is changing is calculated as;

[tex]\frac{dA}{dt} = 2l \frac{dl}{dt} \\\\\frac{dA}{dt} = 2 \times l \times \frac{dl}{dt}\\\\\frac{dA}{dt} = 2 \times 25 \times 3\\\\\frac{dA}{dt} = 150 \ m^2/s[/tex]

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Answer:

Kinetic energy, [tex]E_k=61.74\ J[/tex]

Explanation:

It is given that,

Mass of ball, m = 0.9 kg

It is dropped form a height of 7 m, h = 7 m

When the ball is at certain height it will have only potential energy. As it hits the ground, its potential energy gets converted to kinetic energy as per the law of conservation of energy.

So, [tex]E_k=PE=mgh[/tex]

[tex]E_k=0.9\ kg\times 9.8\ m/s^2\times 7\ m[/tex]

[tex]E_k=61.74\ J[/tex]

So, the kinetic energy when it hits the ground is 61.74 Joules.    

The kinetic energy of a 0.9 kg ball dropped from a height of 7 m is calculated using the conservation of mechanical energy, resulting in 61.74 Joules when it hits the ground.

The kinetic energy of a 0.9 kg ball when it hits the ground after being dropped from a height of 7 m. The principle used to determine this is the conservation of mechanical energy, which states that the sum of the potential and kinetic energies in a system remains constant if only conservative forces are doing work.

When the ball is at the height of 7 m, it posses gravitational potential energy (GPE) which can be calculated using the formula GPE = mgh, where m is the mass of the ball, g is the acceleration due to gravity (9.8 m/s²), and h is the height from which the ball is dropped.

Since the ball is dropped from rest, its initial kinetic energy is 0. When the ball reaches the ground, its potential energy is converted into kinetic energy.

Therefore, the kinetic energy (KE) of the ball when it hits the ground can be calculated using the potential energy formula, assuming no energy is lost:
KE = mgh = 0.9 kg x 9.8 m/s² x 7 m = 61.74 Joules.

This means that the ball will have a kinetic energy of 61.74 Joules when it impacts the ground.

Explanation:

33×10^-6 ×74 ×(0.86 - 0.41)

Final answer:

An additional charge of 2100 μC flows from the battery into the capacitor when it is immersed in transformer oil with a dielectric constant of 4.5.

Explanation:

To determine how much additional charge flows from the battery when a 6.0-μF air capacitor is immersed in transformer oil with a dielectric constant of 4.5, we need to examine the effect of the dielectric on the capacitor's capacitance.

Initially, the capacitance of the air capacitor Cair is 6.0 μF. The charge Qinitial on the capacitor when connected to a 100-V battery is given by:

Qinitial = Cair × Vbattery

Qinitial = 6.0 μF × 100 V

Qinitial = 600 μC

When the capacitor is immersed in oil, the capacitance increases due to the dielectric constant (κ) of the oil:

Coil = κ × Cair

Coil = 4.5 × 6.0 μF = 27.0 μF

Since the battery remains connected, the voltage across the capacitor stays at 100 V, so the new charge Qfinal becomes:

Qfinal = Coil × Vbattery

Qfinal = 27.0 μF × 100 V

Qfinal = 2700 μC

The additional charge ΔQ that flows from the battery is the difference between Qfinal and Qinitial:

ΔQ = Qfinal - Qinitial

ΔQ = 2700 μC - 600 μC

ΔQ = 2100 μC

Therefore, an additional charge of 2100 μC flows from the battery into the capacitor.

Answer:

Maximum speed = 19.81 m/s

Explanation:

Maximum speed can the car have without flying off the road at the top of the hill.

For this condition to occur we have

         Centripetal force ≥ Weight of car.

          [tex]\frac{mv^2}{r}\geq mg[/tex]

For maximum speed without flying we have

        [tex]\frac{mv^2}{r}=mg\\\\\frac{v^2}{r}=g\\\\v=\sqrt{rg}=\sqrt{40\times 9.81}=19.81m/s[/tex]

Maximum speed = 19.81 m/s

The maximum speed of the car on top of hill is 19.8 m/s.

The given parameters;

radius of the hill, r = 40 m

The maximum speed of the car on top of hill is calculated as follows;

the centripetal force must be equal or greater than weight of the car.

[tex]F_c = mg\\\\\frac{mv^2}{r} = mg\\\\\frac{v^2}{r} = g\\\\v^2 = rg\\\\v = \sqrt{rg} \\\\v = \sqrt{40 \times 9.8} \\\\v = 19.80 \ m/s[/tex]

Thus, the maximum speed of the car on top of hill is 19.8 m/s.

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Answer:

[tex]\omega = 2.56 rad/s[/tex]

Explanation:

As the cylinder rotates the centripetal force on all the passengers is due to normal force due to the wall

So here we can say

[tex]N = m\omega^2 R[/tex]

now when floor is removed all the passengers are safe because here friction force on the passenger is counter balanced by the weight of the passengers

so we can say

[tex]F_f = mg[/tex]

[tex]\mu_s F_n = mg[/tex]

[tex]\mu_s (m\omega^2 R) = mg[/tex]

[tex]\mu_s \omega^2 R = g[/tex]

[tex]\omega = \sqrt{\frac{g}{\mu_s R}}[/tex]

for minimum rotational speed we have

[tex]\omega = \sqrt{\frac{9.8}{0.60(2.5)}[/tex]

[tex]\omega = 2.56 rad/s[/tex]

The minimum rotational frequency for the ride to be safe is approximately 8.28 rpm.

To determine the minimum rotational frequency for which the ride is safe, we need to consider the static coefficient of friction between clothing and steel. Since the passengers are standing inside a hollow cylinder, they will experience a centrifugal force pushing them against the wall of the cylinder. To prevent sliding, the static friction force needs to be greater than or equal to the gravitational force pulling them downward. The formula to calculate the static friction force is Fs = μs * N, where μs is the coefficient of static friction and N is the normal force.

For clothing against steel, the coefficient of static friction ranges from 0.60 to 1.0. Assuming the worst-case scenario with μs = 0.60, we can calculate the minimum rotational frequency:

Centrifugal force = m * g = m * ω^2 * R, where m is the mass of the passengers, g is the acceleration due to gravity, ω is the angular velocity in radians per second, and R is the radius of the cylinder.
Static friction force = μs * m * g
Equating these two forces, we get μs * m * g = m * ω^2 * R
Simplifying the equation, we find ω = sqrt(μs * g / R)

Converting the angular velocity to revolutions per minute (rpm), we have rpm = 60 * ω / (2 * π)
Substituting the values, the minimum rotational frequency for the ride to be safe is approximately 8.28 rpm.

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Answer:

10 seconds

Explanation:

x = x₀ + v₀ t + ½ at²

250 = 0 + (0) t + ½ (5) t²

250 = 2.5 t²

t² = 100

t = 10

It takes 10 seconds to land from a height of 250 ft.

Answer:

10 seconds

Explanation:

It takes 10 seconds for an object to fall from an elevation of 250 ft.

250 = 0 + (0) t + ½ (5) t²

Final answer:

Using the Doppler Effect formulae for electromagnetic waves, the difference in frequency (Dopler shift) experienced by radar signals upon hitting and returning from a moving vehicle allows a radar gun to calculate the vehicle's speed. The principle involves the use of shift in frequency and the speed of light to measure the speed at which the vehicle is moving.

Explanation:

The question involves the Doppler Effect in physics, specifically its application in a radar speed trap. To find the frequency shift, we must use the Doppler Effect formulae for electromagnetic waves. However, it seems there might have been a mix-up with the frequencies provided in several example problems. Since those seem to be examples rather than the actual frequencies we are working with, let's focus on finding the principle behind calculating the speed of the vehicle based on a known frequency shift of a radar emission and its return signal.

The frequency shift (Δf) in the Doppler Effect for electromagnetic waves such as radar can be calculated by the formula: Δf = (2 * f * v) / c, where f is the original frequency emitted by the radar gun, v is the speed of the vehicle, and c is the speed of light. The factor of 2 is because the radar signal experiences a frequency shift once when it hits the moving vehicle and another shift when the echo returns. The radar gun's internal processors calculate the difference in frequency (the Dopler shift) to find the speed of the vehicle. For accurate measurement, the radar unit must be able to discern even small frequency shifts to effectively differentiate speeds with fine resolution.

Answer:

[tex]V_{A}= 11.88 m/s[/tex]

Explanation:

given data:

water level at point A = 8 m

diameter of orifice = 0.5 cm

velocity at point A  =  0

h1 =0.8 m

h2 = 8 - h1 = 8 - 0.8 = 7.2 m

Applying Bernoulli  theorem between point A and B

[tex]P_{o}+\rho _{water}gh_{2}+\frac{1}{2}\rho v_{B}^{2}+\rho _{water}gh_{1}=P_{o} +\frac{1}{2}\rho v_{A}^{2}+\rho _{water}gh_{1}[/tex]

[tex]V_{A}=\sqrt{2gh_{2}}[/tex]

[tex]V_{A}=\sqrt{2*9.81*7.2}[/tex]

[tex]V_{A}= 11.88 m/s[/tex]

Answer:

Acceleration of the ship, [tex]a=2.14\times 10^{-7}\ m/s^2[/tex]

Explanation:

It is given that,

Mass of both ships, [tex]m=39000\ metric\ tons=39\times 10^6\ kg[/tex]

Distance between two ships, d = 110 m

The gravitational force between two ships is given by :

[tex]F=G\dfrac{m^2}{d^2}[/tex]

[tex]F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}[/tex]

F = 8.38 N

Let a is the acceleration. Now, using second law of motion as :

[tex]a=\dfrac{F}{m}[/tex]

[tex]a=\dfrac{8.38\ N}{39\times 10^6\ kg}[/tex]

[tex]a=2.14\times 10^{-7}\ m/s^2[/tex]

So, the acceleration of either ship due to the gravitational attraction of the other is [tex]2.14\times 10^{-7}\ m/s^2[/tex]. Hence, this is the required solution.

The gravitational attraction between the two ships can be determined by using Newton's law of gravitation. Afterwards, the acceleration of one ship due to this force is found by applying Newton's second law of motion.

In this scenario, we're dealing with the concept of gravitational attraction between two massive bodies. We can find the gravitational force between the ships using Newton's law of gravitation:

F = G * (m1 * m2) / r²,

where F is the gravitational force, G is the gravitational constant (6.674 x 10^-11 N(m²/kg²)), m1 and m2 are the two masses, and r is the distance between them.

So, let's plug in the values. Each ship mass (m1 = m2) is 39,000 metric tons, equivalent to 3.9 x 10^7 kg. The distance between them (r) is 110 m, which we need to square. Calculating for F, we get an incredibly small value, which evidences the weak nature of gravitational force.

To find the acceleration of either ship (let's say m1), due to the gravitational force of the other, we will apply Newton's second law (F = m*a). Here, F is the gravitational force we found, m is the mass of the ship, and a is the acceleration we're looking for:

a = F / m,

Plugging the values found into this equation will result in the desired acceleration in m/s².

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Answer:

[tex]\Delta U = 0.2072 J[/tex]

Explanation:

Potential difference between two points in constant electric field is given by the formula

[tex]\Delta V = E.\Delta x[/tex]

here we know that

[tex]E = 370 N/C[/tex]

also we know that

[tex]\Delta x = 2.1 - 1.9 = 0.2 m[/tex]

now we have

[tex]\Delta V = 370 (0.2) = 74 V[/tex]

now change in potential energy is given as

[tex]\Delta U = Q\Delta V[/tex]

[tex]\Delta U = (2.80 \times 10^{-3})(74)[/tex]

[tex]\Delta U = 0.2072 J[/tex]

Answer:

D = 230.2 Km

Explanation:

let distance between seismograph and focus of quake is D

From time distance formula we  can calculate the time taken by the S wave

[tex]T_1 =\frac{D}{4.5}[/tex]

From time distance formula we  can calculate the time taken by the P wave

[tex]T_2 =\frac{D}{6.90}[/tex]

It is given in equation both waves are seperated from each other by 17.8 sec

so we have

[tex]T_1 - T_2 = 17.8[/tex]sec

Putting both time value to get distance value

[tex]\frac{D}{4.5} - \frac{D}{6.90} = 17.8[/tex]

D = 230.2 Km

Answer:

Heat generated by the current = 1547.89 J

Explanation:

We have equation for heat energy H = mCΔT

Mass of copper = 0.221 kg

Specific heat of copper = 0.093 kcal/kgC° = 389.112 J/kgC°

ΔT = 38 - 20 = 18°C

Substituting in H = mCΔT

           H = 0.221 x 389.112 x 18 = 1547.89 J

Heat generated by the current = 1547.89 J

The heat generated by the electric current that heated a 221 g copper wire from 20.0 °C to 38.0 °C is calculated to be 1.542 kJ, using the specific heat capacity of copper and the formula Q = mcΔT.

An electric current heats a 221 g (0.221 kg) copper wire from 20.0 °C to 38.0 °C. To calculate the heat generated by the current, we use the formula for heat energy, Q = mcΔT, where m is the mass in kg, c is the specific heat capacity in kcal/kg°C, and ΔT is the change in temperature in °C.

Given:
m = 0.221 kg
c = 0.093 kcal/kg°C
ΔT = (38.0 - 20.0) °C = 18.0 °C

Substituting the values into the formula:
Q = 0.221 kg * 0.093 kcal/kg°C * 18.0 °C = 0.368658 kcal

To convert kcal to joules (since 1 kcal = 4.184 kJ),
Q = 0.368658 kcal * 4.184 kJ/kcal = 1.542 kJ

Therefore, the heat generated by the electric current is 1.542 kJ.

Answer:

The magnification is -6.05.

Explanation:

Given that,

Focal length = 34 cm

Distance of the image =2.4 m = 240 cm

We need to calculate the distance of the object

[tex]\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}[/tex]

Where, u = distance of the object

v = distance of the image

f = focal length

Put the value into the formula

[tex]\dfrac{1}{u}=\dfrac{1}{34}-\dfrac{1}{240}[/tex]

[tex]\dfrac{1}{u}=\dfrac{103}{4080}[/tex]

[tex]u =\dfrac{4080}{103}[/tex]

The magnification is

[tex]m = \dfrac{-v}{u}[/tex]

[tex]m=\dfrac{-240\times103}{4080}[/tex]

[tex]m = -6.05[/tex]

Hence, The magnification is -6.05.

Answer:

209.3 seconds

Explanation:

P = 300 W, m =  0.250 kg, T1 = 10 degree C, T2 = 70 degree C

c = 4186 J / kg C

Heat given to water = mass x specific heat of water x rise in temperature

H = 0.250 x 4186 x (70 - 10)

H = 62790 J

Power = Heat / Time

Time, t = heat / Power

t = 62790 / 300 = 209.3 seconds

It took approximately [tex]\( 209.3 \)[/tex] seconds to heat the water from [tex]\( 10.0^\circ \text{C} \)[/tex] to [tex]\( 70.0^\circ \text{C} \)[/tex].

To determine how many seconds it took to heat the water using the immersion heater, we need to calculate the amount of energy required and then use the power of the heater to find the time.

First, calculate the change in temperature of the water:

[tex]\[ \Delta T = 70.0^\circ \text{C} - 10.0^\circ \text{C} = 60.0^\circ \text{C} \][/tex]

Next, calculate the energy [tex]\( Q \)[/tex] required to heat the water using the specific heat capacity of water [tex]\( C = 4186 \text{ J/kg}^\circ \text{C} \)[/tex]:

[tex]\[ Q = mc\Delta T \][/tex]

where [tex]\( m \)[/tex] is the mass of water and [tex]\( c \)[/tex] is the specific heat capacity of water.

Given:

[tex]\[ m = 0.250 \text{ kg} \][/tex]

[tex]\[ c = 4186 \text{ J/kg}^\circ \text{C} \][/tex]

[tex]\[ \Delta T = 60.0^\circ \text{C} \][/tex]

[tex]\[ Q = 0.250 \times 4186 \times 60.0 \][/tex]

[tex]\[ Q = 62790 \text{ Joules} \][/tex]

Now, calculate the time t required using the power P  of the heater:

[tex]\[ P = 300.0 \text{ W} \][/tex]

The time [tex]\( t \)[/tex] is given by:

[tex]\[ t = \frac{Q}{P} \][/tex]

[tex]\[ t = \frac{62790}{300.0} \][/tex]

[tex]\[ t = 209.3 \text{ seconds} \][/tex]

Final answer:

The peak emf generated by a 1000-turn, 42.0 cm diameter coil initially perpendicular to the Earth's 5.00 × 10^-5 T magnetic field and rotated to be parallel in 12.0 ms is approximately 1.1 V.

Explanation:

Calculating the Peak EMF in a Rotating Coil

To calculate the peak emf generated by a rotating coil in a magnetic field, we can use Faraday's Law of electromagnetic induction. The formula derived from Faraday's Law for a coil with multiple turns is:

emf = -N * (change in magnetic flux)/change in time

The magnetic flux (Φ) is given by the product of the magnetic field (B), the area (A) of the coil, and the cosine of the angle (θ) between the magnetic field and the normal to the surface of the coil:

Φ = B * A * cos(θ)

The question states that the coil with 1000 turns and a 42.0 cm diameter is initially perpendicular to the Earth's magnetic field of 5.00 × 10-5 T, and then rotated to become parallel. This change goes from cos(90°), which is 0, to cos(0°), which is 1, over a time interval of 12.0 ms.

The area A of the coil is π * (radius)2, where radius is half the diameter. The radius is 42.0 cm / 2 = 21.0 cm = 0.21 m. Thus:

A = π * (0.21 m)2

Now, we plug the values into the equation to find the peak emf:

Peak emf = -(1000) * (5.00 × 10-5 T * π * (0.21 m)2 - 0) / (12.0 × 10-3 s)

After calculation:

Peak emf ≈ 1.1 V

This is the maximum emf induced in the coil during its rotation.

The peak emf generated by rotating the given coil in the Earth's magnetic field is approximately 3.63 V.

To find the peak emf generated by rotating a coil in a magnetic field, we use Faraday's Law of Induction. The peak emf (ε) can be calculated using the formula:

ε = NABω

where N is the number of turns, A is the area of the coil, B is the magnetic field strength, and ω is the angular velocity in radians per second. First, we need to find the area (A) of the coil:

A = πr²

Given the diameter of the coil is 42.0 cm, the radius (r) is 21.0 cm or 0.21 m. So,

A = π(0.21)² = 0.1385 m²

Next, we determine the angular velocity (ω) given one full rotation in 12.0 ms:

ω = 2π / T

where T is the period (12.0 ms or 0.012 s),

ω = 2π / 0.012 = 523.6 rad/s

Now, substituting the values into the emf formula:

ε = 1000 × 0.1385 m² × 5.00 × 10⁻⁵ T × 523.6 rad/s = 3.629 V

Thus, the peak emf generated is approximately 3.63 V.

Answer:

424088766.068 m

Explanation:

Radius of the circular orbit that the satellite is 2.6 Earth radii (r) = 2.6 R

R = Radius of earth = 6371000 m (mean radius)

In order to find the distance that the satellite travels in 5.89 hours to complete one complete revolution is the circumference of the circular orbit

Circumference of a circle = 2×π×r

⇒Distance travelled in 5.89 hours = 2×π×2.6 R

⇒Distance travelled in 5.89 hours = 2×π×2.6×6371000

⇒Distance travelled in 5.89 hours = 104078451.3393m

Distance travelled in 1 hour = 104078451.3393/5.89 = 17670365.252 m

∴ Distance travelled in 24 hours = 17670365.252×24 = 424088766.068 m

Final answer:

The satellite orbits Earth at approximately 2.60 Earth radii and travels a total distance of about 4.23×10⁸ meters in one day.

Explanation:

The question pertains to calculating the total distance traveled by a satellite orbiting Earth at a given distance over the duration of one day. To solve this, we need to determine the circumference of the satellite's circular orbit and then calculate how many orbits it completes in a day.

To determine the satellite's orbit circumference, we use the formula for the circumference of a circle, C = 2πr, where r represents the orbit radius. Given that the satellite orbits at a distance of about 2.60 Earth radii, we can express this distance in meters by multiplying the Earth's radius, 6.37×10⁶ meters, by 2.60, yielding a radius of approximately 1.6562×10⁷ meters. Therefore, the satellite's orbit circumference is approximately 1.04×10⁸ meters.

Since the satellite completes one orbit in 5.89 hours, we can now calculate how many orbits it completes in a day (24 hours) by dividing 24 by 5.89, which is approximately 4.07 orbits per day. The total distance traveled in one day is the circumference multiplied by the number of orbits, resulting in a distance of approximately 4.23×10⁸ meters per day.

Explanation:

It is given that,

Number of turns, N = 200

Area of cross section, A = 8.5 cm²

Magnetic field is directed out of the paper and is, B = 0.06 T

The magnetic field is  out of the paper decreases to 0.02 T in 12 milliseconds. We need to find the direction of current induced. The induced emf is given by :

[tex]\epsilon=-N\dfrac{d\phi}{dt}[/tex]

Since, [tex]\epsilon=IR[/tex]

I is the induced current

[tex]I=-\dfrac{N}{R}\dfrac{d\phi}{dt}[/tex]

According to Lenz's law, the direction of induced current is such that it always opposes the change in current that causes it.

Here, the field is directed out of the plane of the paper, this gives the induced current in counterclockwise direction.

Answer:

[tex]w_{damped}= 4.76[/tex]  s^-1

Explanation:

The mathematical relationship is

[tex]w_{damped}=w_{undamped} *\sqrt{1-(\frac{c}{2\sqrt{km}})^{2}}[/tex]

where:

c is the damper constant

k is the spring constant

m is the mass

ω_undamped is the natural frequency

ω_damped is the damped frequency

[tex]w_{undamped} =\sqrt{\frac{k}{m}}=4.79[/tex] s^-1

[tex]w_{damped}= 4.79 *\sqrt{1-(\frac{2}{2\sqrt{295*13}})^{2}}[/tex]

[tex]w_{damped}= 4.76[/tex]  s^-1

Answer:4.08 N

Explanation:

Given data

superball dropped from a height of 10  cm

Mass of ball[tex]\left ( m\right )[/tex]=0.05kg

time of contact[tex]\left ( t\right )[/tex]=34.3[tex]\times 10^{-3}[/tex] s

Now we know impulse =[tex]Force\times time\ of\ contact[/tex]=Change in momentum

[tex]F_{average}\times t[/tex]=[tex]m\left ( v-(-v)\right )[/tex]

and velocity at the bottom is given by

v=[tex]\sqrt{2gh}[/tex]

[tex]F_{average}\times 34.3\times 10^{-3}[/tex]=[tex]0.05\left ( 1.4-(-1.4)\right )[/tex]

[tex]F_{average}[/tex]=4.081N

To find the average force exerted on the superball by the table, we calculate its change in momentum during the bounce and divide by the contact time. The result is an approximate average force of 4.08 N upward.

The question involves a superball and a clay ball, both with the same mass of 0.05 kg, dropped from a height of 10 cm. The superball bounces back to its original height, while the clay ball sticks to the table. We need to calculate the average force exerted on the superball by the table.

Step-by-Step Solution

Calculate the velocity of the superball just before hitting the table: Using the equation for free fall, v = √(2gh), where g is 9.8 m/s² (acceleration due to gravity) and h is 0.10 m.

v = √(2 * 9.8 * 0.10)

v = √(1.96)

v ≈ 1.40 m/s

Determine the change in momentum: Before impact, the momentum is [tex]P_{before[/tex] = m * v. Since the superball bounces back with the same speed, the momentum after impact is [tex]P_{after[/tex] = -m * v because the direction changes.

[tex]P_{before[/tex] = 0.05 kg * 1.40 m/s = 0.07 kg·m/s

[tex]P_{after[/tex] = 0.05 kg * (-1.40 m/s) = -0.07 kg·m/s

Calculate the impulse: Impulse is the change in momentum, so Impulse = [tex]P_{after[/tex] - [tex]P_{before[/tex].

Impulse = -0.07 kg·m/s - 0.07 kg·m/s

Impulse = -0.14 kg·m/s

Calculate the average force: The impulse-momentum theorem states that Impulse = [tex]F_{avg[/tex] * Δt, where [tex]F_{avg[/tex] is the average force and Δt is the time of contact (34.3 ms = 0.0343 s). Solving for [tex]F_{avg[/tex] ,

[tex]F_{avg[/tex] = Impulse / Δt

[tex]F_{avg[/tex] = -0.14 kg·m/s / 0.0343 s

[tex]F_{avg[/tex] ≈ -4.08 N

The magnitude of the average force exerted on the ball by the table is approximately 4.08 N in the upward direction.

Answer:

The moment of inertia of the bar is [tex]45\times10^{-4}\ kg-m^2[/tex]

Explanation:

Given that,

mass of bar = 150 g

Length l = 36 cm

We need to calculate the moment of inertia of the bar

Using formula of moment inertia

[tex]I=\dfrac{1}{12}Ml^2[/tex]

Where,

M = mass of the bar

L = length of the bar

Put the value into the formula

[tex]I=\dfrac{1}{12}\times150\times10^-3\times36\times10^{-2}[/tex]

[tex]I=45\times10^{-4}\ kg-m^2[/tex]

Hence, The moment of inertia of the bar is [tex]45\times10^{-4}\ kg-m^2[/tex]

The moment of inertia of a bar with a mass of 150g and length of 36cm, rotating about its center, is approximately 0.00162 kg·m². The calculation uses the formula I = (1/12) * M * L². First, convert the mass and length to SI units and then substitute them into the formula.

To find the moment of inertia of a uniform bar with a mass of 150g and a length of 36cm, we can use the formula for a rod rotating about its center:

I = (1/12) * M * L²

I = (1/12) * 0.15kg * (0.36m)²

I = (1/12) * 0.15kg * 0.1296m²

I ≈ 0.00162 kg·m²

Therefore, the moment of inertia of the bar rotating about its center is approximately 0.00162 kg·m².

Given:

I = 30dB

P = 66 × [tex]10^{-9}[/tex] Pa

Solution:

Formula used:

I = [tex]20\log_{10}(\frac{P}{P_{o}})[/tex]           (1)

where,

I = intensity of sound

P = absolute pressure

[tex]P_{o}[/tex] = reference pressure

Using Eqn (1), we get:

[tex]30 = 20\log _{10}\frac{66\times 10^{-9}}{P^{o}}[/tex]

[tex]P_{o}[/tex] = [tex]\frac{66\times 10^{-9}}{10^{1.5}}[/tex]

[tex]P_{o}[/tex] = 2.08 × [tex]10^{-9}[/tex] Pa

The reference pressure for a sound intensity level of 0 dB is always 20 micropascals, or 2 x 10^-5 Pa, regardless of the absolute pressure of the sound.

If a sound is 30 dB and its absolute pressure was 66 x 10-9 Pa, we need to find the reference pressure. The reference pressure is known as the threshold of hearing and corresponds to a sound intensity level of 0 dB. In acoustics, 0 dB is quantified relative to a reference which has been set at a sound pressure level of 20 micropascals, equivalent to 2 x 10-5 Pa. The question of what is the reference pressure can be answered easily: the reference pressure is always 20 micropascals or 2 x 10-5 Pa, because the decibel scale is logarithmic and based on this fixed reference.

Answer:

The capacitance is 11 F for half and fully charged capacitor.

Explanation:

Capacitance of capacitor is given by the expression

             [tex]C=\frac{\epsilon A}{d}[/tex]

Where ε is the  vacuum permittivity, A is the area of plates and d is the separation between plates.

So capacitance does not depend upon charge and potential. So capacitance fully and half charged capacitors are same.

Here the capacitance is 11 F for half and fully charged capacitor.

Hi there!

[tex]\boxed{C = 11F}[/tex]

Even though C = Q/V, the capacitance is NEVER changed by the charge or voltage.

The only factors that change the capacitance of a capacitor are those related to its geometry or if a dielectric is inserted. We can look at some examples:

For a parallel plate capacitor:
[tex]C = \frac{\epsilon_0A}{d}[/tex]

C = Capacitance (F)

A = Area of plates (m²)
d = distance between plates (m)

For a spherical capacitor:
[tex]C = 4\pi \epsilon_0 (\frac{r_{outer}r_{inner}}{r_{outer} - r_{inner}}})[/tex]

Notice how the capacitance is strictly determined by its geometric properties. Therefore, changing the charge or voltage has no effect, so the capacitance will remain 11 F.

Answer:

Option B is the correct answer.

Explanation:

Period of a spring is given by the expression

            [tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]

Here, spring constant, k = 110 N/m

         Mass = 0.41 kg

Substituting,

        [tex]T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{0.41}{110}}=0.38s[/tex]

Option B is the correct answer.

Answer:

[tex]t =4.605 *10^{-3}[/tex]

Explanation:

given data:

wavelength of emission =[tex]0.8 \mu m[/tex]

output power = 100 mW

We can deduce value of obsorption coefficient from the graph obsorption coefficient vs wavelength

for wavelength [tex]0.8 \mu m[/tex] the obsorption coefficient value is 10^{3}

intensity can be expressed as a function of thickness as following:

[tex]I(t) = I_{O} *e^{-\lambda *t}[/tex]

putting all value to get thickness

[tex]1*10^{-3} =100*10^{-3}e^{-10^{3}*t}[/tex]

[tex]0.01 = e^{10^{3}t}[/tex]

[tex]t =4.605 *10^{-3}[/tex]

Answer:

Charge, [tex]Q=1.56\times 10^{-8}\ C[/tex]

Explanation:

It is given that,

Electric field strength, E = 180000 N/C

Distance from a small object, r = 2.8 cm = 0.028 m

Electric field at a point is given by :

[tex]E=\dfrac{kQ}{r^2}[/tex]

Q is the charge on an object

[tex]Q=\dfrac{Er^2}{k}[/tex]

[tex]Q=\dfrac{180000\ N/C\times (0.028\ m)^2}{9\times 10^9\ Nm^2/C^2}[/tex]

[tex]Q=1.56\times 10^{-8}\ C[/tex]

So, the charge on the object is [tex]1.56\times 10^{-8}\ C[/tex]. Hence, this is the required solution.

The object's charge q is calculated to be approximately 1.57 × 10⁻⁵ C.

To find the object's charge q, we use the formula for the electric field due to a point charge:

E = kQ/r²

Here, E is the electric field, k is the Coulomb's constant (8.99 × 10⁹ N·m²/C²), Q is the charge, and r is the distance from the charge.

We are provided with:

Electric field, E = 180,000 N/C

Distance, r = 2.8 cm = 0.028 m

Rearranging the formula to solve for Q gives:

Q = Er²/k

Substituting in the given values:

Q = (180,000 N/C) × (0.028 m)² / (8.99 × 10⁹ N·m²/C²)

Q = (180,000) × (0.000784) / (8.99 × 10⁹)

Q ≈  1.57 × 10⁻⁵ C

The object's charge q is approximately  1.57 × 10⁻⁵ C.

For a direct current resistor-capacitor circuit where the capacitor is initially uncharged, the charge stored on one of the capacitor's plates is given by:

Q(t) = Cℰ(1-e^{-t/(RC)})

Q(t) is the charge, t is time, ℰ is the battery's terminal voltage, R is the resistor's resistance, and C is the capacitor's capacitance.

The time constant of the circuit τ is the product of the resistance and capacitance:

τ = RC

Q(t) can be rewritten as:

Q(t) = Cℰ(1-e^{-t/τ})

We want to know how much charge is stored when one time constant has elapsed, i.e. what Q(t) is when t = τ. Let us plug in this time value:

Q(τ) = Cℰ(1-e^{-τ/τ})

Q(τ) = Cℰ(1-1/e)

Q(τ) = Cℰ(0.63)

Given values:

C = 5.0×10⁻⁶F

ℰ = 12V

Plug in these values and solve for Q(τ):

Q(τ) = (5.0×10⁻⁶)(12)(0.63)

Q(τ) = 3.8×10⁻⁵C

The charge on either plate of given capacitor after one time constant has elapsed is 3.8×10⁻⁵C.

The charge stored on one of the capacitor's plates can be calculated by,  

[tex]\bold{Q(t) = C\epsilon(1-e^{-t/(RC)})}[/tex]

Where,

Q(t) - the charge,

t - time,

ℰ- the battery's terminal voltage,

R - the resistor's resistance,

C -  the capacitor's capacitance.

The time constant of the circuit τ is equal to the product of the resistance R and capacitance C :  

τ = RC  

The amount of charge Q(t) when t = τ. put the values,  

Q(τ) = Cℰ(1-e^{-τ/τ})  

Q(τ) = Cℰ(1-1/e)  

Q(τ) = Cℰ(0.63)

Given values:  

C = 5.0×10⁻⁶F  

ℰ = 12V

Put the values in the formula and solve for Q(τ):  

Q(τ) = (5.0×10⁻⁶)(12)(0.63)  

Q(τ) = 3.8×10⁻⁵C

Therefore, the charge on either plate of given capacitor after one time constant has elapsed is 3.8×10⁻⁵C.

To know more about capacitor,

https://brainly.com/question/12883102

Answer:

C They can change the amount of work done on an object. Controlling amount of work done would be down to user / more complex machines.

Explanation:

Simple machines help direct, increase, and affect distance force is applied.

Simple machines do not change the amount of work done but can alter the force applied and distance over which the force is exerted.

Simple machines do not change the amount of work done. Although they cannot alter the work done, they can change the amount of force applied and the distance over which the force is exerted. For instance, a simple machine like a lever can help reduce the force needed to lift an object.

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

[tex]T + mg =\frac{mv^{2}}{r}[/tex]

Inserting the values

[tex]T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}[/tex]

T = 31.1 N

Answer: 41N

Explanation :

T= mv^2/R + mgcos θ

At the highest point on the circle θ=0

Cos 0 = 1

T= mv^2/R + mg

m = 0.5kg

Velocity at the highest point (amplitude)= 12m/s

T = 0.5× 12^2/2 + 0.5×10

0.5×144/2 +5

T = 0.5×72 + 5

T = 36+5

T = 41N