When 0.511 g of sodium metal is added to an excess of hydrochloric acid, 5310 J of heat are produced. What is the enthalpy of the reaction as written?

Answer:

0.35 T

Explanation:

Side, a = 0.132 m, e = 27.1 mV = 0.0271 V, dA / dt = 0.0785 m^2 / s

Use the Faraday's law of electromagnetic induction

e = rate of change of magnetic flux

Let b be the strength of magnetic field.

e = dФ / dt

e = d ( B A) / dt

e = B x dA / dt

0.0271 = B x 0.0785

B = 0.35 T

The magnitude of the magnetic field is 3.45 × 10^-4 T.

The magnitude of the induced emf in a square, single-turn coil can be calculated using the equation: emf = -N * A * (d(B)/dt), where N is the number of turns, A is the area of the coil, and d(B)/dt is the rate of change of the magnetic field. In this case, the area of the coil is decreasing at a rate of 0.0785 m2/s, and the emf is given as 27.1 mV. We can rearrange the equation to solve for the magnitude of the magnetic field (B):

B = -emf / (N * d(A)/dt) = -27.1 mV / (1 * 0.0785 m2/s) = -345.222 T/s = 3.45 × 10^-4 T/s.

Therefore, the magnitude of the magnetic field is 3.45 × 10^-4 T.

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Answer:

[tex]\lambda = 260.9nm[/tex]

Explanation:

As we know that work function is defined as the minimum energy of photons required to produce the photo electric effect

so here we have

[tex]\phi = \frac{hc}[\lambda}[/tex]

now we know that

[tex]h = 6.63\times 10^{-34}[/tex]

[tex]c = 3\times 10^8 m/s[/tex]

[tex]\lambda[/tex] = wavelength of photons

now from above formula

[tex]7.59 \times 10^{-19} = \frac{(6.36 \times 10^{-34})(3\times 10^8)}{\lambda}[/tex]

[tex]\lambda = \frac{(6.36 \times 10^{-34})(3\times 10^8)}{7.59 \times 10^{-19}}[/tex]

[tex]\lambda = 260.9nm[/tex]

Answer:

[tex]R_{79} = 28.91 OHM[/tex]

Explanation:

Resistance ca be determine by using following formula

[tex]R = R_{ref}\left [ 1+ \alpha (T - T_{ref}) \right ][/tex]

where

R = Conductor resistance = 23.7Ω

Rref = conductor reistance at reference temperature,

α = temperature coefficient of resistance for material, for copper 40.41*10^{-4}

T = Conductor temperature in Celcius =  20.3°C

Tref = reference temperature at which α  is specified.

[tex]23.7 = R_{ref}\left [ 1+ 40.41*10^{-4} (20.3) \right ][/tex][tex]R_{ref} =21.92 OHM[/tex]

now for 79 degree celcius

[tex]R_{79} = R_{ref}\left [ 1+ \alpha (T - T_{ref}) \right ][/tex]

[tex]R_{79} =21.92\left [ 1+ 40.41*10^{-4} (79) \right ][/tex]

[tex]R_{79} = 28.91 OHM[/tex]

Answer:

0.012 m

Explanation:

m = mass of the marble = 0.0265 kg

M = mass of the pendulum = 0.250 kg

v = initial velocity of the marble before collision = 5.05 m/s

V = final velocity of marble-pendulum combination after the collision = ?

using conservation of momentum

m v = (m + M) V

(0.0265) (5.05) = (0.0265 + 0.250) V

V = 0.484 m/s

h = height gained by the marble-pendulum combination

Using conservation of energy

Potential energy gained by the combination = Kinetic energy of the combination just after collision

(m + M) gh = (0.5) (m + M) V²

gh = (0.5) V²

(9.8) h = (0.5) (0.484)²

h = 0.012 m

Using the concept of inelastic collison and principle of conservation of momentum, the height of swing made by the pendulum would be 0.012 meters

Collison is inelastic :

(0.0265 × 5.05) = (0.0265 + 0.250)v

0.133825 = 0.2765v

v = (0.133825 ÷ 0.2765)

v = 0.484 m/s

Final velocity after collision = 0.484 m/s

Assuming linear momentum is conserved :

0.5mv² = mgh

Mass cancels out

0.5v² = gh

0.5(0.484)² = 9.8h

9.8h = 0.1171

h = (0.1171) ÷ 9.8

h = 0.0119

Therefore, height of the swing made by the pendulum would be 0.012 meters.

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Answer:

    4.4 square meters = 47 square foot

Explanation:

We have

    1 meter = 3.28084 foot

    1 square meter = 3.28084 x 3.28084 square foot = 10.76 square foot

    4.4 square meters = 4.4 x 10.76 = 47.36 square foot = 47 square foot

    4.4 square meters = 47 square foot

Answer:

a)The parachutist in the air for 12.63 seconds.

b)The parachutist falls from a height of 293 meter.

Explanation:

Vertical motion of  parachutist:

 Initial speed, u = 0m/s

 Acceleration, a = 9.81 m/s²

 Displacement, s = 50 m

 We have equation of motion, v² = u² + 2as

 Substituting

   v² = 0² + 2 x 9.81 x 50

    v = 31.32 m/s

 Time taken for this

          31.32 = 0 + 9.81 x t

            t = 3.19 s

 After 50m we have

 Initial speed, u = 31.32m/s

 Acceleration, a = -2 m/s²

 Final speed , v = 3 m/s

 We have equation of motion, v² = u² + 2as

 Substituting

   3² = 31.32² - 2 x 2 x s

    s = 243 m

 Time taken for this

          3 = 31.32 - 2 x t

            t = 9.44 s

a) Total time = 3.19 +  9.44 = 12.63 s

    The parachutist in the air for 12.63 seconds.

b) Total height = 50 + 243 = 293 m

    The parachutist falls from a height of 293 meter.

Answer:

0.4 times the radius of moon

Explanation:

gravity on moon is equal to the one sixth of gravity on earth.

g' = g / 6

where, g' is the gravity on moon and g be the gravity on earth.

As the earth shrinks, the mass of earth remains same.

The acceleration due to gravity is inversely proportional to the square of radius of planet.

g' ∝ 1/R'²   .....(1)

Where, R' is the radius of moon.

g ∝ 1/R²      ..... (2)

Where, R be the radius of earth.

Divide equation (1) by (2)

g / g' = R'² / R²

Put g' = g / 6

6 = R'² / R²

2.5 = R' / R

R = R' / 2.5 = 0.4 R'

Thus, the radius of earth should be 0.4 times the radius of moon.

Answer:

[tex]F = \frac{Qq}{2\pi \epsilon_0 L d}[/tex]

Explanation:

As we know that if a charge q is distributed uniformly on the line then its linear charge density is given by

[tex]\lambda = \frac{q}{L}[/tex]

now the electric field due to long line charge at a distance d from it is given as

[tex]E = \frac{2k\lambda}{d}[/tex]

[tex]E = \frac{q}{2\pi \epsilon_0 d}[/tex]

now the force on the other charge in this electric field is given as

[tex]F = QE[/tex]

[tex]F = \frac{Qq}{2\pi \epsilon_0 L d}[/tex]

Answer:

29.3 rad/s

Explanation:

Moment of inertia in straight position, I1 = I

He takes 2 rotations in 1.5 second

So, Time period, T = 1.5 / 2 = 0.75 second

w1 = 2 x 3.14 / T = 2 x 3.14 / 0.75 = 8.37 rad/s

Moment of inertia in tucked position, I2 = I / 3.5

Let the new angular speed is w2.

By the use of conservation of angular momentum, if no external torque is applied, then angular momentum is constant.

L1 = L2

I1 w1 = I2 w2

I x 8.37 = I / 3.5 x w2

w2 = 29.3 rad/s

To find the angular speed of the diver in the straight position, we can use the conservation of angular momentum. The moment of inertia decreases by a factor of 3.5 when changing from the tuck position to the straight position. By applying the conservation of angular momentum equation, we can calculate the angular speed in the straight position.

To find the angular speed of the diver in the straight position, we can use the conservation of angular momentum. According to the conservation of angular momentum, the product of the initial moment of inertia and initial angular speed is equal to the product of the final moment of inertia and final angular speed.

Let's assume the initial angular speed in the tuck position is denoted by w' and the final angular speed in the straight position is denoted by w. We are given that the moment of inertia decreases by a factor of 3.5 when changing from the tuck position to the straight position. Therefore, the final moment of inertia (I) is 3.5 times greater than the initial moment of inertia (I').

The conservation of angular momentum equation can be written as:

I' * w' = I * w

Since the moment of inertia decreases by a factor of 3.5 when changing from the tuck position to the straight position, we have:
w = w' * (I' / I)

Using the given values, we can calculate the angular speed in the straight position.

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Answer:

6.4 J

Explanation:

m = mass of the bullet = 10 g = 0.010 kg

v = initial velocity of bullet before collision = 1.8 km/s = 1800 m/s

v'  = final velocity of the bullet after collision = 1 km/s = 1000 m/s

M = mass of the block = 5 kg

V = initial velocity of block before collision = 0 m/s

V'  = final velocity of the block after collision = ?

Using conservation of momentum

mv + MV = mv' + MV'

(0.010) (1800) + (5) (0) = (0.010) (1000) + (5) V'

V' = 1.6 m/s

Kinetic energy of the block after the collision is given as

KE = (0.5) M V'²

KE = (0.5) (5) (1.6)²

KE = 6.4 J

Final answer:

Using the conservation of momentum, and given that the bullet exits the wooden block with a lower speed, we calculate the block's final speed to be 7.5 m/s after the bullet exits.

Explanation:

To solve this problem, we need to apply the law of conservation of momentum. The law states that if no external forces act on a system, the total momentum of the system remains constant, even though individual objects within the system might change velocities. In this case, the system is made up of the bullet and the wooden block. Before the collision, the block is at rest, so its initial momentum is 0. The bullet’s initial momentum is its mass times its velocity (momentum = mass × velocity). After the collision, the bullet exits the block with a lower speed, indicating that some of its momentum has been transferred to the block.

Step 1: Calculate the initial momentum of the bullet:
Initial momentum of the bullet = mass of bullet × initial velocity of bullet
= 0.05 kg × 500 m/s
= 25 kg·m/s

Step 2: Calculate the final momentum of the bullet:
Final momentum of the bullet = mass of bullet × final velocity of bullet
= 0.05 kg × 200 m/s
= 10 kg·m/s

Step 3: Calculate the change in momentum of the bullet:
Change in momentum = Initial momentum - Final momentum
Change in momentum = 25 kg·m/s - 10 kg·m/s
= 15 kg·m/s

Step 4: Calculate the final momentum of the block:
Since the block was initially at rest, its final momentum is equal to the change in momentum of the bullet (due to conservation of momentum). Therefore, the final momentum of the block is 15 kg·m/s.

Step 5: Calculate the final velocity of the block:
Using the final momentum and the mass of the block, we can find the final velocity:
Final velocity of the block = Final momentum of block / mass of block
= 15 kg·m/s / 2 kg
= 7.5 m/s

Thus, the speed of the block after the bullet has come out the other side is 7.5 m/s.

Answer:

6.889 ft/s^2

Explanation:

1 m = 3.28 feet

So, 2.1 m/s^2 = 2.1 × 3.28 ft/s^2

= 6.889 ft/s^2

In this exercise we have to use the knowledge of finance to calculate the equilibrium value and the quantity, so we have:

Organizing the information given in the statement we have:

So equating the two given equations we have:

[tex]-x^2 - 3x + 80 = 7x+5\\x^2 +3x + 7x + 5 - 80 = 0\\x^2 + 10x - 75 = 0\\x^2- 5x + 15x -75 = 0\\x(x-5) + 15(x-5) = 0[/tex]

So we can see that the roots will be x = 5 and x = -15 since the quantity cannot be in negative therefore, the equilibrium quantity will be = 5 So replace that value at:

[tex]p = -(5)^2-3(5) + 80 = 40\\p = 7(5) + 5 = 40[/tex]

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To calculate the change in entropy during the approach to equilibrium, use the formula ΔS = mcΔT. For the aluminum block, ΔS_aluminum = m_aluminum * c_aluminum * ΔT_aluminum. For the water, ΔS_water = m_water * c_water * ΔT_water. Set ΔT_aluminum equal to ΔT_water and solve for T_water. Add the change in entropy for the aluminum and water to find the total change in entropy. Simplify the equation by converting cal to J and kg to g. Set the total change in entropy equal to zero and solve for T_aluminum to find the initial temperature of the aluminum.

To calculate the change in entropy during the approach to equilibrium, we need to use the formula for entropy change:

ΔS = mcΔT

where ΔS is the change in entropy, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

In this case, the aluminum block is dropped into the water, so the final temperature will be the same for both the aluminum and the water. We can calculate the change in entropy for each substance separately and then add them together to find the total change in entropy.

For the aluminum block:

ΔS_aluminum = m_aluminum * c_aluminum * ΔT_aluminum

ΔS_aluminum = 2.00 kg * 0.22 cal/(g∙K) * (50.0 °C - T_aluminum)

For the water:

ΔS_water = m_water * c_water * ΔT_water

ΔS_water = 5.00 kg * 1 cal/(g∙K) * (T_water - 20.0 °C)

Since the final temperature is the same for both substances, we can set ΔT_aluminum equal to ΔT_water and solve for T_water:

50.0 °C - T_aluminum = T_water - 20.0 °C

70.0 °C - T_aluminum = T_water

Substituting this value into the equation for ΔS_water:

ΔS_water = 5.00 kg * 1 cal/(g∙K) * [(70.0 °C - T_aluminum) - 20.0 °C]

Now we can add the change in entropy for the aluminum and water:

ΔS_total = ΔS_aluminum + ΔS_water

ΔS_total = (2.00 kg * 0.22 cal/(g∙K) * (50.0 °C - T_aluminum)) + (5.00 kg * 1 cal/(g∙K) * [(70.0 °C - T_aluminum) - 20.0 °C])

To simplify the equation, we can convert cal to J by multiplying by 4.184 and divide both sides by 1000 to convert kg to g:

ΔS_total = (2.00 * 4.184 J/(g∙K) * (50.0 °C - T_aluminum)) + (5.00 * 4.184 J/(g∙K) * [(70.0 °C - T_aluminum) - 20.0 °C])

We can now solve for T_aluminum by setting ΔS_total equal to 0:

0 = (2.00 * 4.184 J/(g∙K) * (50.0 °C - T_aluminum)) + (5.00 * 4.184 J/(g∙K) * [(70.0 °C - T_aluminum) - 20.0 °C])

Simplifying the equation:

0 = (8.368 J/(g∙K) * (50.0 °C - T_aluminum)) + (20.92 J/(g∙K) * (50.0 °C - T_aluminum)) - (104.6 J/(g∙K) * (70.0 °C - T_aluminum))

Combining like terms:

0 = (29.288 J/(g∙K) * (50.0 °C - T_aluminum)) - (104.6 J/(g∙K) * (70.0 °C - T_aluminum))

Simplifying further:

0 = (29.288 J/(g∙K) * (50.0 °C - T_aluminum) - 104.6 J/(g∙K) * (70.0 °C - T_aluminum)

Expanding the equation:

0 = (29.288 J/(g∙K) * 50.0 °C - 29.288 J/(g∙K) * T_aluminum) - (104.6 J/(g∙K) * 70.0 °C - 104.6 J/(g∙K) * T_aluminum)

Simplifying the equation further:

0 = (1464.4 J - 29.288 J/(g∙K) * T_aluminum) - (7322 J - 104.6 J/(g∙K) * T_aluminum)

Combining like terms:

0 = 1464.4 J - 29.288 J/(g∙K) * T_aluminum - 7322 J + 104.6 J/(g∙K) * T_aluminum

Simplifying the equation:

0 = -5857.6 J + 75.312 J/(g∙K) * T_aluminum

Isolating T_aluminum:

5857.6 J = 75.312 J/(g∙K) * T_aluminum

T_aluminum = (5857.6 J) / (75.312 J/(g∙K))

T_aluminum ≈ 77.7 °C

Therefore, the initial temperature of the aluminum was approximately 77.7 °C.

Explanation:

It is given that,

Mass of the object, m = 3.66 kg

Charge, q = -19 μC = -19 × 10⁻⁶ C

It is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground such that,

[tex]F_g=F_e[/tex]

[tex]F_g\ and\ F_e[/tex] are gravitational and electrostatic forces respectively

[tex]mg=qE[/tex]

[tex]E=\dfrac{mg}{q}[/tex]

[tex]E=\dfrac{3.66\ kg\times 9.8\ m/s^2}{-19\times 10^{-6}\ C}[/tex]

E = −1887789.47 N/C

[tex]E=-1.89\times 10^6\ N/C[/tex]

Negative sign shows that the electric field is in the opposite direction of the electric force. Since, the weight of the object is in downward direction and its electric force (which is balancing its weight) is in upward direction. So, we can say that the electric field is in downward direction.

Answer:

ΔT / Δx = 771 K/m

ΔT = 771 x 0.0475 = 36.62 k

Explanation:

P = 31700 W, A = 0.819 m^2, Δx = 0.0475 m, K = 50.2 W /m k

Use the formula of conduction of heat

H / t = K A x ΔT / Δx

So, ΔT / Δx = P / K A

ΔT / Δx = 31700 / (50.2 x 0.819)

ΔT / Δx = 771 K/m

Now

ΔT = 771 x 0.0475 = 36.62 k

Answer:

5706.5

Explanation:

B = 7.171 T, i = 1000 A, μ0 = 1.25664 x 10^-6 T m/A, l = 1 m

Let the number of turns be N.

The magnetic field due to a current carrying solenoid is given by

B = μ0 x N x i / l

B x l / (μ0 x i) = N

N = 7.171 x 1 / (1.25664 x 10^-6 x 1000)

N = 5706.5

To design a superconducting solenoid that generates an interior field of 7.171 T with a maximum current of 1000 A, approximately 5709 turns per meter are required on a 1-meter length of the solenoid.

To determine the number of windings required on a 1-meter solenoid to achieve a magnetic field of 7.171 T with a maximum current of 1000 A, we can use the formula for the magnetic field inside a solenoid, which is given by:

B = μ_0 * n * I

Where:

Rearranging the formula to solve for n, we get:

n = B / (μ_0 * I)

Substituting the given values:

n = 7.171 T / (1.25664 × 10⁻⁶ T · m/A × 1000 A)

Now, calculating the value:

n = 7.171 / (1.25664 × 10⁻⁶ × 1000)

n = 7.171 / (1.25664 × 10⁻³)

n = 5708.57 turns per meter

To achieve the desired magnetic field, the solenoid must have approximately 5709 turns per meter to accommodate the rounding to the nearest whole number, since we can't have a fraction of a turn.

Answer:

The pendulum clock does it loses 83.0304 seconds a day at the new location.

Explanation:

T1= 1 s

T2= ?

g1= 9.8135 m/s

g2= 9.7943 m/s

L=?

L= (T1/2π)²  * g1=

L= 0.24857m

T2= 2π * √(L/g2)

T2= 1.000961 s

ΔT= T2 - T1

ΔT = 0.000961 s

seconds lost a day = 24 * 3600 * ΔT

seconds lost a day= 24 * 3600 * 0.000961 s

seconds lost a day= 83.0304 s

A pendulum clock loses accuracy when it is moved from one location to another due to changes in the acceleration due to gravity. The period of a pendulum is dependent on its length and the acceleration due to gravity. To calculate how many seconds a day the clock loses at the new location, we can subtract the new period from the original period and convert it to seconds.

A pendulum clock loses accuracy when it is moved from one location to another due to changes in the acceleration due to gravity, not changes in the length of the pendulum. In this case, the clock was moved from a location where g = 9.8135 m/s2 to a new location where g = 9.7943 m/s2. The clock loses accuracy because the period of a pendulum is dependent on its length and the acceleration due to gravity.

The period of a pendulum can be calculated using the equation: T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

So, to calculate how many seconds a day the clock loses at the new location, we can subtract the new period from the original period and convert it to seconds. Since the original period is 2.00000 s, we have:

ΔT = Toriginal - Tnew

ΔT = 2.00000 s - Tnew

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Answer:

0.273 m/s

Explanation:

Momentum is conserved:

mu = (m + M) v

(0.200 kg) (0.750 m/s) = (0.200 kg + 0.350 kg) v

v = 0.273 m/s

The momentum of the conservation said that the momentum before collision is equal to the momentum before collision.The final velocity will be 0.273 m/sec.

The momentum of the conservation said that the momentum before collision is equal to the momentum before collision.

The given data in the problem is;

m is the mass of clay model before collision= 0.200 Kg

u is the sliding velocity before collision=0.75 m/sec

M is the mass after collision= 0.350 Kg

v is the velocity after collision=?

On applying the law of conservation of momentum we get;

[tex]\rm mu = (m + M) v \\\\\ (0.200 kg) (0.750 m/s) = (0.200 kg + 0.350 kg) v \\\\ v = 0.273 m/s[/tex]

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Answer:

18.7 m

Explanation:

[tex]v_{o}[/tex] = initial speed of the shot = 13 m/s

θ = angle of launch from the horizontal = 47 deg

Consider the motion along the vertical direction

[tex]v_{oy}[/tex] = initial velocity along vertical direction = 13 Sin47 = 9.5 m/s

[tex]a_{y}[/tex] = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 1.80 m

t = time of travel

Using the kinematics equation

[tex]y=v_{oy} t+(0.5)a_{y} t^{2}[/tex]

- 1.80 = (9.5) t + (0.5) (- 9.8) t²

t = 2.11 s

Consider the motion along the horizontal direction

x = horizontal displacement of the shot

[tex]v_{ox}[/tex] = initial velocity along horizontal direction = 13 Cos47 = 8.87 m/s

[tex]a_{x}[/tex] = acceleration along the horizontal direction = 0 m/s²

t = time of travel =  2.11 s

Using the kinematics equation

[tex]x=v_{ox} t+(0.5)a_{x} t^{2}[/tex]

x = (8.87) (2.11) + (0.5) (0) (2.11)²

x = 18.7 m

The horizontal distance traveled by the shotput when launched at a velocity of 13 m/s is 18.7 m.

A projectile motion is a curved motion that is projected near the surface of the earth such that it falls under the influence of gravity only.

Given to us

The initial velocity of the shotput, u = 13 m/s

The angle from the horizontal, θ = 47.0°

The height from which shotput was thrown h = 1.80 m

As we know that when the shotput will be thrown it will be in a projectile motion, therefore, the distance that will be covered by the shot put will be the range of the projectile motion,

[tex]R = \dfrac{u\ cos\theta}{g} [u\cdot sin\theta + \sqrt{u^2sin^2\theta+2gh}][/tex]

Substitute the values,

[tex]R = \dfrac{13\ cos47^o}{9.81} [13\cdot sin47 + \sqrt{13^2(sin47)^2+2(9.81)(1.80)}][/tex]

R = 18.7 m

Hence, the horizontal distance traveled by the shotput when launched at a velocity of 13 m/s is 18.7 m.

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Answer:

He made it, he across safely the river without falling ing.

Explanation:

Vt= 7.8 m/s

r= 11m

m= 80.5 kg

Vt= ω * r

ω= 0.71 rad/s

ac= ω² * r

ac= 5.54 m/s²

F= m * ac

F= 445.97 N  < 1,000 N

Answer:

Time taken, t = 3 s

Explanation:

It is given that,

Initial velocity of the particle, u = 50 m/s

Final velocity, v = 20 m/s

Distance covered, s = 105 m

Firstly we will find the acceleration of the particle. It can be calculated using third equation of motion as :

[tex]v^2-u^2=2as[/tex]

[tex]a=\dfrac{v^2-u^2}{2s}[/tex]

[tex]a=\dfrac{(20\ m/s)^2-(50\ m/s)^2}{2\times 105\ m}[/tex]

[tex]a=-10\ m/s^2[/tex]

So, the particle is decelerating at the rate of 10 m/s². Let t is the time taken for the particle to slow down. Using first equation of motion as :

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

[tex]t=\dfrac{20\ m/s-50\ m/s}{-10\ m/s^2}[/tex]

t = 3 s

So, the time taken for the particle to slow down is 3 s. Hence, this is the required solution.

Answer:

[tex]\frac{dB}{dt}[/tex] = 591.45 T/s

Explanation:

i = induced current in the loop = 0.367 A

R = Resistance of the loop = 117 Ω

E = Induced voltage

Induced voltage is given as

E = i R

E = (0.367) (117)

E = 42.939 volts

[tex]\frac{dB}{dt}[/tex] = rate of change of magnetic field

A = area of loop = 7.26 x 10⁻² m²

Induced emf is given as

[tex]E = A\frac{dB}{dt}[/tex]

[tex]42.939 = (7.26\times 10^{-2})\frac{dB}{dt}[/tex]

[tex]\frac{dB}{dt}[/tex] = 591.45 T/s

Answer:

[tex]i_2 = 978750 A[/tex]

Since the force between wires is attraction type of force so current must be flowing in upward direction

Explanation:

Force per unit length between two current carrying wires is given by the formula

[tex]F = \frac{\mu_0 i_1 i_2}{2 \pi d}[/tex]

here we know that

[tex]F = 7.83 \times 10 N/m[/tex]

[tex]d = 7.0 cm = 0.07 m[/tex]

[tex]i_1 = 28 A[/tex]

now we will have

[tex]F = \frac{4\pi \times 10^{-7} (28.0)(i_2)}{2\pi (0.07)}[/tex]

[tex]7.83 \times 10 = \frac{2\times 10^{-7} (28 A)(i_2)}{0.07}[/tex]

[tex]i_2 = 978750 A[/tex]

Since the force between wires is attraction type of force so current must be flowing in upward direction

Answer:

The two balls meet in 1.47 sec.

Explanation:

Given that,

Height = 25 m

Initial velocity of ball= 0

Initial velocity of another ball = 17 m/s

We need to calculate the ball

Using equation of motion

[tex]s=ut+\dfrac{1}{2}gt^2+h[/tex]

Where, u = initial velocity

h = height

g = acceleration due to gravity

Put the value in the equation

For first ball

[tex]s_{1}=0-\dfrac{1}{2}gt^2+25[/tex]....(I)

For second ball

[tex]s_{2}=17t-\dfrac{1}{2}gt^2+0[/tex]....(II)

From equation (I) and (II)

[tex]-\dfrac{1}{2}gt^2+25=17t-\dfrac{1}{2}gt^2+0[/tex]

[tex]t=\dfrac{25}{17}[/tex]

[tex]t=1.47\ sec[/tex]

Hence, The two balls meet in 1.47 sec.

The two balls will meet at a height of approximately 16.465 meters above the ground.

To find the height at which the two balls meet, we need to consider their respective equations of motion and solve for the height when they intersect.

For the tennis ball thrown from the building:

[tex]\[ h_1(t) = 25 - \frac{1}{2} g t^2 \][/tex]

For the ball thrown from the ground:

[tex]\[ h_2(t) = 17t - \frac{1}{2} g t^2 \][/tex]

To find the time when they meet, we'll set:

[tex]\[ 25 - \frac{1}{2} g t^2 = 17t - \frac{1}{2} g t^2 \]\[ 25 = 17t \]\[ t = \frac{25}{17} \][/tex]

Now, substitute this value of \( t \) into either equation to find the height at which they meet:

[tex]\[ h_1\left(\frac{25}{17}\right) = 25 - \frac{1}{2} g \left(\frac{25}{17}\right)^2 \]\[ h_1\left(\frac{25}{17}\right) = 25 - \frac{1}{2} g \times \frac{625}{289} \]\[ h_1\left(\frac{25}{17}\right) = 25 - \frac{1}{2} \times 9.8 \times \frac{625}{289} \]\[ h_1\left(\frac{25}{17}\right) = 25 - \frac{1}{2} \times 17.07 \]\[ h_1\left(\frac{25}{17}\right) = 25 - 8.535 \]\[ h_1\left(\frac{25}{17}\right) = 16.465 \][/tex]

Answer:

106.83

Explanation:

N = 332, l = 14 cm = 0.14 m, i = 0.88 A, B = 0.28 T

Let ur be the relative permeability

B = u0 x ur x n x i

0.28 = 4 x 3.14 x 10^-7 x ur x 332 x 0.88 / 0.14       ( n = N / l)

ur = 106.83

Answer:

0.25

Explanation:

A = area of each plate = 30 cm² = 30 x 10⁻⁴ m²

d = separation between the plates = 5 mm = 5 x 10⁻³ m

[tex]C_{air}[/tex] = Capacitance of capacitor when there is air between the plates

k = dielectric constant = 4

[tex]C_{dielectric}[/tex] = Capacitance of capacitor when there is dielectric between the plates

Capacitance of capacitor when there is air between the plates is given as

[tex]C_{air} = \frac{\epsilon _{o}A}{d}[/tex]                            eq-1

Capacitance of capacitor when there is dielectric between the plates is given as

[tex]C_{dielectric} = \frac{k \epsilon _{o}A}{d}[/tex]                            eq-2

Dividing eq-1 by eq-2

[tex]\frac{C_{air}}{C_{dielectric}}=\frac{\frac{\epsilon _{o}A}{d}}{\frac{k \epsilon _{o}A}{d}}[/tex]

[tex]\frac{C_{air}}{C_{dielectric}}=\frac{1}{k}[/tex]

[tex]\frac{C_{air}}{C_{dielectric}}=\frac{1}{4}[/tex]

[tex]\frac{C_{air}}{C_{dielectric}}=0.25[/tex]

Charge stored in the capacitor when there is air is given as

[tex]Q_{air}=C_{air}V[/tex]                             eq-3

Charge stored in the capacitor when there is dielectric is given as

[tex]Q_{dielectric}=C_{dielectric}V[/tex]                             eq-4

Dividing eq-3 by eq-4

[tex]\frac{Q_{air}}{Q_{dielectric}}=\frac{C_{air}V}{C_{dielectric} V}[/tex]

[tex]\frac{Q_{air}}{Q_{dielectric}}=\frac{C_{air}}{C_{dielectric}}[/tex]

[tex]\frac{Q_{air}}{Q_{dielectric}}=0.25[/tex]

To convert the thermal conductivity value of 0.3 Btu/(h ft°F) to W/(m °C), multiply by 694.7. To convert the surface heat transfer coefficient value of 105 Btu/(h ft2 oF) to W/(m2 °C), multiply by 11,545.

To convert the thermal conductivity value of 0.3 Btu/(h ft°F) to W/(m °C), we can use the following conversion factors:

Using these conversion factors, we can convert the thermal conductivity value as follows:

0.3 Btu/(h ft°F) = (0.3 * 1055.06 J)/(h * 0.3048 m * 0.5556 °C) = 694.7 W/(m °C)

Similarly, to convert the surface heat transfer coefficient value of 105 Btu/(h ft2 oF) to W/(m2 °C), we can use the following conversion factors:

Using these conversion factors, we can convert the surface heat transfer coefficient value as follows:

105 Btu/(h ft2 oF) = (105 * 1055.06 J)/(h * 0.0929 m2 * 0.5556 °C) = 11,545 W/(m2 °C)

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The conversion of the thermal conductivity value yields approximately 0.52 W/(m°C), and the surface heat transfer coefficient conversion yields approximately 596.15 W/(m²°C). Conversion factors for BTU to J, ft to m, and °F to °C are critical in these calculations.

To convert thermal conductivity from BTU/(h ft°F) to W/(m°C):

To convert surface heat transfer coefficient from BTU/(h ft²°F) to W/(m²°C):

Final answer:

The weight of the piece when made of lead is 992.6 lb. The weight of the piece when made of aluminum is 236.6 lb. By using aluminum instead of lead, a weight of 756 lb is saved.

Explanation:

(a) Weight of the piece when it is made of lead:

To find the weight of the piece when it is made of lead, we need to know the density of lead. The density of lead is approximately 709 lb/ft³. We can use the formula:

Weight = Density x Volume

Given that the volume of the piece is 1.4 ft³, the weight of the piece when it is made of lead can be calculated as:

Weight = 709 lb/ft³ x 1.4 ft³ = 992.6 lb

(b) Weight of the piece when made of aluminum, and weight saved:

To find the weight of the piece when it is made of aluminum, we need to know the density of aluminum. The density of aluminum is approximately 169 lb/ft³. Using the same formula as before:

Weight = Density x Volume

Given that the volume of the piece is still 1.4 ft³, the weight of the piece when it is made of aluminum can be calculated as:

Weight = 169 lb/ft³ x 1.4 ft³ = 236.6 lb

The weight saved by using aluminum instead of lead can be determined by subtracting the weight of the aluminum piece from the weight of the lead piece:

Weight saved = Weight of lead piece - Weight of aluminum piece

Weight saved = 992.6 lb - 236.6 lb = 756 lb

Answer:

P₁ = 192 W and P₂ = 384 W

Explanation:

It is given that,

Resistor 1, [tex]R_1=12\ \Omega[/tex]

Resistor 2, [tex]R_1=6\ \Omega[/tex]

Voltage, V = 48 V

Power dissipated by resistor 1 is given by :

[tex]P_1=\dfrac{V^2}{R_1}[/tex]

[tex]P_1=\dfrac{(48)^2}{12}[/tex]

P₁ = 192 watts

Power dissipated by resistor 2 is given by :

[tex]P_2=\dfrac{V^2}{R_1}[/tex]

[tex]P_2=\dfrac{(48)^2}{6}[/tex]

P₂ = 384 watts

So, the power dissipated by both the resistors is 192 watts and 384 watts respectively. Hence, this is the required solution.