Suppose that you wish to construct a simple ac generator having 64 turns and an angular velocity of 377 radians/second (this is the frequency point of 60 Hz). A uniform magnetic field of 0.050 T is available. If the area of the rotating coil is 0.01 m 2, what is the maximum output voltage?

Answer:

B

Explanation:

Newton's third law of motion states that to every action there is equal an opposite reaction. Momentum is always conserved provided there is no net force on the body. Considering the experiment: the momentum is conserved as expected but the  kinetic energy is not conserved meaning that the collision is not elastic; some energy is converted into internal energy. When the collision is elastic the total kinetic energy before will equal total kinetic energy after and when the body stick together, they move with a common velocity and that described a perfectly inelastic collision.  5J ≠ 4J

Final answer:

The scenario describes an inelastic collision because while the momentum is conserved, the kinetic energy decreases from 5J to 4J, indicating that not all kinetic energy is conserved in the collision.

Explanation:

When examining collisions in physics, an elastic collision is one in which both momentum and kinetic energy are conserved, whereas an inelastic collision is one where momentum is conserved but kinetic energy is not. Given that in the scenario the total momentum of the system remains the same before and after the collision, that part of the conservation law is satisfied in both cases. However, since the kinetic energy of the system decreases from 5J to 4J, this means that some of the kinetic energy has been transformed into other forms of energy, like heat or sound, which typically happens during an inelastic collision.

Therefore, the correct answer is (b): This describes an inelastic collision (and it could NOT be elastic) since the total kinetic energy of the system is not the same before and after the collision. An elastic collision would have required that the kinetic energy also remain constant.

Answer:

e. 1.2 x 10²³

Explanation:

According to the problem, The current equation is given by:

[tex]I(t)=0.88e^{-t/6\times3600s}[/tex]

Here time is in seconds.

Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.

The relation between current and number of charge carriers is:

[tex]q=\int\limits {I} \, dt[/tex]

Here the limits of integration is from 0 to infinite. So,

[tex]q=\int\limits {0.88e^{-t/6\times3600s}}\, dt[/tex]

[tex]q=0.88\times(-6\times3600)(0-1)[/tex]

q = 1.90 x 10⁴ C

Consider N be the total number of charge carriers. So,

q = N e

Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.

N = q/e

Substitute the suitable values in the above equation.

[tex]N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}[/tex]

N = 1.2 x 10²³

Answer:

(b) 4.775 kN

Explanation:

see the attached file

Answer:

[tex]w=694.575\ m[/tex]

Explanation:

Given:

velocity of the sound, [tex]v=343\ m.s^{-1}[/tex]

time lag in echo form one wall of the valley, [tex]t= 1.55\ s[/tex]

time lag in echo form the other wall of the valley, [tex]t'=2.5\ s[/tex]

distance travelled by the sound in the first case:

[tex]d=v.t[/tex]

[tex]d=343\times 1.55[/tex]

[tex]d=531.65\ m[/tex]

Since this is the distance covered by the sound while going form the source to the walls and then coming back from the wall to the source so the distance of the wall form the source will be half of the distance obtained above.

[tex]s=\frac{d}{2}[/tex]

[tex]s=\frac{531.65}{2}[/tex]

[tex]s=265.825\ m[/tex]

distance travelled by the sound in the second case:

[tex]d'=v.t'[/tex]

[tex]d'=343\times 2.5[/tex]

[tex]d'=857.5\ m[/tex]

Since this is the distance covered by the sound while going form the source to the walls and then coming back from the wall to the source so the distance of the wall form the source will be half of the distance obtained above.

[tex]s'=\frac{d'}{2}[/tex]

[tex]s'=\frac{857.5}{2}[/tex]

[tex]s'=428.75\ m[/tex]

Now the width of valley:

[tex]w=s+s'[/tex]

[tex]w=265.825+428.75[/tex]

[tex]w=694.575\ m[/tex]

To determine the width of the valley with vertical walls using echo timings.

To find the width of the valley:

Calculations:

Distance to first wall = 343 m/s * 1.55 s = 531.85 m

Distance to second wall = 343 m/s * 2.5 s = 857.5 m

Width of the valley: 857.5 m - 531.85 m = 325.65 m

Answer:

14.3kg

Explanation:

Given parameters:

Quantity of heat = 149000J

Change in temperature = 5.23°C

specific heat of the ice = 2000J/kg°C

Unknown:

Mass of the ice in the bag = ?

Solution:

The heat capacity of a substance is given as:

            H = m c Ф

H is the heat capacity

m is the mass

c is the specific heat

Ф is the temperature change;

  since m is the unknown, we make it the subject of the expression;

                    m = H/ mФ

                  m = [tex]\frac{149000}{2000 x 5.23}[/tex]  = 14.3kg

Answer: 14.2447

Explanation:

Answer:

-3413 ft/s2

Explanation:

We need to know the velocity with which he landed on the snow.

He 'dropped' from 512 feet. This is the displacement. His initial velocity is 0 and the acceleration of gravity is 32 ft/s2.

We use the equation of mition

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

v and u are the initial and final velocities, a is the acceleration and s is the displacement. Putting the appropriate values

[tex]v^2 = 0^2 + 2\times32\times512[/tex]

[tex]v = \sqrt{2\times32\times512} = 128\sqrt{2}[/tex]

This is the final velocity of the fall and becomes the initial velocity as he goes into the snow.

In this second motion, his final velocity is 0 because he stops after a displacement of 4.8 ft. We use the same equation of motion but with different values. This time, [tex]u=128\sqrt{2}[/tex], v = 0 and s = 4.8 ft.

[tex]0^2 = (128\sqrt{2})^2 + 2a\times4.8[/tex]

[tex]a = -\dfrac{2\times128^2}{2\times4.8} = -3413[/tex]

Note that this is negative because it was a deceleration, that is, his velocity was decreasing.

Answer with Explanation:w

a.We are given that

Potential difference, V=48 V

[tex]R_1=24\Omega[/tex]

[tex]R_2=96\Omega[/tex]

Equivalent resistance when R1 and R2 are connected in series

[tex]R_{eq}=R_1+R_2[/tex]

Using the formula

[tex]R_{eq}=24+96=120\Omega[/tex]

We know that

[tex]I=\frac{V}{R_{eq}}=\frac{48}{120}=0.4 A[/tex]

In series combination, current passing through each resistor is  same and potential difference across each resistor is different.

Power, P=[tex]I^2 R[/tex]

Using the formula

Power,[tex]P_1=I^2R_1=(0.4)^2\times 24=3.84 W[/tex]

Power, [tex]P_2=I^2 R_2=(0.4)^2(96)=15.36 W[/tex]

b.

In parallel combination, potential difference remains same across each resistor and current passing through each resistor is different..

Current,[tex]I=\frac{V}{R}[/tex]

Using the formula

[tex]I_1=\frac{V}{R_1}=\frac{48}{24}=2 A[/tex]

[tex]I_2=\frac{V}{R_2}=\frac{48}{96}=0.5 A[/tex]

[tex]P_1=\frac{V^2}{R_1}=\frac{(48)^2}{24}=96 W[/tex]

[tex]P_2=\frac{V^2}{R_2}=\frac{(48)^2}{96}=24 W[/tex]

When two resistors are in series, the current through each is the same and is calculated by dividing the total voltage by the total resistance. The power through each resistor is the current squared times the resistance of each resistor. When resistors are in parallel, the total resistance decreases and the power increases. The current remains the same.

When the 24-ohm and 96-ohm resistors are connected in series, the total resistance can be calculated using the formula R = R1 + R2. Here, R1 is the resistance of the first resistor (24 ohms) and R2 is the resistance of the second resistor (96 ohms). R equals 120 ohms. The current, I, is calculated using Ohm's Law, I = V / R. Therefore, the current flowing through the circuit is I = 48.0 V / 120 Ω = 0.4 A. The power, P, through each resistor is calculated using the formula P = I^2 * R. Therefore, for the 24-ohm resistor, P is (0.4 A)^2 * 24 Ω = 3.84 W, and for the 96-ohm resistor, P is (0.4 A)^2 * 96 Ω = 15.36 W. When the resistors are connected in parallel, the total resistance is calculated using the formula 1/R = 1/R1 + 1/R2. The current through each resistor remains the same (0.4 A) and the power for each resistor can be calculated in the same way as described above.

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Answer:

Torque = 0.47 Nm

Explanation:

Torque = BIAsin∅

Since the plane is parallel to the magnetic field, ∅ = 90⁰

sin 90 = 1

Magnetic field, B = 0.30 T

current, I = 2.0 A

Radius, R = 0.50 m

A = πR²

A = π(0.5)² = 0.785 m²

Torque = 0.3 * 2 * 0.785

Torque = 0.47 Nm

Answer:

The magnetic torque when the plane of the loop is parallel to the magnetic field is zero.

Explanation:

Given;

radius of wire, r = 0.50 m

strength of magnetic field, B = 0.3 T

current in the wire, I = 2.0 A

Dipole moment, μ = I x A

where;

A is the area of the circular loop = πr² = π(0.5)² = 0.7855 m²

Dipole moment, μ = 2A x 0.7855m² = 1.571 Am²

magnetic torque, τ = μBsinθ

when the plane of the loop is parallel to the magnetic field, θ = 0

τ = 1.571 Am² x 0.3sin0

τ = 0

Therefore, the magnetic torque when the plane of the loop is parallel to the magnetic field is zero.

Answer:

dB = (-5 × 10⁻⁷k) T

Magnitude of dB = (-5 × 10⁻⁷) T; magnitude is actually (5 × 10⁻⁷) T in the mathematical sense of what magnitude is, but this question instructs to include the negative of the direction of dB is negative.

Direction is in the negative z-direction as evident from the sign on dB's vector notation.

Explanation:

From Biot Savart's relation, the magnetic field is given by

dB = (μ₀I/4πr³) (dL × r)

μ₀ = (4π × 10⁻⁷) H/m

I = 5.0 A

r = (2î) m

Magnitude of r = 2

dL = (4j)

(dL × r) is the vector product of both length of current carrying wire vector and the vector position of the point where magnetic field at that point is needed.

(dL × r) = (4j) × (2î)

​|i j k|

|0 4 0|

|2 0 0|

(dL × r) = (0î + 0j - 8k) = (-8k)

(μ₀I/4πr³) = (4π × 10⁻⁷ × 5)/(4π×2³)

(μ₀I/4πr³) = (6.25 × 10⁻⁸)

dB = (μ₀I/4πr³) (dL × r) = (6.25 × 10⁻⁸) × (-8k)

dB = (-5 × 10⁻⁷k) T

Magnitude of dB = (-5 × 10⁻⁷) T; magnitude is actually (5 × 10⁻⁷) T in the mathematical sense of what magnitude is, but this question instructs to include the negative of the direction of dB is negative.

Direction is in the negative z-direction as evident from the sign on dB's vector notation.

Hope this Helps!!!

When the distance is 4 m, the magnetic field strength is 2.5 x 10⁻⁷ T.

When the distance is 2 m, the magnetic field strength is 5 x 10⁻⁷ T.

The magnitude of magnetic field at any distance from a wire is determined by applying Biot-Savart law,

[tex]B = \frac{\mu_o I}{2\pi r}[/tex]

Where;

r is the distance from the conductor

When the distance is 4 m, the magnetic field strength is calculated as;

[tex]B = \frac{(4\pi \times 10^{-7}) \times5 }{2\pi \times 4} \\\\B = 2.5 \times 10^{-7} \ T[/tex]

When the distance is 2 m, the magnetic field strength is calculated as;

[tex]B = \frac{(4\pi \times 10^{-7} \times 5}{2\pi \times 2} \\\\B = 5 \times 10^{-7} \ T[/tex]

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Answer:

Explanation:.

Given that

First car velocity is 2m/s

Second car velocity is 9m/s.

At a certain time the first car is 8m from intersection

And at the same time second car is 6m from intersection.

The rate of change of distance, i.e dx/dt, which is the speed of the car.

Using Pythagoras theorem, their distance apart is given as

Z²=X²+Y²

Z²=6²+8²

Z²=36+64

Z²=100

Z=√100

Z=10m

Let assume the direction,

Let assume the first car is moving in positive x direction, then

dx/dt=2m/s

And also second car will be moving in negative y direction

dy/dt=-9m/s

Now, to know dz/dt, let use the Pythagoras formulae above

x²+y²=z²

differentiate with respect to t

2dx/dt+2dy/dt=2dz/dt

Divide through by 2

dz/dt=dx/dt+dy/dt

dz/dt=2-9

dz/dt=-7m/s

The rate of change of distance between the two body is -7m/s

Option B is correct

Answer:

Rate of change of the distance between the cars = -7 m/s

Explanation:

Let the distance between the first car and the intersection be p = 8 meters

Let the distance between the second car and the intersection be q = 6 meters

velocity of the first car, dp/dt = -2 m/s

velocity of the second car, dq/dt = -9 m/s

We can get the distance between the two cars using pythagora's theorem

s² = p² + q²...................................(1)

s² = 8² + 6²

s = √(8² + 6²)

s = 10 m

Differentiating equation (1) through with respect to t

2s ds/dt = 2p dp/dt + 2q dq/dt

(2*10*ds/dt) = (2*8*(-2)) + (2*6*(-9))

20 ds/dt = -32 - 108

20 ds/dt = -140

ds/dt = -140/20

ds/dt = -7 m/s

Answer:

a) 138 ft

b) 4.62 s

c) 1.375 s

d) 168.25 ft

Explanation:

The height of a rock (thrown from the top of a bridge) above the level of water surface as it varies with time when thrown is given in the question as

h = f(t) = -16t² + 44t + 138

with t in seconds, and h in feet

a) The bridge's height above the water.

At t=0 s, the rock is at the level of the Bridge's height.

At t = 0,

h = 0 + 0 + 138 = 138 ft

b) How many seconds after being thrown does the rock hit the water?

The rock hits the water surface when h = 0 ft. Solving,

h = f(t) = -16t² + 44t + 138 = 0

-16t² + 44t + 138 = 0

Solving this quadratic equation,

t = 4.62 s or t = -1.87 s

Since time cannot be negative,

t = 4.62 s

c) How many seconds after being thrown does the rock reach its maximum height above the water?

At maximum height or at the maximum of any function, the derivative of that function with respect to the independent variable is equal to 0.

At maximum height,

(dh/dt) = f'(t) = (df/dt) = 0

h = f(t) = -16t² + 44t + 138

(dh/dt) = (df/dt) = -32t + 44 = 0

32t = 44

t = (44/32)

t = 1.375 s

d) What is the rock's maximum height above the water?

The maximum height occurs at t = 1.375 s,

Substituting this for t in the height equation,

h = f(t) = -16t² + 44t + 138

At t = 1.375 s, h = maximum height = H

H = f(1.375) = -16(1.375²) + 44(1.375) + 138

H = 168.25 ft

Hope this Helps!!!

Answer:

observer see the leading edges of the two ships pass each other at time 0.136 s

Explanation:

given data

spaceship length = 100 m

speed of 0.700 Co = [tex]0.700\times 3 \times 10^8[/tex]  m/s

distance d = 58,000 km = 58000 × 10³ m

solution

as here distance will be half because both spaceship travel with same velocity

so they meet at half of distance

Distance Da = [tex]\frac{d}{2}[/tex]   ............1

Distance Da = [tex]\frac{58000\times 10^3}{2}[/tex]  

Distance Da = 29 ×[tex]10^{6}[/tex] m

and

time at which observer see leading edge of 2 spaceship pass

Δ time = [tex]\frac{Da}{v}[/tex]   ..........2

Δ time = [tex]\frac{29 \times 10^6}{0.700\times 3 \times 10^8}[/tex]

Δ time = 0.136 s

Answer:

The time at which the observer see the leading edges of the two ships pass each other is 0.138 sec.

Explanation:

Given that,

Length = 100 m

Speed = 0.700 c

Distance = 58000 km

The distance should be halved because the spaceships both travel the same speed.

So they will meet at the middle of the distance

We need to calculate the distance

Using formula for distance

[tex]d'=\dfrac{d}{2}[/tex]

Put the value into the formula

[tex]d'=\dfrac{58000}{2}[/tex]

[tex]d'=29000\ km[/tex]

We need to calculate the time at which the observer see the leading edges of the two ships pass each other

Using formula of time

[tex]\Delta t=\dfrac{d}{V}[/tex]

Put the value into the formula

[tex]\Delta t=\dfrac{29\times10^{6}}{0.700\times3\times10^{8}}[/tex]

[tex]\Delta t=0.138\ sec[/tex]

Hence, The time at which the observer see the leading edges of the two ships pass each other is 0.138 sec.

Explanation:

Below is an attachment containing the solution.

Answer:

The frequency is 1.3 kHz.

Explanation:

Given that,

Capacitance of the capacitor, [tex]C=1\ \mu F=10^{-6}\ C[/tex]

Inductance of the inductor, [tex]L=15\ mH=15\times 10^{-3}\ H[/tex]

We need to find the frequency when the inductive reactance equal the capacitive reactance such that :

[tex]X_c=X_L[/tex]

[tex]2\pi fL=\dfrac{1}{2\pi fC}[/tex]

[tex]f=\dfrac{1}{2\pi \sqrt{LC} }[/tex]

[tex]f=\dfrac{1}{2\pi \sqrt{15\times 10^{-3}\times 10^{-6}} }[/tex]

f = 1299.49 Hz

[tex]f=1.29\times 10^3\ Hz[/tex]

or

[tex]f=1.3\ kHz[/tex]

So, the frequency is 1.3 kHz. Therefore, the correct option is (d).

Answer: 9.59° and 350.41°

Explanation: The formulae that relates the force F exerted on a moving charge q with velocity v in a magnetic field of strength B is given as

F =qvB sin x

Where x is the angle between the strength of magnetic field and velocity of the charge.

q = 1.609×10^-19 C

v = 4×10³ m/s

B = 1.25 T

F = 1.40×10^-16 N

By substituting the parameters, we have that

1.40×10^-16 = 1.609×10^-19 × 4×10³ × 1.25 × sinx

sin x = 1.40×10^-16/ 1.609×10^-19 × 4×10³ × 1.25

sin x = 1.40×10^-16 /8.045*10^(-16)

sin x = 0.1666

x = 9.59°

The value of sin x is positive in first and fourth quadrant.

Hence to get the second value of x, we move to the 4th quadrant of the trigonometric quadrant which is 360 - x

Hence = 360 - 9.59 = 350.41°

Final answer:

The angle between the velocity of the electron and the magnetic field is approximately 6.27° or 173.73°.

Explanation:

To find the angle between the velocity of the electron and the magnetic field, we can use the formula:

F = q(vsinθ)B

Where F is the magnetic force, q is the charge of the electron, v is the velocity of the electron, θ is the angle between the velocity and the magnetic field, and B is the magnetic field strength.

In this case, the magnetic force is given as 1.40 × 10^-16 N, the charge of an electron is 1.6 × 10^-19 C, the velocity of the electron is 4.00 × 10³ m/s, and the magnetic field strength is 1.25 T.

Plugging in these values, we can solve for θ:

1.40 × 10^-16 N = (1.6 × 10^-19 C)(4.00 × 10³ m/s)(sinθ)(1.25 T)

Solving for sinθ:

sinθ = (1.40 × 10^-16 N) / [(1.6 × 10^-19 C)(4.00 × 10³ m/s)(1.25 T)]

sinθ ≈ 0.109

Taking the inverse sine of 0.109, we find that θ ≈ 6.27° or θ ≈ 173.73°.

Explanation:

It is known that the relation between speed and distance is as follows.

               velocity = [tex]\frac{distance}{time}[/tex]

As it is given that velocity is 6 m/s and distance traveled by the bear is (d + 29). Therefore, time taken by the bear is calculated as follows.

         [tex]t_{bear} = \frac{(d + 29)}{6 m/s}[/tex] ............. (1)

As the tourist is running in a car at a velocity of 4.2 m/s. Hence, time taken by the tourist is as follows.

              [tex]t_{tourist} = \frac{d}{4.2}[/tex] ............. (2)

Now, equation both equations (1) and (2) equal to each other we will calculate the value of d as follows.

              [tex]t_{bear} = t_{tourist}[/tex]

       [tex]\frac{(d + 29)}{6 m/s}[/tex] = [tex]\frac{d}{4.2}[/tex]

                   4.2d + 121.8 = 6d

                         d = [tex]\frac{121.8}{1.8}[/tex]

                            = 67.66

Thus, we can conclude that the maximum possible value for d is 67.66.

the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is approximately [tex]\( 0.47 \, \text{N} \cdot \text{m} \)[/tex].

The magnetic torque [tex]\( \tau \)[/tex] on a current-carrying loop in a uniform magnetic field can be calculated using the formula:

[tex]\[ \tau = N \cdot I \cdot A \cdot B \cdot \sin(\theta) \][/tex]

Where:

- N is the number of turns in the loop (1 in this case, as there is a single loop),

- I is the current flowing through the loop (2.0 A),

- A is the area of the loop (πr² for a circular loop, where \( r \) is the radius),

- B is the magnetic field strength (0.30 T),

- [tex]\( \theta \)[/tex] is the angle between the normal to the loop's plane and the magnetic field (90° in this case, as the loop is perpendicular to the magnetic field).

Substituting the given values:

[tex]\[ \tau = (1) \cdot (2.0 \, \text{A}) \cdot (\pi \times (0.50 \, \text{m})^2) \cdot (0.30 \, \text{T}) \cdot \sin(90°) \][/tex]

[tex]\[ \tau = 2.0 \cdot \pi \cdot (0.50)^2 \cdot 0.30 \][/tex]

[tex]\[ \tau = 2.0 \cdot \pi \cdot 0.25 \cdot 0.30 \][/tex]

[tex]\[ \tau = 0.15 \cdot \pi \][/tex]

[tex]\[ \tau \approx 0.47 \, \text{N} \cdot \text{m} \][/tex]

So, the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is approximately [tex]\( 0.47 \, \text{N} \cdot \text{m} \)[/tex].

Therefore, the correct answer choice is [tex]\(\boxed{\text{0.47 N*m}}\)[/tex].

The complete Question is given below:

A flat circular loop of wire of radius 0.50 m that is carrying a 2.0-A current is in a uniform magnetic field of 0.30 T. What is the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field?

Answer choices.

.41N*m

.00N*m

.47N*m

.58N*m

.52N*m

The given question is incomplete. The complete question is as follows.

An oxygen molecule is adsorbed onto a small patch of the surface of a catalyst. It's known that the molecule is adsorbed on 1 of 36 possible sites for adsorption. Calculate the entropy of this system.

Explanation:

It is known that Boltzmann formula of entropy is as follows.

             s = k ln W

where,   k = Boltzmann constant

              W = number of energetically equivalent possible microstates or configuration of the system

In the given case, W = 36. Now, we will put the given values into the above formula as follows.

                  s = k ln W

                    = [tex]1.38 \times 10^{-23} ln (36)[/tex]        

                    = [tex]4.945 \times 10^{-23} J/K[/tex]

Thus, we can conclude that the entropy of this system is [tex]4.945 \times 10^{-23} J/K[/tex].

The entropy of the given system is [tex]4.954 \times 10^{-23 } \rm\; J/K[/tex]  in which an oxygen molecule is absorbed in one of the 36 possible states.

From the Boltzmann formula of entropy,

[tex]s = k \times ln W[/tex]

Where,  

[tex]k[/tex]= Boltzmann constant = [tex]1.38\times 10^{-23} \rm\; J/K[/tex]

[tex]W[/tex] = Number of energetically equivalent possible microstates of the system = 36

Put the values in the formula  

[tex]s =1.38\times 10^{-23} \times ln (36)\\\\s = 4.954 \times 10^{-23 } \rm\; J/K[/tex]

Therefore, the entropy of the given system is [tex]4.954 \times 10^{-23 } \rm\; J/K[/tex]

To know more about entropy,

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Answer:

a. Increases

Explanation:

By the definition we know that the acceleration is the rate of change in velocity and here we have acceleration component in the direction of motion of the block.

Mathematically:

when the body is moving down:

[tex]v=u+gt[/tex]

where:

[tex]v=[/tex] final velocity of the block

[tex]u=[/tex] initial velocity of the block

[tex]g=[/tex] acceleration due to gravity

[tex]t=[/tex] time of observation during the instance of motion

As we know that kinetic energy is given as:

[tex]KE=\frac{1}{2} \times m.v^2[/tex]

where:

[tex]m=[/tex] mass of the block which remains constant (macroscopically)

[tex]v=[/tex] velocity of the block (which increases here as the body descends)

Explanation:

Thermodynamic Laws -

Standard Electronic potential  -

Aerobic grow is much more efficient at making ATP because of the presence of oxygen the cycles in the respiration are carried out at an efficient rate which forces the cell to undergo a much large amount of ATP production.

Answer:

Velocity at 64 mm is 0.532 m/s

Maximum displacement = 0.082 m or 82 mm

Explanation:

The maximum displacement or amplitude is determined by the applied force from Hooke's law

[tex]F = kA[/tex]

[tex]A=\dfrac{F}{k}=\dfrac{14 \text{ N}}{170 \text{ N/m}}=0.082 \text{ m}[/tex]

The velocity at at any point, x, is given by

[tex]v=\sqrt{\dfrac{k}{m}(A^2 - x^2)}[/tex]

m is the mass of the load, here the cylinder.

In fact, the expression [tex]\sqrt{\dfrac{k}{m}}[/tex] represents the angular velocity, [tex]\omega[/tex].

Substituting given values,

[tex]v=\sqrt{\dfrac{170}{1.5}(0.082^2 - 0.065^2)} = 0.532 \text{ m/s}[/tex]

Answer:

0.1111 W/m²

Explanation:

If all other parameters are constant, sound intensity is inversely proportional to the square of the distance of the sound. That is,

I ∝ (1/r²)

I = k/r²

Since k can be the constant of proportionality. k = Ir²

We can write this relation as

I₁ × r₁² = I₂ × r₂²

I₁ = 0.25 W/m²

r₁ = 16 m

I₂ = ?

r₂ = 24 m

0.25 × 16² = I₂ × 24²

I₂ = (0.25 × 16²)/24²

I₂ = 0.1111 W/m²

The sound intensity at a distance of 28 m from the generator will be "0.1111 W/m²".

According to the question,

Distance, r₁ = 16 m

                r₂ = 28 m

Sound intensity, I₁ = 0.25 W/m²

We know the relation,

→ I ∝ ([tex]\frac{1}{r^2}[/tex])

or,

  I ∝ [tex]\frac{k}{r^2}[/tex]

Now,

→ I₁ × r₁² = I₂ × r₂²

By substituting the values,

0.25 × (16)² = I₂ × (24)²

                I₂ = [tex]\frac{0.25\timers (16)^2}{(24)^2}[/tex]

                   = 0.1111 W/m²

Thus the above answer is correct.

Find out more information about sound intensity here:

https://brainly.com/question/26672630

The fundamental frequency of a string fixed at both ends is f₁ = c/2L, where 'c' is calculated from the tension and linear mass density of the string. Using the given tension and density, one can find the wave speed 'c' first, then substitute into the formula to get the fundamental frequency.

To calculate the fundamental frequency of a string fixed at both ends, you can use the formula for the fundamental frequency of a string, which is given by f1 = c/2L, where f1 is the fundamental frequency, c is the speed of the wave on the string, and L is the length of the string.

To find the speed of the wave on the string, we use the formula c = \/(T/μ), where c is the speed of the wave, T is the tension in the string, and μ is the linear mass density of the string. Substituting the given values, c = \/(357 N / (0.02 g/cm * 100 cm/m)) = \/(357 / 0.0002 kg/m), we can calculate c and then use the result to find f1.

Answer:

[tex]4.725 m/s^{2}[/tex]

Explanation:

We know that from Newton's second law of motion, F=ma hence making acceleration the subject then [tex]a=\frac {F}{m}[/tex]  where a is acceleration, F is force and m is mass

Also making mass the subject of the formula [tex]m=\frac {F}{a}[/tex]

For [tex]m1= \frac {F}{13.5}[/tex] and [tex]m2=\frac {F}{3.5}[/tex] hence [tex]F=(m2-m1)a= (\frac {F}{3.5}-\frac {F}{13.5})a=0.2116402116\\\frac {1}{a}=0.2116402116\\a=4.725 m/s^{2}[/tex]

Answer:

(a) strikes the plane at the same time as the other body

Explanation:

Two bodies are falling with negligible air resistance, side by side, above a horizontal plane near Earth's surface. If one of the bodies is given an additional horizontal acceleration during its descent, it strikes the plane at the same time as the other body

The velocity of the ball just before it hits the ground, thrown upward from 2 meters with an initial velocity of 10 m/s, can be calculated using kinematic equations and is approximately 11.8 m/s.

To determine the velocity of the ball just before it hits the ground, we can use kinematic equations. One equation that relates initial velocity, acceleration due to gravity, and final velocity is:

v² = u² + 2as

Where:

In this case, the ball is tossed upward so we consider the acceleration due to gravity as negative (-9.8 m/s²). The displacement, in this case, is 2 m (the height from which the ball is thrown).

Insert the known values into the equation:

v² = (10 m/s)² + 2 × (-9.8 m/s²) × (-2 m)

The negative signs cancel for the displacement and acceleration due to the upward throw and the downward acceleration, which will give us a positive number.

v² = 100 + (19.6 × 2)

v² = 100 + 39.2

v² = 139.2

v = √139.2

v ≈ 11.8 m/s

Therefore, the velocity of the ball just before it hits the ground is approximately 11.8 m/s.

Answer:

13.875 T

Explanation:

Parameters given:

Length of solenoid, L = 2.5 cm = 0.025 m

Radius of solenoid, r = 0.75 cm = 0.0075 m

Number of turns, N = 25 turns

Current, I = 1.85 A

Magnetic field, B, is given as:

B = (N*r*I) /L

B = (25 * 0.0075 * 1.85)/0.025

B = 13.875 T

Answer:

0.308 m/s2 at an angle of 13.5° below the horizontal

Explanation:

The parallel acceleration to the roadway is the tangential acceleration on the rise.

The normal acceleration is the centripetal acceleration due to the arc. This is given by

[tex]a_N = \dfrac{v^2}{r} = \dfrac{36^2}{500}=0.072[/tex]

The tangential acceleration, from the question, is

[tex]a_T = 0.300[/tex]

The magnitude of the total acceleration is the resultant of the two accelerations. Because these are perpendicular to each other, the resultant is given by

[tex]a^2 =a_T^2 + a_N^2 = 0.300^2 + 0.072^2[/tex]

[tex]a = 0.308[/tex]

The angle the resultant makes with the horizontal is given by

[tex]\tan\theta=\dfrac{a_N}{a_T}=\dfrac{0.072}{0.300}=0.2400[/tex]

[tex]\theta=13.5[/tex]

Note that this angle is measured from the horizontal downwards because the centripetal acceleration is directed towards the centre of the arc

Final answer:

The magnitude of the force exerted on the right cube by the middle cube, when three identical cubes accelerate together on a frictionless surface, is calculated using Newton's second law (F = ma) and is found to be 16.0 Newtons.

Explanation:

The student's question pertains to Newton's second law of motion and the concept of force and acceleration. The problem involves three identical cubes on a frictionless surface accelerating due to a force applied to the first cube. To find the magnitude of the force exerted on the right cube by the middle cube, we use the formula F = ma, where F is the force, m is the mass, and a is the acceleration.

Since the cubes are identical and accelerate together, each 4.0-kg cube has an acceleration of 4.0 m/s². Therefore, the force exerted by one cube on another can be found by multiplying one cube's mass by the given acceleration:

The force exerted on the right cube by the middle cube is 16.0 Newtons.