High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbed from a climber's body under the following conditions. The specific heat of ice is 2100 J/kg⋅C∘, the latent heat of fusion is 333 kJ/kg, the specific heat of water is 4186 J/kg⋅C∘

Answer:

μk = 0.408

Explanation:

Given:

m=0.50 Kg,

Let compressed distance x = 0.20 m, and

stretched distance after releasing y = 1.00 m

K = 100 N/M

Sol:

Law of conservation of energy

Energy dissipation due to friction = P.E stores in the spring

Ff * y = 1/2 K x ²   (Ff = μk Fn)  And (Fn = mg) so

μk mgy  =   1/2 K x ²

μk = 1/2 K x ² /mgy   Putting values

μk = (1/2 ) (100 N/M) (0.20 m)² / (0.50 Kg x 9.8 m/s² x 1 m)

μk = 0.408

0.4

(i) Since the mass is forced against the spring, an elastic energy ([tex]E_{E}[/tex]) due to the compression of the spring by the force is produced and is given according to Hooke's law by;

[tex]E_{E}[/tex] = [tex]\frac{1}{2}[/tex] k c²            --------------------------------(i)

Where;

k = spring's constant

c = compression caused on the spring.

From the question;

k = 100N/m

c = 0.20m

Substitute these values into equation (i) as follows;

[tex]E_{E}[/tex] = [tex]\frac{1}{2}[/tex] x 100 x 0.20²

[tex]E_{E}[/tex] = 2J

(ii) Now, when the mass is released, it causes the block to move some distance until it stops thereby doing some work within that distance. This means that the elastic energy is converted to workdone. i.e

[tex]E_{E}[/tex] = W          ------------(ii)

The work done (W) is given by the product of the net force(F) on the block and the distance covered(s). i.e

W = F x s           -----------------(iii)

But, the only force acting on the body as it moves is the frictional force ([tex]F_{R}[/tex]) acting to oppose its motion. i.e

F = [tex]F_{R}[/tex]

Where;

[tex]F_{R}[/tex] = μN      [μ = coefficient of kinetic friction, N = mg = normal reaction between the block and the tabletop, m = mass of the block, g = gravity]

[tex]F_{R}[/tex] = μ x mg

Substitute F = [tex]F_{R}[/tex] = μ x mg into equation (iii) as follows;

W = μ x mg x s             ----------------(iv)

Now substitute the value of W into equation (ii) as follows;

[tex]E_{E}[/tex] = μ x mg x s                ------------------(v)

Where;

[tex]E_{E}[/tex] = 2 J               [as calculated above]

m = 0.50 kg

s = distance moved by block = 1.00m

g = 10m/s²            [a known constant]

Substitute these values into equation (v) as follows;

2 = μ x 0.50 x 10 x 1

2 = 5μ

μ = 2 / 5

μ = 0.4

Therefore, the coefficient of kinetic friction between the block and the tabletop is 0.4

Answer:

The magnitude of the net electric field is [tex]6.57\times10^{5}\ N/C[/tex]

Explanation:

Given that,

Charge density [tex]\lambda = 4.54\ \mu C/m[/tex]

Charge density [tex]\lambda' = -2.58\ \mu C/m[/tex]

Distance [tex]y_{1}= 0.384\ m[/tex]

Distance [tex]y_{2}= 0.204\ m[/tex]

We need to calculate the magnitude of the net electric field

Using formula of electric field

[tex]E=E_{1}+E_{2}[/tex]

[tex]E=\dfrac{1}{2\pi\epsilon_{0}}(\dfrac{\lambda}{r}+\dfrac{\lambda'}{r'})[/tex]

Put the value into the formula

[tex]E=\dfrac{1}{2\pi\times8.85\times10^{-12}}(\dfrac{4.54\times10^{-6}}{0.204}+\dfrac{2.58\times10^{-6}}{0.384-0.204})[/tex]

[tex]E=6.57\times10^{5}\ N/C[/tex]

Hence, The magnitude of the net electric field is [tex]6.57\times10^{5}\ N/C[/tex]

Answer:

maximum value of the magnetic field  B  = 1 ×[tex]10^{-8}[/tex] T

average intensity of the light = 0.011937 W/m²

power of source =  14.40 J

Explanation:

given data

maximum electric field E = 3.0 V/m

distance from a point source r = 9.8 m

solution

first we get here maximum value of the magnetic field  

maximum value of the magnetic field   = [tex]\frac{E}{C}[/tex]   .........1

maximum value of the magnetic field   = [tex]\frac{3}{3 \times 10^8}[/tex]

maximum value of the magnetic field  B  = 1 ×[tex]10^{-8}[/tex] T

and

now we get average intensity of the light that is

average intensity of the light =  [tex]\frac{EB}{2\mu _o}[/tex]   .........2

average intensity of the light = [tex]\frac{3 \times 1 \times 10^{-8}}{2 \times 4\pi \times 10^{-7}}[/tex]  

average intensity of the light = 0.011937 W/m²

and

now we get power of source that is express as

power of source = average intensity × 4×π×r²   ..........3

power of source = 0.011937 × 4×π×9.8²

power of source =  14.40 J

Answer:

(a) the magnitude of the magnetic moment of the loop is  0.75 Am²

(b) the torque the magnetic field exerts on the loop is 0.028 N.m

Explanation:

Given;

number of turns, N = 15

width of the loop, w = 10 cm = 0.1 m

length of loop, L = 20 cm = 0.2 m

current through the loop, I  = 2.5 A

strength of the magnetic field, B = 0.037 T

Area of the loop, A = L x w = 0.2 x 0.1 = 0.02 m²

Part (a) the magnitude of the magnetic moment of the loop

μ = NIA

where;

μ is the magnitude of the magnetic moment of the loop

μ = 15 x 2.5 x 0.02 = 0.75 Am²

Part (b) the torque the magnetic field exerted on the loop

τ = μB

where;

τ is the torque the magnetic field exerts on the loop

τ = μB = 0.75 x 0.037 = 0.028 N.m

Given Information:

Magnetic field = B =  0.037 T

Current = I = 2.5 A

Number of turns = N = 15 turns

Length of rectangular coil = L = 20 cm = 0.20 m

Width of rectangular coil = W = 10 cm = 0.10 m

Required Information:

(a) Magnetic moment = µ = ?

(b) Torque = τ = ?

Answer:

(a) Magnetic moment = 0.75 A.m ²

(b) Torque = 0.0277 N.m

Explanation:

(a) The magnetic moment µ is given by

µ = NIA

Where µ is the magnetic, N is the number of turns, I is the current, A is the area of rectangular loop and is given by

A = W*L

A = 0.10*0.20

A = 0.02 m²  

µ = 15*2.5*0.02

µ = 0.75 A.m²

(b) The toque τ exerted on current carrying loop with A area in the presence of a magnetic field B is given by

τ = NIAB

τ = 15*2.5*0.02*0.037

τ = 0.0277 N.m

Alternatively,

τ = µB

τ = 0.75*0.037

τ = 0.0277 N.m

Explanation:

Below is an attachment containing the solution.

Answer:

the question is incomplete, below is the complete question

"Determine the power required for a 1150-kg car to climb a 100-m-long uphill road with a slope of 30 (from horizontal) in 12 s (a) at a constant velocity, (b) from rest to a final velocity of 30 m/s, and (c) from 35 m/s to a final velocity of 5 m/s. Disregard friction, air drag, and rolling resistance."

a. [tex]47KW[/tex]

b. 90.1KW

c. -10.5KW

Explanation:

Data given

mass m=1150kg,

length,l=100m

angle, =30

time=12s

Note that the power is define as the rate change of the energy, note we consider the potiential and the

Hence

[tex]Power,P=\frac{energy}{time} \\P=\frac{mgh+1/2mv^{2}}{t}[/tex]

to determine the height, we draw the diagram of the hill as shown in the attached diagram

a. at constant speed, the kinetic energy is zero.Hence the power is calculated as

[tex]p=\frac{mgh}{t} \\p=\frac{1150*9.81*100sin30}{12}\\p=47*10^{3}W\\[/tex]

b.

for a change in velocity of 30m/s

we have the power to be

[tex]P=\frac{mgh}{t} +\frac{1/2mv^2}{t}\\ P=47Kw+\frac{ 0.5*1150*30^2}{12}\\ P=47kw +43.1kw\\P=90.1KW[/tex]

c. when i decelerate, we have the power to be

[tex]P=\frac{mgh}{t} +\frac{1/2mv^2}{t}\\ P=47Kw+\frac{ 0.5*1150*\\((5^2)-(35^2))}{12}\\ P=47kw -57.5kw\\P=-10.5KW[/tex]

Answer:

2.

Explanation:

v=6ft/sec

To determine the velocity of the block when it moves a distance of 30 ft, we can use the equations of motion and consider the force applied and the friction acting on the block. First, let's find the acceleration of the block using the force applied. Next, we need to consider the frictional force acting on the block. Finally, we can calculate the net force and use the equation of motion to determine the velocity of the block.

To determine the velocity of the block when it moves a distance of 30 ft, we can use the equations of motion and consider the force applied and the friction acting on the block.

First, let's find the acceleration of the block using the force applied. The force applied is given by F = 8t^2 lb, where t is the time in seconds. So, at t = 0, the force is F = 0 lb, and at t = 1, the force is F = 8 lb.

We can now use the equations of motion to find the acceleration. Since the mass of the block is not given, we can assume it to be 10 lb (as mentioned in the question). From the second law of motion, F = ma, where F is the net force, m is the mass, and a is the acceleration. Therefore, 8 = 10a. Solving for a, we get a = 0.8 ft/s^2.

Next, we need to consider the frictional force acting on the block. The coefficient of kinetic friction is given as μs = 0.2. The frictional force, Ff, can be calculated using Ff = μs * N, where N is the normal force. The normal force is equal to the weight of the block, N = mg, where g is the acceleration due to gravity, approximately 32 ft/s^2. So, N = 10 * 32 = 320 lb. Substituting the values, we get Ff = 0.2 * 320 = 64 lb.

Now, we can calculate the net force acting on the block. The net force, Fn, is given as Fn = F - Ff, where F is the force applied. Substituting the values, Fn = 8t^2 - 64 lb.

Finally, we can use the equation of motion, v^2 = u^2 + 2as, to determine the velocity of the block when it moves a distance of 30 ft. Here, v is the final velocity, u is the initial velocity (which is 0 ft/s because the block starts from rest), a is the acceleration, and s is the distance traveled. Substituting the values, v^2 = 0 + 2 * 0.8 * 30. Solving for v, we find

v = 6 ft/s.

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Taking moments about B

N x L'' = W x L'

Here the moment is = force x perpendicular distance between the axis of rotation and the point of applied force .

Here L' = L/3 and L'' = 9

Thus from figure

471 x 9 = W x [tex]\frac{L}{3}[/tex]

But L'' = [tex]\frac{1}{2}[/tex]( L - [tex]\frac{L}{3}[/tex] ) = [tex]\frac{L}{3}[/tex] = 9

Thus W = 471 N

The value of the weight ( W ) is ; 471 N

First step : take moments about B

N * L" = W * L'

Where : L' = L/3,  L" = 9

From the figure

471 * 9 = W * [tex]\frac{L}{3}[/tex]

also:

L" = 1/2 ( L - L/3 ) = L/3 = 9

Hence ; W = 471 N

Hence we can conclude that The value of the weight ( W ) is ; 471 N

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Answer:46.05 mm

Explanation:

Given

[tex]Power=260\ hp\approx 260\times 746=193.96\ KW[/tex]

speed [tex]N=550\ rev/min[/tex]

allowable shear stress [tex](\tau )_{max}=15\ kpsi\approx 103.421\ MPa[/tex]

Power is given by

[tex]P=\frac{2\pi NT}{60}[/tex]

[tex]193.96=\frac{2\pi 550\times T}{60}[/tex]

[tex]T=3367.6\ N-m[/tex]

From Torsion Formula

[tex]\frac{T}{J}=\frac{\tau }{r}-----1[/tex]

where J=Polar section modulus

T=Torque

[tex]\tau [/tex]=shear stress

For square cross section

[tex]r=\frac{a}{2}[/tex]

where a=side of square

[tex]J=\frac{a^4}{6}[/tex]

Substituting the values in equation 1

[tex]\frac{3376.6}{\frac{a^4}{6}}=\frac{103.421\times 10^6}{\frac{a}{2}}[/tex]

[tex]a=0.04605\ m[/tex]

[tex]a=46.05\ mm[/tex]

A square cross-section solid steel shaft of approximately 5 inches on each side can transmit 260 hp at 550 rev/min without exceeding a maximum allowable shear stress of 15 kpsi.

The size of the steel shaft can be calculated by using the power-transmitting capacity of a shaft equation, which states:

P = (16πNT) / (60 * 33000), where P is power in hp, N is rotational speed in rev/min, and T is torque in lb-ft.

We also know that the shear stress (τ) is given by the equation: τ = (16T) / (πd^3), where d is the diameter in inches.

To find the correct torque, we start by rearranging the power equation for T: T = (P * 60 * 33000) / (16πN).

Substituting in the given power and rotational speed, we find that the torque is approximately 13400 lb-ft.

We then substitute this value and the allowable shear stress into the shear stress equation, solving for d to get d ≈ 5 inches.

So, a square cross-section solid steel shaft about 5 inches on each side should be able to transmit the given power at the stated speed without exceeding the maximum allowable shear stress.

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Answer:

d. There is no way of knowing the phase composition without more information.

Explanation:

There is no way to ascertain the phases present in the container when equilibrium is established because we are not furnished with enough information.

Final answer:

The force exerted on the water in the MHD drive utilizing a 25.0 cm diameter tube, with a 125 A current and a 1.95 T magnetic field, is 60.94 N.

Explanation:

To find the force exerted on the water in the MHD drive, we can use the formula for the magnetic force:

F = ILB

Where F is the force, I is the current, L is the length of the wire (in this case, the 25.0-cm diameter tube), and B is the magnetic field strength.

Plugging in the given values, we have:

F = (125 A)(0.25 m)(1.95 T)

Solving for F, we get:

F = 60.94 N

Therefore, the force exerted on the water in the MHD drive is 60.94 N.

Final answer:

The force exerted on the water in an MHD drive utilizing a 25.0 cm diameter tube, with a 125 A current and a 1.95 T magnetic field, is approximately 118.13 N.

Explanation:

The force exerted on the water in an MHD drive can be calculated using the formula:

Force (N) = Current (A) * Magnetic Field (T) * Area (m²)

Given that the diameter of the tube is 25.0 cm and the current is 125 A, we can calculate the area as follows:

Area = π * (radius)²

Area = π * (0.125 m)²

Substituting the values into the formula:

Force = 125 A * 1.95 T * (π * (0.125 m)²)

Force ≈ 118.13 N

Therefore, the force exerted on the water in the MHD drive is approximately 118.13 N.

Answer: (a) peak emf = 13.38V

(b) frequency = 591.78Hz

Explanation: Please see the attachments below

Final answer:

To find the distance L that the block will travel up the incline before stopping, we apply conservation of energy, accounting for the initial spring potential energy, the gravitational potential energy, and the work done against kinetic friction. By setting up the energy equation and substituting the given values, we can solve for L. Therefore, the value of L is approximately [tex]\( 0.25069 \)[/tex] m

Explanation:

To determine the distance L that the block will travel up the incline before coming to rest, we need to use the conservation of energy principle. The mechanical energy conserved will be the initial potential energy stored in the spring when compressed and the final kinetic energy of the block up the slope, taking into account the work done against friction.

Initially, the spring's potential energy (Us) is given by Us = 1/2 k d^2, where k is the spring constant and d is the compression distance. As the spring pushes the block up the incline, the block gains gravitational potential energy (Ug = mgh), does work against friction (Wf), and could have some residual kinetic energy (which is zero at the highest point).

The work done against friction can be found by Wf = μk N L, where μk is the coefficient of kinetic friction, N is the normal force (N=m*g*cos(θ)), and L is the distance traveled. Since we are considering the point where the block comes to rest, we set the total mechanical energy at this position equal to the initial energy.

Equating the initial and final energies we get: (1/2 k d^2) = mgh + μk m*g*cos(θ)L. We can now solve for L by plugging in the known values: m = 2.5 kg, k = 940 N/m, d = 0.13 m, μk = 0.11, g = 9.8 m/s^2, and θ = 29 degrees.

Substituting these values, we solve for L:

L = [(1/2) * 940 N/m * (0.13 m)^2 - 2.5 kg * 9.8 m/s^2 * sin(29 degrees)] / [2.5 kg * 9.8 m/s^2 * cos(29 degrees)* 0.11]

let's break it down step by step.

Given:

[tex]- Spring constant: \(k = 940 \, \text{N/m}\)\\- Displacement: \(x = 0.13 \, \text{m}\)\\- Mass: \(m = 2.5 \, \text{kg}\)\\- Gravitational acceleration: \(g = 9.8 \, \text{m/s}^2\)\\- Angle: \(\theta = 29^\circ\)\\- Length: \(l = 0.11 \, \text{m}\)[/tex]

Let's calculate it step by step:

1. Calculate the potential energy stored in the spring using the formula:

[tex]\[ U_{\text{spring}} = \frac{1}{2} k x^2 \][/tex]

Substitute the given values:

[tex]\[ U_{\text{spring}} = \frac{1}{2} \times 940 \, \text{N/m} \times (0.13 \, \text{m})^2 \]\[ U_{\text{spring}} = \frac{1}{2} \times 940 \times 0.0169 \, \text{N} \cdot \text{m}^2 \]\[ U_{\text{spring}} = 7.963 \, \text{J} \][/tex]

2. Calculate the gravitational potential energy using the formula:

[tex]\[ U_{\text{gravitational}} = mgh \]\\where \( h \) is the vertical height.\[ h = l \sin(\theta) \]\[ h = 0.11 \, \text{m} \times \sin(29^\circ) \]\[ h \approx 0.055 \, \text{m} \][/tex]

Substitute the values into the gravitational potential energy formula:

[tex]\[ U_{\text{gravitational}} = 2.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.055 \, \text{m} \]\[ U_{\text{gravitational}} = 1.3525 \, \text{J} \][/tex]

3. Now, plug these values into the given expression:

[tex]\[ L = \frac{(7.963 \, \text{J} - 1.3525 \, \text{J})}{(2.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \cos(29^\circ) \times 0.11 \, \text{m})} \][/tex]

Let's evaluate the denominator first:

[tex]\[ 2.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \cos(29^\circ) \times 0.11 \, \text{m} \]\[ = 2.5 \times 9.8 \times \cos(29^\circ) \times 0.11 \]\[ \approx 26.368 \][/tex]

Now, let's plug it back into the expression:

[tex]\[ L = \frac{(7.963 \, \text{J} - 1.3525 \, \text{J})}{26.368} \]\[ L = \frac{6.6105}{26.368} \]\[ L \approx 0.25069 \][/tex]

Therefore, the value of L is approximately [tex]\( 0.25069 \)[/tex] m.

Answer:

d = <23, 33, 0> m ,    F_W = <0, -9.8, 0> ,   W = -323.4 J

Explanation:

We can solve this exercise using projectile launch ratios, for the x-axis the displacement is

         x = vox t

Y Axis  

         y = [tex]v_{oy}[/tex] t - ½ g t²

It's displacement is

      d = x i ^ + y j ^ + z k ^

Substituting

      d = (23 i ^ + 33 j ^ + 0) m

Using your notation

   d = <23, 33, 0> m

The force of gravity is the weight of the body

         W = m g

        W = 1  9.8 = 9.8 N

In vector notation, in general the upward direction is positive

         W = (0 i ^  - 9.8 j ^ + 0K ^) N

         W = <0, -9.8, 0>

Work is defined

           W = F. dy

             W = F dy cos θ

In this case the force of gravity points downwards and the displacement points upwards, so the angle between the two is 180º

          Cos 180 = -1

           W = -F y

           W = - 9.8 (33-0)

           W = -323.4 J

Answer:

Explanation:

Check attachment for solution

Answer:

E = I(R + r)

Making I the subject of the formular by dividing both sides by R + r,

I = E/(R + r)

E = 4V, r = 0.7Ohm, R = 0

I = 4/(0 + 0.7) = 4/0.7

I = 5.174285714A

Explanation:

For a cell of emf E, internal resistance r, connected to an external resistance R, the current flowing through the circuit will be given as:

I = E/(R + r). I is measured in Amperes(A), emf in volts(V), R in Ohms and internal resistance r also in ohms

(a) The baseball's speed must be approximately 6.89 m/s.

(b) The bullet has greater kinetic energy.

To determine the baseball's speed, we use the principle of conservation of momentum, which states that the total momentum of an isolated system remains constant. The momentum of the bullet before the pitch must equal the combined momentum of the baseball and the pitcher afterward. By equating the momenta and solving for the baseball's speed, we find it to be approximately 6.89 m/s.

Now, to compare the kinetic energies of the baseball and the bullet, we use the kinetic energy formula, which is proportional to the square of the velocity. Despite the baseball having a larger mass, the bullet's significantly higher velocity results in greater kinetic energy. This is due to the quadratic relationship between velocity and kinetic energy.

In conclusion, the baseball must travel at around 6.89 m/s to match the claimed momentum. However, the bullet still possesses greater kinetic energy due to its higher speed, highlighting the importance of velocity in determining kinetic energy.

a) The speed of the baseball must be [tex]\( {31.0 \, \text{m/s}} \)[/tex] to match the momentum of the bullet.

b) The bullet has significantly greater kinetic energy [tex](\(3375 \, \text{J}\))[/tex] compared to the baseball [tex](\(69.86 \, \text{J}\))[/tex].

To solve the problem, we will use the principles of momentum and kinetic energy.

Part (a): Speed of the Baseball

First, we need to calculate the momentum of the bullet and then find the speed at which the baseball must be thrown to have the same momentum.

1. Momentum of the Bullet:

  - Mass of the bullet [tex](\(m_b\))[/tex]: [tex]\(3.00 \, \text{g} = 0.003 \, \text{kg}\)[/tex]

  - Speed of the bullet [tex](\(v_b\))[/tex]: [tex]\(1.50 \times 10^3 \, \text{m/s}\)[/tex]

  Momentum [tex](\(p\))[/tex] is given by:

 [tex]\[ p = m_b \cdot v_b \][/tex]

  [tex]\[ p = 0.003 \, \text{kg} \times 1.50 \times 10^3 \, \text{m/s} \][/tex]

 [tex]\[ p = 4.50 \, \text{kg} \cdot \text{m/s} \][/tex]

2. Speed of the Baseball:

  - Mass of the baseball [tex](\(m_{bb}\)): \(0.145 \, \text{kg}\)[/tex]

  - Let the speed of the baseball be [tex]\(v_{bb}\).[/tex]

  We want the baseball to have the same momentum as the bullet:

 [tex]\[ p = m_{bb} \cdot v_{bb} \][/tex]

 [tex]\[ 4.50 \, \text{kg} \cdot \text{m/s} = 0.145 \, \text{kg} \cdot v_{bb} \][/tex]

Solving for [tex]\(v_{bb}\):[/tex]

 [tex]\[ v_{bb} = \frac{4.50 \, \text{kg} \cdot \text{m/s}}{0.145 \, \text{kg}} \][/tex]

  [tex]\[ v_{bb} = 31.034 \, \text{m/s} \][/tex]

So, the baseball must be thrown with a speed of approximately [tex]\( \boxed{31.0 \, \text{m/s}} \).[/tex]

Part (b): Kinetic Energy Comparison

To compare the kinetic energies of the baseball and the bullet, we use the kinetic energy formula:

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

1. Kinetic Energy of the Bullet:

  - Mass [tex](\(m_b\)): \(0.003 \, \text{kg}\)[/tex]

  - Speed [tex](\(v_b\)): \(1.50 \times 10^3 \, \text{m/s}\)[/tex]

 [tex]\[ KE_b = \frac{1}{2} \cdot 0.003 \, \text{kg} \cdot (1.50 \times 10^3 \, \text{m/s})^2 \][/tex]

 [tex]\[ KE_b = \frac{1}{2} \cdot 0.003 \, \text{kg} \cdot 2.25 \times 10^6 \, \text{m}^2/\text{s}^2 \][/tex]

 [tex]\[ KE_b = 0.0015 \cdot 2.25 \times 10^6 \][/tex]

  [tex]\[ KE_b = 3375 \, \text{J} \][/tex]

2. Kinetic Energy of the Baseball:

  - Mass [tex](\(m_{bb}\)): \(0.145 \, \text{kg}\)[/tex]

  - Speed [tex](\(v_{bb}\)): \(31.034 \, \text{m/s}\)[/tex]

[tex]\[ KE_{bb} = \frac{1}{2} \cdot 0.145 \, \text{kg} \cdot (31.034 \, \text{m/s})^2 \][/tex]

 [tex]\[ KE_{bb} = \frac{1}{2} \cdot 0.145 \, \text{kg} \cdot 963.1 \, \text{m}^2/\text{s}^2 \][/tex]

  [tex]\[ KE_{bb} = 0.0725 \cdot 963.1 \][/tex]

 [tex]\[ KE_{bb} = 69.86 \, \text{J} \][/tex]

Answer:

Theta1 = 12° and theta2 = 168°

The solution procedure can be found in the attachment below.

Explanation:

The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).

In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.

Answer:

Current (I) = 3 x 10^-2 A

Explanation:

As we know, [tex]B = 4\pi 10^-7 *l/ 2\pi r[/tex]

By putting up the values needed from the data...

Current (I) = 2 x 3.14 x (3.0 x 10^-6) (2.0 x 10^-3) / 4 x 3.14 x 10^-7 = 3 x 10^-2 A

Answer: 0.03002A

Explanation: The formulae that relates the magnetic field strength B at a point (r) away from the center of a conductor carrying a current of value (I) is given below as

B = Uo×I/2πr

From our question, B =2.0×10^-3 T, r = 3.0×10^-6m

I =?, Uo = permeability of free space = 1.256×10^-6 mkg/s²A².

By substituting the parameters, we have that

2×10^-3 = 1.256×10^-6 × I/2π(3.0×10^-6)

2×10^-3 × 2π(3.0×10^-6) = 1.256×10^-6 × I

3.77×10^-8 = 1.256×10^-6 × I

I = 3.77×10^-8/ 1.256×10^-6

I = 3.002×10^-2 = 0.03002A

Final answer:

The ball will go higher up the hill if the hill has enough friction to prevent slipping.

Explanation:

The ball will go higher up the hill if the hill has enough friction to prevent slipping. This is because in the case where there is enough friction, the ball can convert some of its kinetic energy to rotational energy, allowing it to roll up the hill. The conservation of energy statement can be used to explain this:

When the ball rolls without slipping, its total mechanical energy is conserved.

As the ball rolls up the hill, its potential energy increases and its kinetic energy decreases.

In the case where there is enough friction, some of the kinetic energy is converted to rotational energy, allowing the ball to reach a higher height on the hill.

Therefore, the ball will go higher up the hill if the hill has enough friction to prevent slipping.

Answer:

The current in the primary is 0.026 A

Explanation:

Using the formula

I1 = (V1/V2)*I2

we have

I1 = (6.4/120)*0.500

I1 = 0.026 A

To find the current in the primary coil of a transformer, we use the power equivalence between the primary and secondary sides, and solve for the primary current using the given voltages and secondary current. This results in a primary current of 26.7 mA.

Calculating the Primary Current in a Transformer

To determine the primary current in a transformer, we need to use the relationship between the power in the primary and secondary circuits. The power delivered by an ideal transformer is the same on both sides, which means Powerprimary = Powersecondary. This can be written as:

Given parameters are:

We first calculate the power on the secondary side:

For an ideal transformer, this power must be equal to the power on the primary side:

Thus, we have:

Solving for [tex]Iprimary[/tex], we get:

Therefore, the current in the primary coil is 26.7 mA.

Answer:

59.55 cm

Explanation:

Note: A meter stick has a length of 100 cm, and it is balanced at 50 cm.

From the principle of moment,

Sum of clockwise moment = sum of anti clockwise moment

Taking moment about the center

mg(50-x) = m'(y-50)g.................. Equation 1

Where m = first mass, m' = second mass, x = position of the first mass, y = position of the second mass, g = acceleration

make y the subject of the equation

y = (m(50-x)/m')+50.................... Equation 2

y = (0.11(50-17)/0.38)+50

y = (0.11(33)/0.38)+50

y = 9.55+50

y = 59.55 cm

Answer:102.84 N

Explanation:

Given

Mass of board [tex]m=12.8\ kg[/tex]

Length of board [tex]l=6.1\ m[/tex]

First support is at one end and second support is at a distance of 2.38 m from the other end

Suppose [tex]R_1[/tex] and [tex]R_2[/tex] are the reactions at the two support

Taking moment about the first end we can write

[tex]mg\times \dfrac{6.1}{2}-R_2(6.1-2.38)=0[/tex]

[tex]R_2=\dfrac{12.8\times 9.8 \times 3.05}{3.72}[/tex]

[tex]R_2=102.84\ N[/tex]    

Answer:

work done is -150 kJ

Explanation:

given data

volume v1 = 2 m³

pressure p1 = 100 kPa

pressure p2 = 200 kPa

internal energy = 10 kJ

heat is transferred  = 150 kJ

solution

we know from 1st law of thermodynamic is

Q = du +W    ............1

put here value and we get

-140 = 10 + W

W = -150 kJ

as here work done is -ve so we can say work is being done on system

Final answer:

The work done by the gas during the compression where 2 m³ of an ideal gas is compressed from 100 kPa to 200 kPa, with an internal energy increase of 10 kJ, and 150 kJ of heat transferred to the surroundings, is -160 kJ.

Explanation:

The student has asked how much work was done by the gas during a compression process in which 2 m³ of an ideal gas is compressed from 100 kPa to 200 kPa, the internal energy increases by 10 kJ, and 150 kJ of heat is transferred to the surroundings. To find the work done by the gas, we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W). The formula is ΔU = Q - W.

In this scenario, we have ΔU as +10 kJ (because the internal energy increases) and Q as -150 kJ (because heat is transferred to the surroundings, meaning it is leaving the system, thus it's a negative value). Plugging these values into the first law of thermodynamics gives us:

10 kJ = -150 kJ - W

When we rearrange the equation to solve for W, it becomes:

W = -150 kJ - 10 kJ

W = -160 kJ

Since work done by the system is a negative value in this case, it indicates that 160 kJ of work has been done on the gas by the surroundings during the compression. Thus, the work done by the gas itself is -160 kJ.

Answer:

[tex]u_x=55.208\ m.s^{-1}[/tex]

Explanation:

Given:

horizontal distance form the point of shooting where the arrow hits ground, [tex]s=107\ ft[/tex] [tex]=32.614\ m[/tex]

angle below the horizontal form the point of release of arrow where it hits ground, [tex]\theta=3^{\circ}[/tex]

So the height above the ground from where the arrow was shot:

[tex]\tan3^{\circ}=\frac{h}{107}[/tex]

[tex]h=5.6076\ ft=1.71\ m[/tex]

[tex]v_y=\sqrt{2g.h}[/tex]

[tex]v_y=\sqrt{2\times 9.8\times 1.71}[/tex]

[tex]v_y=5.789\ m.s^{-1}[/tex]

Using equation of motion:

[tex]v_y=u_y+g.t[/tex]

where:

t = time taken

[tex]5.789=0+9.8\times t[/tex]

[tex]t=0.591\ s[/tex]

[tex]u_x=\frac{s}{t}[/tex]

[tex]u_x=\frac{32.614}{0.591}[/tex]

[tex]u_x=55.208\ m.s^{-1}[/tex]

Answer:

a. 0.21 rad/s2

b. 2.205 N

Explanation:

We convert from rpm to rad/s knowing that each revolution has 2π radians and each minute is 60 seconds

200 rpm = 200 * 2π / 60 = 21 rad/s

180 rpm = 180 * 2π / 60 = 18.85 rad/s

r = d/2 = 30cm / 2 = 15 cm = 0.15 m

a)So if the angular speed decreases steadily (at a constant rate) from 21 rad/s to 18.85 rad/s within 10s then the angular acceleration is

[tex]\alpha = \frac{\Delta \omega}{\Delta t} = \frac{21 - 18.85}{10} = 0.21 rad/s^2[/tex]

b) Assume the grind stone is a solid disk, its moment of inertia is

[tex]I = mR^2/2[/tex]

Where m = 28 kg is the disk mass and R = 0.15 m is the radius of the disk.

[tex] I = 28*0.15^2/2 = 0.315 kgm^2[/tex]

So the friction torque is

[tex]T_f = I\alpha = 0.315*0.21 = 0.06615 Nm[/tex]

The friction force is

[tex]F_f = T_f/R = 0.06615 / 0.15 = 0.441 N[/tex]

Since the friction coefficient is 0.2, we can calculate the normal force that is used to press the knife against the stone

[tex]N = F_f/\mu = 0.441/0.2 = 2.205 N[/tex]

Answer:

Explanation:

Mass of car (M)=1200kg

Initial velocity (u)=20m/s

Stop after time (t)=3sec.

Come to stop implies that the final velocity is zero, v=0m/s

Using newton second law of motion

F=m(v-u)/t

Ft=m(v-u)

Since impulse is Ft

I=Ft

Then, I=Ft=m(v-u)

I=m(v-u)

I=1200(0-20)

I=1200×-20

I=-24,000Ns

The impulse delivered to the car by static friction is -24,000Ns

The  impulse delivered to the car by static friction is -24,000Ns

Since A 1200-kg car, initially moving at 20 m/s, comes to a stop at a red light over a time of 3 s from the moment the driver hit the brakes.

Now here we used second law of motion of newton.

F=m(v-u)/t

Ft=m(v-u)

Since impulse is Ft

So,

I=Ft

Now

, I=Ft=m(v-u)

So,

I=m(v-u)

I=1200(0-20)

I=1200×-20

I=-24,000Ns

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Answer:

a) (vᵣₘₛ₁/vᵣₘₛ₂) = 1.00429

where vᵣₘₛ₁ represents the vᵣₘₛ for 235-UF6 and vᵣₘₛ₂ represents the vᵣₘₛ for 238-UF6.

b) T = 767.34 K

c) The answers point to a difficult seperation technique, as the two compounds of the different isotopes have very close rms speeds and to create a difference of only 1 m/s In their rms speeds would require a high temperature of up to 767.34 K.

Explanation:

The vᵣₘₛ for an atom or molecule is given by

vᵣₘₛ = √(3RT/M)

where R = molar gas constant = J/mol.K

T = absolute temperature in Kelvin

M = Molar mass of the molecules.

₁₂

Let the vᵣₘₛ of 235-UF6 be vᵣₘₛ₁

And its molar mass = M₁ = 349.0 g/mol

vᵣₘₛ₁ = √(3RT/M₁)

√(3RT) = vᵣₘₛ₁ × √M₁

For the 238-UF6

Let its vᵣₘₛ be vᵣₘₛ₂

And its molar mass = M₂ = 352.0 g/mol

√(3RT) = vᵣₘₛ₂ × √M₂

Since √(3RT) = √(3RT)

vᵣₘₛ₁ × √M₁ = vᵣₘₛ₂ × √M₂

(vᵣₘₛ₁/vᵣₘₛ₂) = (√M₂/√M₁) = [√(352)/√(349)]

(vᵣₘₛ₁/vᵣₘₛ₂) = 1.00429

b) Recall

vᵣₘₛ₁ = √(3RT/M₁)

vᵣₘₛ₂ = √(3RT/M₂)

(vᵣₘₛ₁ - vᵣₘₛ₂) = 1 m/s

[√(3RT/M₁)] - [√(3RT/M₂)] = 1

R = 8.314 J/mol.K, M₁ = 349.0 g/mol = 0.349 kg/mol, M₂ = 352.0 g/mol = 0.352 kg/mol, T = ?

√T [√(3 × 8.314/0.349) - √(3 × 8.314/0.352) = 1

√T (8.4538 - 8.4177) = 1

√T = 1/0.0361

√T = 27.7

T = 27.7²

T = 767.34 K

c) The answers point to a difficult seperation technique, as the two compounds of the different isotopes have very close rms speeds and to create a difference of only 1 m/s In their rms speeds would require a high temperature of up to 767.34 K.

Answer:

The proton was moving at a speed of 1.64 x 10⁶ m/s

Explanation:

Given;

strength of magnetic field, B = 0.280 T

circular radius, R = 6.12 cm

mass of proton, m = 1.67 x 10⁻²⁷ kg

charge of electron. q = 1.602 x 10⁻¹⁹ C

When the proton follows a circular arc, then magnetic force will be equal to the centripetal force.

Magnetic force, Fm = qvB

Centripetal force, Fr = mv²/R

Fm = Fr

qvB = mv²/R

[tex]qvB = \frac{Mv^2}{R} \\\\\frac{qBR}{M} = \frac{v^2}{v} \\\\v =\frac{qBR}{M} = \frac{1.602*10^{-19} *0.28*0.0612}{1.67*10^{-27}} =1.64 *10^6 \ m/s[/tex]

Therefore, the proton was moving at a speed of 1.64 x 10⁶ m/s

Given Information:

Magnetic field = B = 0.280 T

Radius = r = 6.12 cm = 0.0612 m

Required Information:  

Speed of proton = v = ?

Answer:

Speed of proton = v = 1641.77x10³ m/s

Explanation:

Since the proton is moving in a circular path we can model it in terms of magnetic force and centripetal force.

The magnetic force is given by

F = qvB

The centripetal force is given by

F = mv²/r

equating the both equations yields,

mv²/r = qvB

mv = qBr

v = qBr/m

Where q = 1.6x10⁻¹⁹ C is the of the proton, m = 1.67x10⁻²⁷ kg is the mass of proton, B is the magnetic field and r is the radius o circular arc around which proton is moving.

v = 1.6x10⁻¹⁹*0.280*0.0612/1.67x10⁻²⁷

v = 1641772.45 m/s

v = 1641.77x10³ m/s

Therefore, the proton is moving at the speed of 1641.77x10³ m/s.

Answer:

the tractor's power output = 1318.47 watts

Explanation:

Weight of bale = (mg) = (150 kg * 9.81 m/s²) = 1471.5 newtons

Resolve that weight force into its components at right angles to the ramp and parallel to the slope.

Down the slope we have 1471.5sin(15°) = 380.85 N

At right angles to the ramp we have 1471.5cos(15°) = 1421.35 N

OK, now we need the relationship between the coefficient of friction (μ), the friction force (Ff) and the normal force (Fn) {The normal force is the force at right angles to the friction surface.  In this case Fn is equal in magnitude to the component of weight force at right angles to the surface.}

Ff = μ * Fn = (0.40 * 1421.35) = 568.54 N

The bale is not accelerating, so the "pull force" up the incline = component of weight + friction force down

= (380 N + 568.54 N)

= 948.54N

We need the speed in m/s not km/h

5.0 km/h = 5000 m/h

= (5000/3600)

= 1.39 m/s

Power = (948.54 N * 1.39m/s)

= 1318.47 N.m/s

= 1318.47 watts

The power output of the tractor is 1318.47 W

According to the question the weight of bale

mg = 150 × 9.81 = 1471.5 N

Resolving the weight we get:

The normal reaction perpendicular to the ramp is given by:

N = 1471.5cos(15°) = 1421.35 N

The force of friction is given by:

[tex]f = \mu N = 0.40 \times 1421.35\\\\f = 568.54 N[/tex]

The bale is not accelerating, so the force up the incline = component of weight + friction force down

F = mgsin(15°) + f

F = 380 N + 568.54 N

F = 948.54N

Now, the engine speed is:

v = 5.0 km/h = 5000 m/h

v = (5000/3600)

v = 1.39 m/s

The power is defined as the product of force and speed, so:

Power P = Fv

P = 948.54 × 1.39

P = 1318.47 W

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