The probability of nuclear fusion is greatly enhanced when the appropriate nuclei are brought close together, but their mutual coulomb repulsion must be overcome. This can be done using the kinetic energy of high temperature gas ions or by accelerating the nuclei toward one another.

Required:
a. Calculate the potential energy of two singly charged nuclei separated by 1.00×10^−12m
b. At what temperature will atoms of a gas have an average kinetic energy equal to this needed electrical potential energy?

Answer:

a. 6.41 m/s

b. 120.85 m/s^2

Explanation:

The computation is shown below:

a. Pebble speed is

As we know that according to the tangential speed,

[tex]v = r \times \omega[/tex]

[tex]= \frac{0.68}{2} \times 18.84[/tex]

= 6.41 m/s

The 18.84 come from

[tex]= 2 \times 3.14 \times 3[/tex]

= 18.84

b. The pebble acceleration is

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

[tex]= \frac{6.41^2}{0.34}[/tex]

= 120.85 m/s^2

We simply applied the above formulas so that the pebble speed and the pebble acceleration could come and the same is to be considered

Answer:

(a) B = 2.85 × [tex]10^{-6}[/tex] Tesla

(b) I =  I = 0.285 A

Explanation:

a. The strength of magnetic field, B, in a solenoid is determined by;

r = [tex]\frac{mv}{qB}[/tex]

⇒ B = [tex]\frac{mv}{qr}[/tex]

Where: r is the radius, m is the mass of the electron, v is its velocity, q is the charge on the electron and B is the magnetic field

B = [tex]\frac{9.11*10^{-31*1.0*10^{4} } }{1.6*10^{-19}*0.02 }[/tex]

  = [tex]\frac{9.11*10^{-27} }{3.2*10^{-21} }[/tex]

B = 2.85 × [tex]10^{-6}[/tex] Tesla

b. Given that; N/L = 25 turns per centimetre, then the current, I, can be determined by;

B = μ I N/L

⇒    I = B ÷ μN/L

where B is the magnetic field,  μ is the permeability of free space = 4.0 ×[tex]10^{-7}[/tex]Tm/A, N/L is the number of turns per length.

I = B ÷ μN/L

 = [tex]\frac{2.85*10^{-6} }{4*10^{-7} *25}[/tex]

I = 0.285 A

Answer:

Explanation:

Given that,

The length of the beam L = 3.10m

The mass of the steam beam [tex]m_1[/tex] = 430kg

The mass of worker [tex]m_2[/tex] = 69.0kg

The distance from  the fixed point to centre of gravity of beam = [tex]\frac{L}{2}[/tex]

and our length of beam is 3.10m

so the distance from  the fixed point to centre of gravity of beam is

[tex]\frac{3.10}{2}=1.55m[/tex]

Then the net torque is

[tex]=-W_sL'-W_wL\\\\=-(W_sL'+W_wL)[/tex]

[tex]W_s[/tex] is the weight of steel rod

[tex]=430\times9.8=4214N[/tex]

[tex]W_w[/tex] is the weight of the worker

[tex]=69\times9.8\\\\=676.2N[/tex]

Torque can now be calculated

[tex]-(4214\times1.55+676.2\times3.9)Nm\\\\-(6531.7+2637.18)Nm\\\\-(9168.88)Nm[/tex]

≅ 9169Nm

Answer: Part 1: Propellant Fraction (MR) = 8.76

Part 2: Propellant Fraction (MR) = 1.63

Explanation: The Ideal Rocket Equation is given by:

Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]

Where:

[tex]v_{ex}[/tex] is relationship between exhaust velocity and specific impulse

[tex]\frac{m_{f}}{m_{e}}[/tex] is the porpellant fraction, also written as MR.

The relationship [tex]v_{ex}[/tex] is: [tex]v_{ex} = g_{0}.Isp[/tex]

To determine the fraction:

Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]

[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]

Knowing that change in velocity is Δv = 9.6km/s and [tex]g_{0}[/tex] = 9.81m/s²

Note: Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.

Part 1: Isp = 450s

[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]

ln(MR) = [tex]\frac{9.6.10^{3}}{9.81.450}[/tex]

ln (MR) = 2.17

MR = [tex]e^{2.17}[/tex]

MR = 8.76

Part 2: Isp = 2000s

[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]

ln (MR) = [tex]\frac{9.6.10^{3}}{9.81.2.10^{3}}[/tex]

ln (MR) = 0.49

MR = [tex]e^{0.49}[/tex]

MR = 1.63

-- As she lands on the air mattress, her momentum is (m v)

Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down

-- As she leaves it after the bounce,

Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up

-- The impulse (change in momentum) is

Change = (60 kg-m/s up) - (300 kg-m/s down)

Magnitude of the change = 360 km-m/s

The direction of the change is up /\ .

The direction of a body or object's movement is defined by its velocity.In its basic form, speed is a scalar quantity.In essence, velocity is a vector quantity.It is the speed at which distance changes.It is the displacement change rate.

Solve the problem ?

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Answer:

θ = 12.60°

Explanation:

In order to calculate the angle below the horizontal for the velocity of the hockey puck, you need to calculate both x and y component of the velocity of the puck, and also you need to use the following formula:

[tex]\theta=tan^{-1}(\frac{v_y}{v_x})[/tex]       (1)

θ: angle below he horizontal

vy: y component of the velocity just after the puck hits the ground

vx: x component of the velocity

The x component of the velocity is constant in the complete trajectory and is calculated by using the following formula:

[tex]v_x=v_o[/tex]

vo: initial velocity = 28.0 m/s

The y component is calculated with the following equation:

[tex]v_y^2=v_{oy}^2+2gy[/tex]         (2)

voy: vertical component of the initial velocity = 0m/s

g: gravitational acceleration = 9.8 m/s^2

y: height

You solve the equation (2) for vy and replace the values of the parameters:

[tex]v_y=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(2.00m)}=6.26\frac{m}{s}[/tex]

Finally, you use the equation (1) to find the angle:

[tex]\theta=tan^{-1}(\frac{6.26m/s}{28.0m/s})=12.60\°[/tex]

The angle below the horizontal is 12.60°

The angle below the horizontal of the velocity of the puck just before it hits the ground is 12.60°.

Given the following data:

To find the angle below the horizontal of the velocity of the puck just before it hits the ground:

First of all, we would determine the horizontal and vertical components of the hockey puck.

For horizontal component:

[tex]V_y^2 = U_y^2 + 2aS\\\\V_y^2 = 0^2 + 2(9.81)(2)\\\\V_y^2 = 39.24\\\\V_y = \sqrt{39.24} \\\\V_y = 6.26 \; m/s[/tex]

For vertical component:

[tex]V_x = U_x\\\\V_x = 28.0 \;m/s[/tex]

Now, we can find the angle by using the formula:

[tex]\Theta = tan^{-1} (\frac{V_y}{V_x} )[/tex]

Substituting the values, we have:

[tex]\Theta = tan^{-1} (\frac{6.26}{28.0} )\\\\\Theta = tan^{-1} (0.2236)\\\\\Theta = 12.60[/tex]

Angle = 12.60 degrees.

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Answer:

   θ = 4.78º

with respect to the vertical or 4.78 to the east - north

Explanation:

This is a velocity compound exercise since it is a vector quantity.

The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed

                  v_fly² = v_nort² + v_air²

                  v_nort² = v_fly² + - v_air²

Let's use trigonometry to find the direction of the plane

        sin θ = v_air / v_fly

        θ = sin⁻¹ (v_air / v_fly)

let's calculate

        θ = sin⁻¹ (10/120)

         θ = 4.78º

with respect to the vertical or 4.78 to the north-east

Given that,

Height =1.5 m

Angle = 45°

We need to find the greater speed of the ball

Using conservation of energy

[tex]P.E_{i}+K.E_{f}=P.E_{f}+K.E_{f}[/tex]

[tex]mgh+\dfrac{1}{2}mv_{i}^2=mgh+\dfrac{1}{2}mv_{f}^2[/tex]

Here, initial velocity and final potential energy is zero.

[tex]mgh=\dfrac{1}{2}mv_{f}^2[/tex]

Put the value into the formula

[tex]9.8\times1.5=\dfrac{1}{2}v_{f}^2[/tex]

[tex]v_{f}^2=2\times9.8\times1.5[/tex]

[tex]v_{f}=\sqrt{2\times9.8\times1.5}[/tex]

[tex]v_{f}=5.42\ m/s[/tex]

Hence, the greater speed of the ball is 5.42 m/s.

Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

[tex]\Sigma F = F - F' = M\cdot a[/tex]

Box with mass 2M

[tex]\Sigma F = F' - F'' = 2\cdot M \cdot a[/tex]

Box with mass 3M

[tex]\Sigma F = F'' = 3\cdot M \cdot a[/tex]

On the third equation, acceleration can be modelled in terms of F'':

[tex]a = \frac{F''}{3\cdot M}[/tex]

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

[tex]F' = 2\cdot M \cdot a + F''[/tex]

[tex]F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''[/tex]

[tex]F' = \frac{5}{3}\cdot F''[/tex]

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

[tex]F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)[/tex]

[tex]F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''[/tex]

[tex]F = 2\cdot F''[/tex]

[tex]F'' = \frac{1}{2}\cdot F[/tex]

Afterwards, F' as function of the external force can be obtained by direct substitution:

[tex]F' = \frac{5}{6}\cdot F[/tex]

The net forces of each block are now calculated:

Box with mass M

[tex]M\cdot a = F - \frac{5}{6}\cdot F[/tex]

[tex]M\cdot a = \frac{1}{6}\cdot F[/tex]

Box with mass 2M

[tex]2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F[/tex]

[tex]2\cdot M \cdot a = \frac{1}{3}\cdot F[/tex]

Box with mass 3M

[tex]3\cdot M \cdot a = \frac{1}{2}\cdot F[/tex]

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.

Answer:

   μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

           W₁  x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

             cos 60 = x / L

where L is the length of the ladder

              x = L cos 60

            sin 60 = y / L

           y = L sin60

the horizontal distance of man is

            cos 60 = x2 / 7.0

            x2 = 7 cos 60

we substitute

         m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

         fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

         fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

         fr = (735 + 2450) / 8.66

         fr = 367.78 N

the friction force has the expression

         fr = μ N

write the translational equilibrium equation

         N - W₁ -W₂ = 0

         N = m₁ g + W₂

         N = 30 9.8 + 700

         N = 994 N

we clear the friction force from the eucacion

        μ = fr / N

        μ = 367.78 / 994

        μ = 0.37

Answer:

If the initial speed is doubled the time is also doubled

Explanation:

You have that a car with velocity v is decelerated by a constant acceleration a in a time t.

You use the following equation to establish the previous situation:

[tex]v'=v-at[/tex]     (1)

v': final speed of the car  = 0m/s

v: initial speed of the car

From the equation (1) you solve for t and obtain:

[tex]t=\frac{v-v'}{a}=\frac{v}{a}[/tex]     (2)

To find the new time that car takesto stop with the new initial velocity you use again the equation (1), as follow:

[tex]v'=v_1-at'[/tex]     (3)

v' = 0m/s

v1: new initial speed = 2v

t': new time

You solve the equation (3) for t':

[tex]0=2v-at'\\\\t'=\frac{2v}{a}=2t[/tex]

If the initial speed is doubled the time is also doubled

Answer:

a) [tex]W_{in} = 214.286\,J[/tex], b) [tex]W_{in} = 428.571\,J[/tex]

Explanation:

a) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:

[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]

Where:

[tex]T_{L}[/tex] - Temperature of surroundings, measured in Kelvin.

[tex]T_{H}[/tex] - Temperature of the house, measured in Kelvin.

Given that [tex]T_{H} = 294\,K[/tex] and [tex]T_{L} = 273\,K[/tex]. The Coefficient of Performance is:

[tex]COP_{HP} = \frac{294\,K}{294\,K-273\,K}[/tex]

[tex]COP_{HP} = 14[/tex]

Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.

[tex]COP_{HP} = \frac{Q_{H}}{W_{in}}[/tex]

The input work to deliver a determined amount of heat to the house:

[tex]W_{in} = \frac{Q_{H}}{COP_{HP}}[/tex]

If [tex]Q_{H} = 3000\,J[/tex] and [tex]COP_{HP} = 14[/tex], the input work that is needed is:

[tex]W_{in} = \frac{3000\,J}{14}[/tex]

[tex]W_{in} = 214.286\,J[/tex]

b) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:

[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]

Where:

[tex]T_{L}[/tex] - Temperature of surroundings, measured in Kelvin.

[tex]T_{H}[/tex] - Temperature of the house, measured in Kelvin.

Given that [tex]T_{H} = 294\,K[/tex] and [tex]T_{L} = 252\,K[/tex]. The Coefficient of Performance is:

[tex]COP_{HP} = \frac{294\,K}{294\,K-252\,K}[/tex]

[tex]COP_{HP} = 7[/tex]

Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.

[tex]COP_{HP} = \frac{Q_{H}}{W_{in}}[/tex]

The input work to deliver a determined amount of heat to the house:

[tex]W_{in} = \frac{Q_{H}}{COP_{HP}}[/tex]

If [tex]Q_{H} = 3000\,J[/tex] and [tex]COP_{HP} = 7[/tex], the input work that is needed is:

[tex]W_{in} = \frac{3000\,J}{7}[/tex]

[tex]W_{in} = 428.571\,J[/tex]

Answer:

22.66Nm²/C

Explanation:

Flux through an electric field is expressed as ϕ = EAcosθ

When a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25N.m^2/c. If the paper is turned 25 degree with respect to the field the flux through it can be calculated using the formula.

From the formula above where:

EA = 25N.m^2/C

θ = 25°

ϕ = 25cos 25°

ϕ = 22.66Nm²/C

Answer:

blue  θ₂ = 22.26º

red    θ₂ = 22.79º

Explanation:

When a light beam passes from one material medium to another, it undergoes a deviation from the path, described by the law of refraction

         n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the incident and transmitted media refractive indices and θ are the angles in the media

let's apply this equation to each wavelength

λ = blue

in this case n₁ = 1, n₂ = 1,645

       sin θ₂ = n₁/ n₂ sin₂ θ₁

let's calculate

       sin θ₂ = 1 / 1,645 sint 38.55

       sin θ₂ = 0.37884

       θ₂ = sin⁻¹ 0.37884

       θ₂ = 22.26º

λ = red

n₂ = 1,609

         sin θ₂ = 1 / 1,609 sin 38.55

         sin θ₂ = 0.3873

         θ₂ = sim⁻¹ 0.3873

         θ₂ = 22.79º

the refracted rays are between these two angles

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}[/tex]

k = 1.4

[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K[/tex]

Work done is given as;

[tex]W = \frac{1}{2} *m*(v_i^2 - v_e^2)[/tex]

inlet velocity is negligible;

[tex]v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s[/tex]

Therefore, the exit velocity is 629.41 m/s

Answer:

Dear user,

Answer to your query is provided below

The angle for which the rope will break θ = 41.8°

Explanation:

Explanation of the same is attached in image

A bag is gently pushed off the top of a wall at A and swings in a vertical plane at the end of a rope of length l. The angle θ for which the rope will break, is 41.81°

The tension is a kind of force which acts on linear objects when subjected to pull.

The maximum tension Tmax =2W

From the work energy principle,

T₂ = 1/2 mv²

Total energy before and after pushing off

0+mglsinθ = 1/2 mv²

v² = 2gflsinθ..............(1)

From the equilibrium of forces, we have

T= ma +mgsinθ = mv²/l +mgsinθ

2mg = mv²/l +mgsinθ

2g = v²/l +gsinθ

Substitute the value of v² ,we get the expression for θ

θ = sin⁻¹(2/3)

θ =41.81°

Hence, the angle θ for which the rope will break, is 41.81°

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Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

μ_k = 0.1773

The coefficient of kinetic friction between the hockey puck and ice is equal to 0.178

Given the following data:

Scientific data:

To determine the coefficient of kinetic friction between the hockey puck and ice:

First of all, we would calculate the acceleration of the hockey puck by using the third equation of motion.

[tex]V^2 = U^2 + 2aS\\\\0^2 =20^2 + 2a(115)\\\\-400=230a\\\\a=\frac{-400}{230}[/tex]

Acceleration, a = -1.74 [tex]m/s^2[/tex]

Note: The negative signs indicates that the hockey puck is slowing down or decelerating.

From Newton's Second Law of Motion, we have:

[tex]\sum F_x = F_k + F_n =0\\\\F_k =- F_n\\\\\mu mg =-ma\\\\\mu = \frac{-a}{g}\\\\\mu = \frac{-(-1.74)}{9.8}\\\\\mu = \frac{1.74}{9.8}[/tex]

Coefficient of kinetic friction = 0.178

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Answer:

The bottom two lines.

Explanation:

They need their own line of voltage quantity. A parallel circuit has the definition of 'two or more paths for current to flow through.' The voltage does stay the same in each line.

Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

                              [tex]F = k\frac{|q_1|.|q_2|}{r^2}[/tex]

Where,

                     k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.  

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

                             [tex]\frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1[/tex] ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

                             [tex]q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}[/tex]

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

                          [tex]\frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2} = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 = \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6[/tex]  .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

                         [tex]-\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123[/tex]

                          [tex]q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\[/tex]

Answer:

a) WT = 137.5 J

b) v2 = 2.34 m/s

Explanation:

a) The total work done on the block is given by the following formula:

[tex]W_T=F_pd-F_fd=(F_p-F_f)d[/tex]          (1)

Fp: force parallel to the displacement of the block = 150N

Ff: friction force

d: distance = 5.0 m

Then, you first calculate the friction force by using the following relation:

[tex]F_f=\mu_k N=\mu_k Mg[/tex]        (2)

μk: coefficient of kinetic friction = 0.25

M: mass of the block = 50kg

g: gravitational constant = 9.8 m/s^2

Next, you replace the equation (2) into the equation (1) and solve for WT:

[tex]W_T=(F_p-\mu_kMg)d=(150N-(0.25)(50kg)(9.8m/s^2))(5.0m)\\\\W_T=137.5J[/tex]

The work done over the block is 137.5 J

b) If the block started from rest, you can use the following equation to calculate the final speed of the block:

[tex]W_T=\Delta K=\frac{1}{2}M(v_2^2-v_1^2)[/tex]     (3)

WT: total work = 137.5 J

v2: final speed = ?

v1: initial speed of the block = 0m/s

You solve the equation (3) for v2:

[tex]v_2=\sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(137.5J)}{50kg}}=2.34\frac{m}{s}[/tex]

The final speed of the block is 2.34 m/s

"1 watt" means 1 joule of energy per second.

75 W means 75 joules/sec .

Energy = (75 Joule/sec) x (12 min) x (60 sec/min)

Energy = (75 x 12 x 60) (Joule-min-sec / sec-min)

Energy = 54,000 Joules

Answer:

9.6 Ns

Explanation:

Note: From newton's second law of motion,

Impulse = change in momentum

I = m(v-u).................. Equation 1

Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.

Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)

Substitute into equation 1

I = 2.4[2.5-(-1.5)]

I = 2.4(2.5+1.5)

I = 2.4(4)

I = 9.6 Ns

The magnitude of impulse will be "9.6 Ns".

According to the question,

Mass,

Final velocity,

Initial velocity,

By using Newton's 2nd law of motion, we get

→ Impulse, [tex]I = m(v-u)[/tex]

By substituting the values, we get

                     [tex]= 2.4[2.5-(1.5)][/tex]

                     [tex]= 2.4(2.5+1.5)[/tex]

                     [tex]= 2.4\times 4[/tex]

                     [tex]= 9.6 \ Ns[/tex]

Thus the above answer is right.    

Learn more about Impulse here:

https://brainly.com/question/15495020

Answer:

D

Explanation:

Answer:

It is D

Explanation: No cap

Answer:

A. a (-321.393, 383.022) b (-76.40, -64.278)

B. (-397.991, 318.744)

C. a. resulting speed 509.9mph  b. bearing of the plane = 51.6°

Explanation:

Answer:

mass 20 times of an amazing and all its motion

Answer:

-67 m/s

Explanation:

We are given that

Mass of ball,m=0.4 kg

Initial speed,u=14 m/s

Impulse,I=-32.4 N-s

Time,t=27 ms=[tex]27\times 10^{-3} s[/tex]

We have to find the mass's velocity in the x- direction.

We know that

[tex]Impulse=mv-mu[/tex]

Substitute the values

[tex]-32.4=0.4v-0.4(14)[/tex]

[tex]-32.4+0.4(14)=0.4 v[/tex]

[tex]-26.8=0.4v[/tex]

[tex]v=\frac{-26.8}{0.4}=-67m/s[/tex]

Answer:

B = 0.024T positive z-direction

Explanation:

In this case you consider that the direction of the motion of the electron, and the direction of the magnetic field are perpendicular.

The magnitude of the magnetic force exerted on the electron is given by the following formula:

[tex]F=qvB[/tex]     (1)

q: charge of the electron = 1.6*10^-19 C

v: speed of the electron = 1.6*10^7 m/s

B: magnitude of the magnetic field = ?

By the Newton second law you also have that the magnetic force is equal to:

[tex]F=qvB=ma[/tex]       (2)

m: mass of the electron = 9.1*10^-31 kg

a: acceleration of the electron = 7.0*10^16 m/s^2

You solve for B from the equation (2):

[tex]B=\frac{ma}{qv}\\\\B=\frac{(9.1*10^{-31}kg)(7.0*10^{16}m/s^2)}{(1.6*10^{-19}C)(1.6*10^7m/s)}\\\\B=0.024T[/tex]

The direction of the magnetic field is found by using the right hand rule.

The electron moves upward (+^j). To obtain a magnetic forces points to the positive x-direction (+^i), the direction of the magnetic field has to be to the positive z-direction (^k). In fact, you have:

-^j X ^i = ^k

Where the minus sign of the ^j is because of the negative charge of the electron.

Then, the magnitude of the magnetic field is 0.024T and its direction is in the positive z-direction

Answer:

5.95m/s to 2 decimal places

Explanation:

In physics speed is measured in metres per second so convert 8mins to seconds

8x60=420 seconds

The formula needed:

Speed (m/s)= Distance (m)/Time (s)

2500/420=5.95m/s