A projectile is fired at an upward angle of 29.7° from the top of a 108-m-high cliff with a speed of 130-m/s. What will be its speed (in m/s) when it strikes the ground below?

Answer:

The magnitude of the electric force on an electron in a uniform electric is [tex]2.4\times10^{-16}\ N[/tex] to the west.

Explanation:

Given that,

Electric field strength = 1500 N/C

We need to calculate the electric force

Using formula of electric field

[tex]F = Eq[/tex]

E = electric field strength

q = charge of electron

Electron has negative charge.

Put the value into the formula

[tex]F=1500\times(-1.6\times10^{-19})[/tex]

[tex]F=-2.4\times10^{-16}\ N[/tex]

Negative sign shows the opposite direction of the field

Hence, The magnitude of the electric force on an electron in a uniform electric is [tex]2.4\times10^{-16}\ N[/tex] to the west.

The magnitude of the electric force on an electron in a uniform electric field of strength 1500 N/C that points due east is 2.4x10⁻¹⁶ C.

We know that electric force is given by the formula,

[tex]F = E \times q[/tex]

It is given that the electric field, E = 1500 N/C,

We also know that an electron is negatively charged and has a charge of 1.60217662 × 10⁻¹⁹ C.

[tex]F = E \times q\\\\F = 1500 \times 1.6 \times 10^{-19}\\\\F = 2.4 \times 10^{-16}\rm\ N[/tex]

Hence, the magnitude of the electric force on an electron in a uniform electric field of strength 1500 N/C that points due east is 2.4x10⁻¹⁶ C.

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Answer:

(a) 4 rad/s^2

(b) 18 rad

Explanation:

w0 = 0, w = 12 rad/s, t = 3 s

(a) Let α be the angular acceleration.

w = w0 + α t

12 = 0 + 3 α

α = 4 rad/s^2

(b) Let θ be the angle rotated

θ = w0 t + 1/2 α t^2

θ = 0 + 0.5 x 4 x 9

θ = 18 rad

Answer:

1.5 m/s²

Explanation:

m = mass of the clarinet case = 3.450 kg

F = upward force on the clarinet through the air = 38.92 N

W = weight of the clarinet case in down direction

weight of the clarinet case is given as

W = mg

W = (3.450) (9.8)

W = 33.81 N

a = acceleration of the case

Force equation for the motion of the case is given as

F - W = ma

38.92 - 33.81 = (3.450) a

a = 1.5 m/s²

Answer:

The speed when it strikes the ground below is V= 157.64 m/s < -56.92º .

Explanation:

V= 150m/s

α= 55º

hi= 120m

Vy= V*sinα

Vy= 122.87 m/s

Vx= V * cos α

Vx= 86.03 m/s

h= hi + Vy * t - g*t²/2

clearing t we get the total flying time of the projectile:

t(total fly)= 26.01 sec

0= Vy - g*t

clearing t we get the maximum height time:

t(max height)= 12.53 sec

to get the fall time:

t(fall)= t(total fly) - t(max height)

t(fall)= 13.48 sec

Vy'= g* t(fall)

Vy'= 132.1 m/s

V'= √(Vx² +Vy'²)

V'= 157.64 m/s

α'= tg⁻¹ (Vy'/Vx)

α'= -56.92º

By using the conservation of energy principle, we find the speed of the projectile when it hits the ground to be roughly 165 m/s (rounded to 3 significant figures).

Given that a projectile is fired upward at an angle of 55° from the top of a 120 m cliff at a speed of 150 m/s, it is asked what its speed will be when it strikes the ground. To answer such a question, we employ the principle of conservation of energy, which states that the total mechanical energy (kinetic energy + potential energy) of an isolated system remains constant if non-conservative forces like air resistance are negligible.

Now, the energy of the projectile at the top of the cliff is equal to its kinetic energy (as it's launched) and its potential energy (due to its height). When the projectile hits the ground, all its potential energy will be converted into kinetic energy as the body has descended from the height, which the body will possess as its speed.

Let's calculate:

Initial Energy = Final Energy

1/2 * mass * (initial speed)² + mass * g * height = 1/2 * mass * (final speed)²

Solving for final speed, -(initial speed)² - 2gh = - (final speed)²

finalSpeed = sqrt((initial speed)² + 2*g*h)

Given that the initial speed = 150m/s, g (acceleration due to gravity) = 9.81m/s², and the height = 120m, the final speed would amount to 164.833 m/s. Therefore, when rounding to 3 significant figures, the speed when it hits the ground will be 165 m/s.

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Answer:

The change in internal energy of the system is -17746.78 J

Explanation:

Given that,

Pressure [tex]P=4.63\times10^{4}\ Pa[/tex]

Remove heat [tex]\Delta U= -1.95\times10^{4}\ J[/tex]

Radius = 0.272 m

Distance d = 0.163 m

We need to calculate the internal energy

Using thermodynamics first equation

[tex]dU=Q-W[/tex]...(I)

Where, dU = internal energy

Q = heat

W = work done

Put the value of W in equation (I)

[tex]dU=Q-PdV[/tex]

Where, W = PdV

Put the value in the equation

[tex]dU=-1.95\times10^{4}-(4.63\times10^{4}\times3.14\times(0.272)^2\times(-0.163))[/tex]

[tex]dU=-17746.78\ J[/tex]

Hence, The change in internal energy of the system is -17746.78 J

Answer:

velocity of red ball is either 0 or 1 m/s and then the velocity of green ball is 10 m/s or 8 m/s.

Explanation:

mass of red ball, m1 = 0.05 kg u 1 = 10 m/s

mass of green ball, m2 = 0.1 kg, u2 = 0

Let the velocity of red ball and green ball after the collision is v1 and v2, respectively.

By use of conservation of momentum

m1 u1 + m2 u2 = m1 v1 + m2 v2

0.05 x 10 + 0 = 0.05 x v1 + 0.1 x v2

0.5 = 0.05 (v1 + 2 v2)

v1 + 2v2 = 10 ...... (1)

Now use conservation of kinetic energy

1/2 m1 x u1^2 + 1/2 x m2 x u2^2 = 1/2 m1 v1^2 + 1/2 m2 x v2^2

0.05 x 10 x 10 + 0 = 0.05 x v1^2 + 01. x v2^2

5 = 0.05(v1^2 + 2v2^2)

10 = v1^2 + 2 v2^2 .....(2)

Put teh value of v1 from equation (1) in equation (2)

10 = 10 - 2 v2 + 2 v2^2

0 = v2^2 - v2

v2 =0 , 1 m/s

So, v1 = 10 - 2 x 0 = 10  m/s

or v1 = 10 - 2 x 1 = 8 m/s

Thus, the velocity of red ball is either 0 or 1 m/s and then the velocity of green ball is 10 m/s or 8 m/s.

Explanation:

It is given that,

Speed of the plane, v = 110 m/s

Radius of circle, r = 2850 m

(1) Let t is the time taken by a plane. Speed of the plane is given by :

[tex]v=\dfrac{d}{t}=\dfrac{2\pi r}{t}[/tex]

[tex]t=\dfrac{2\pi r}{v}[/tex]

[tex]t=\dfrac{2\pi \times 2850}{110}[/tex]

t = 162.79 s

(2) Mass of a driver, m = 50 kg

Height, h = 6 m

As the driver reaches ground, its potential energy point to zero. So, the change in potential energy is given by :

[tex]\Delta PE=PE_f-PE_i[/tex]

[tex]\Delta PE=0-mgh[/tex]

[tex]\Delta PE=-50\times 9.8\times 6[/tex]

[tex]\Delta PE=-2940\ J[/tex]

So, the change in potential energy is 2940 joules.

Let v is the speed. As the driver reaches water surface, its potential energy is converted to kinetic energy as :

[tex]\dfrac{1}{2}mv^2=PE[/tex]

[tex]v=\sqrt{\dfrac{2\times PE}{m}}[/tex]

[tex]v=\sqrt{\dfrac{2\times 2940}{50}}[/tex]

v = 10.84 m/s

So, the speed of the reactors the water is 10.84 m/s. Hence, this is the required solution.

It takes approximately 52.3 seconds for a plane, traveling at a constant speed of 110 m/s, to fly once around a circle with a radius of 2850 m. The change in potential energy for a driver of mass 50 kg jumping off a 6 m high cliff is -2940 J. The driver's velocity as they reach the water is approximately 10.8 m/s.

To find the time it takes for a plane to fly once around a circle, we need to calculate the circumference of the circle and divide it by the plane's speed. The formula for the circumference of a circle is C = 2πr, where r is the radius. In this case, the radius is given as 2850 m, so the circumference is 2π(2850) = 5700π m. Dividing the circumference by the plane's speed of 110 m/s gives us the time it takes to complete one lap around the circle: (5700π)/(110) ≈ 52.3 s.

When a driver jumps off a cliff, the force of gravity causes her potential energy to change into kinetic energy as she falls. The formula for potential energy is PE = mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height. In this case, the mass is given as 50 kg and the height is 6 m. Plugging these values into the formula, we get PE = (50)(9.8)(6) = 2940 J. The change in potential energy is simply the negative of the initial potential energy, since it decreases when the driver jumps off the cliff. Therefore, the change in potential energy is -2940 J.

As the driver reaches the water, all of her potential energy has been converted into kinetic energy. The formula for kinetic energy is KE = 0.5mv², where m is the mass and v is the velocity. Since the potential energy at the end of the fall is 0, the entire initial potential energy is now kinetic energy. Plugging in the mass of 50 kg, we can solve for the velocity using the formula 2940 = 0.5(50)v². Simplifying this equation, we get v² = 2940/25 = 117.6. Taking the square root of both sides, we find that v ≈ 10.8 m/s.

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(a) [tex]2.98\cdot 10^{10} J[/tex]

The change in energy of the transferred charge is given by:

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

where

q is the charge transferred

[tex]\Delta V[/tex] is the potential difference between the ground and the clouds

Here we have

[tex]q=31 C[/tex]

[tex]\Delta V = 0.96\cdot 10^9 V[/tex]

So the change in energy is

[tex]\Delta U = (31 C)(0.96\cdot 10^9 V)=2.98\cdot 10^{10} J[/tex]

(b) 7921 m/s

If the energy released is used to accelerate the car from rest, than its final kinetic energy would be

[tex]K=\frac{1}{2}mv^2[/tex]

where

m = 950 kg is the mass of the car

v is the final speed of the car

Here the energy given to the car is

[tex]K=2.98\cdot 10^{10} J[/tex]

Therefore by re-arranging the equation, we find the final speed of the car:

[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2.98\cdot 10^{10})}{950}}=7921 m/s[/tex]

Answer:

Part a)

F = 0.051 N

Part b)

Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.

Explanation:

Part a)

Electrostatic force between two charged spherical balls is given as

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we will have

[tex]q_1 = q_2 = 50 nC[/tex]

here the distance between the center of two balls is given as

[tex]r = 2.1 cm = 0.021 m[/tex]

now we will have

[tex]F = \frac{(9\times 10^9)(50 \times 10^{-9})(50 \times 10^{-9})}{0.021^2}[/tex]

[tex]F = 0.051 N[/tex]

Part b)

Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.

Answer:

24 m/s

Explanation:

Let's say that under very icy conditions, there is no friction.

Draw a free body diagram.  There are 2 forces acting on the van.  Gravity straight down, and normal force perpendicular to the surface.

Sum of the forces in the radial (+x) direction:

∑F = ma

N sin θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ − mg = 0

Solve for N in the second equation:

N cos θ = mg

N = mg / cos θ

Substitute into the first equation:

(mg / cos θ) sin θ = m v² / r

mg tan θ = m v² / r

g tan θ = v² / r

v = √(gr tan θ)

Given g = 9.8 m/s², r = 130 m, and θ = 24.8°:

v = √(9.8 m/s² × 130 m × tan 24.8°)

v = 24.3 m/s

Rounded to two significant figures, the maximum velocity is 24 m/s (approximately 54 mph).

Answer:

24 m/s

Explanation:

The maximum speed that a minivan can have and still follow the curve safely under very icy conditions is 24 m/s.

g = 9.8 m/s², r = 130 m, and θ = 24.8°:

v = √(9.8 m/s² × 130 m × tan 24.8°)

v = 24.3 m/s

Answer:

(a): The frequency of the waves is f= 0.16 Hz

Explanation:

T/4= 1.5 s

T= 6 sec

f= 1/T

f= 0.16 Hz (a)

(a) The frequency of the waves is ¹/₆ Hz ≈ 0.167 Hz

(b) The speed of the waves is 5¹/₃ m/s ≈ 5.33 m/s

[tex]\texttt{ }[/tex]

Let's recall the speed of wave and intensity of wave formula as follows:

[tex]\large {\boxed {v = \lambda f}}[/tex]

f = frequency of wave ( Hz )

v = speed of wave ( m/s )

λ = wavelength ( m )

[tex]\texttt{ }[/tex]

[tex]\large {\boxed {I = 2 \pi^2 A^2 f^2 \rho v}}[/tex]

I = intensity of wave ( W/m² )

A = amplitude of wave ( m )

f = frequeny of wave ( Hz )

ρ = density of medium ( kg/m³ )

v = speed of wave ( m/s )

Let's tackle the problem!

[tex]\texttt{ }[/tex]

Given:

time taken = t = 1.50 seconds

distance covered = d = 8.00 m

Asked:

(a) frequency of the waves = ?

(b) speed of the waves = ?

Solution:

[tex]t = \frac{1}{4}T[/tex]

[tex]1.50 = \frac{1}{4}T[/tex]

[tex]T = 4 \times 1.50[/tex]

[tex]T = 6 \texttt{ seconds}[/tex]

[tex]\texttt{ }[/tex]

[tex]f = \frac{1}{T}[/tex]

[tex]f = \frac{1}{6} \texttt{ Hz}[/tex]

[tex]\texttt{ }[/tex]

[tex]d = \frac{1}{4}\lambda[/tex]

[tex]8.00 = \frac{1}{4}\lambda[/tex]

[tex]\lambda = 8.00 \times 4[/tex]

[tex]\lambda = 32 \texttt{ m}[/tex]

[tex]\texttt{ }[/tex]

[tex]v = \lambda f[/tex]

[tex]v = 32 \times \frac{1}{6}[/tex]

[tex]v = 5\frac{1}{3} \texttt{ m/s}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: College

Subject: Physics

Chapter: Sound Waves

Explanation:

It is given that,

Mass of concrete pilling, m = 50 kg

Diameter of wire, d = 1 mm

Radius of wire, r = 0.0005 m

Length of wire, L = 11.2

Young modulus of steel, [tex]Y=20\times 10^{10}\ N/m^2[/tex]

The young modulus of a wire is given by :

[tex]Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}[/tex]

[tex]Y=\dfrac{F.L}{A\Delta L}[/tex]

[tex]\Delta L=\dfrac{F.L}{A.Y}[/tex]

[tex]\Delta L=\dfrac{50\ kg\times 9.8\ m/s^2\times 11.2\ m}{\pi (0.0005\ m)^2\times 20\times 10^{10}\ N/m^2}[/tex]

[tex]\Delta L=0.034\ m[/tex]

So, the wire will stretch 0.034 meters. Hence, this is the required solution.

Answer:

Work is done by the gas  = 5.92 x 10⁵ J = 592 kJ

Explanation:

Work done at fixed pressure, W = PΔV

Pressure, P = 3.7 x 10⁵ Pa

Change in volume, ΔV = 1.6 m³

Substituting the values of pressure and change in volume we will get

Work done at fixed pressure, W = PΔV =  3.7 x 10⁵ x 1.6 = 5.92 x 10⁵ J

Work is done by the gas  = 5.92 x 10⁵ J = 592 kJ

Answer:

The contraction in the rod is 71 mm.

Explanation:

Given that,

original length L'= 2.99 m

Speed [tex]v= 6.49\times10^{7}\ m/s[/tex]

We need to calculate the length

Using expression for length contraction

[tex]L'=\gamma L[/tex]

[tex]L=\dfrac{L'}{\gamma}[/tex]

Where,

[tex]\gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}[/tex]

[tex]L=\sqrt{1-\dfrac{v^2}{c^2}}L'[/tex]

Where, v = speed of observer

c = speed of the light

Put the value into the formula

[tex]L=\sqrt{1-\dfrac{(6.49\times10^{7})^2}{(3\times10^{8})^2}}\times2.99[/tex]

[tex]L=2.919\ m[/tex]

The expression for the contraction in the rod

[tex]d =L'-L[/tex]

[tex]d=2.99-2.919 [/tex]

[tex]d=0.071[/tex]

[tex]d= 71\ mm[/tex]

Hence, The contraction in the rod is 71 mm.

Answer:

The magnetic field strength of the proton is 0.026 Tesla.

Explanation:

It is given that,

Speed of the proton, [tex]v=0.25\times 10^7\ m/s[/tex]

The radius of circular path, r = 0.975 m

It is moving perpendicular to a magnetic field such that the magnetic force is balancing the centripetal force.

[tex]qvB\ sin90=\dfrac{mv^2}{r}[/tex]

[tex]B=\dfrac{mv}{qr}[/tex]

q = charge on proton

[tex]B=\dfrac{1.67\times 10^{-27}\ kg\times 0.25\times 10^7\ m/s}{1.6\times 10^{-19}\ C\times 0.975\ m}[/tex]

B = 0.026 Tesla

So, the magnetic field strength of the proton is 0.026 Tesla.

Explanation:

The motion of satellite is an example of uniform circular motion. In this type of motion, the velocity of object varies at each and every point but its speed is constant. The satellite in a circular orbit travel at a constant speed.

The gravitational force of earth is balanced by the centripetal force such that,

[tex]F_g=F_c[/tex]

A satellite travel at a constant speed because there is no force acting in the direction of motion of satellite i.e. there is no resistive force that opposite the motion of the satellite.

Answer:

No answer is correct.

Explanation:

given data:

Q1 = 400 gpm

P1 = 28 psi

Q2 = ?

P2 = 30 psi

Change in pressure in a pipe is given as

[tex]\Delta P = \frac{32\mu vl}{D^{2}}[/tex]

where v is velocity and it is given as [tex]v = \frac{Q}{A}[/tex]

[tex]\Delta P = \frac{32\mu Ql}{AD^{2}}[/tex]

Therefore, change in pressure is directly proportional to flow

thus we have

[tex]\frac{P_{1}}{Q_{1}}=\frac{P_{2}}{Q_{2}}[/tex]

[tex]Q_{2}=\frac{P_{2}}{P_{1}}*Q_{1}[/tex]

[tex]Q_{2}=\frac{30}{28}*400 = 428.57 gpm[/tex]

[tex]Q_{2} =428.57 gpm[/tex]

no answer is correct

Answer:

428.5 gmp

Explanation:

Given that,

A piping system is operating at quantity = 400 gpm

Pressure = 28 psi

Increased pressure = 30 psi

We need to calculate the resultant flow rate

We know that,

The flow rate is directly proportional to pressure

[tex]Q\propto P[/tex]

Therefore,

[tex]\dfrac{Q_1}{Q_2}=\dfrac{P_1}{P_2}[/tex]

where,

[tex]Q_1\rightarrow 400\text{ gpm}[/tex]

[tex]Q_2\rightarrow x\text{ gpm}[/tex]

[tex]P_1\rightarrow 28\text{ psi}[/tex]

[tex]P_2\rightarrow 30\text{ psi}[/tex]

By substituting into formula

[tex]\dfrac{400}{x}=\dfrac{28}{30}[/tex]

[tex]x=\dfrac{12000}{28}\approx 428.5\text{ gpm}[/tex]

Hence, The resultant flow rate will be 428.5 gmp

Answer: 135 grams

Explanation:

[tex]Q_{absorbed}=Q_{released}[/tex]

As we know that,  

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex]    

where,

[tex]m_1[/tex] = mass of ice = 100 g

[tex]m_2[/tex] = mass of aluminium cup =? g

[tex]T_{final}[/tex] = final temperature  =[tex]20^0C[/tex]

[tex]T_1[/tex] = temperature of ice = [tex]-10^oC[/tex]

[tex]T_2[/tex] = temperature of aluminium cup= [tex]70^oC[/tex]

[tex]c_1[/tex] = specific heat of ice= [tex]2.03J/g^0C[/tex]

[tex]c_2[/tex] = specific heat of aluminium cup = [tex] 0.902 J/g^0C[/tex]

Now put all the given values in equation (1), we get

[tex][100\times 2.03\times (20-(-10))]=-[m_2\times 0.902\times (20-70)][/tex]

[tex]m_2=135g[/tex]

Therefore, the mass of the aluminium cup was 135 g.

By calculating the heat transfer between a 100 g ice cube and an aluminum cup, the mass of the cup is found to be approximately 978.3 g.

To solve this, we will use principles of thermal equilibrium and specific heat capacities. Let's go through the steps:

Step 1: Calculate the heat required to warm the ice to 0 degrees Celsius.

The specific heat capacity of ice is 2.1 J/g°C. The formula for heat is:

Q = m * c * ΔT

m = 100.0 g (mass of ice)

c = 2.1 J/g°C (specific heat capacity of ice)

ΔT = (0°C - (-10°C)) = 10°C

Q₁ = 100 g * 2.1 J/g°C * 10°C = 2100 J

Step 2: Calculate the heat required to melt the ice at 0 degrees Celsius.

The enthalpy of fusion of ice ,as we know ,is 334 J/g.

Q₂ = m * L

m = 100.0 g

L = 334 J/g

Q₂ = 100 g * 334 J/g = 33400 J

Step 3: Calculate the heat required to warm the melted ice from 0°C to 20°C.

The specific heat capacity of water is 4.18 J/g°C.

Q₃ = m * c * ΔT

m = 100.0 g

c = 4.18 J/g°C

ΔT = (20°C - 0°C) = 20°C

Q₃ = 100 g * 4.18 J/g°C * 20°C = 8360 J

Step 4: Calculate the total heat gained by the ice.

[tex]Q_{total[/tex] = Q₁ + Q₂ + Q₃ = 2100 J + 33400 J + 8360 J = 43860 J

Step 5: Calculate the heat lost by the aluminum cup.

The specific heat capacity of aluminum is 0.897 J/g°C. Since we need the heat lost, we use:

[tex]Q_{lost[/tex] = m * c * ΔT

[tex]Q_{lost[/tex] = 43860 J

c = 0.897 J/g°C

ΔT = (70°C - 20°C) = 50°C

Rearranging for m:

m = [tex]Q_{lost[/tex] / (c * ΔT)

m = 43860 J / (0.897 J/g°C * 50°C) = 978.3 g

Therefore, the mass of the aluminum cup is approximately 978.3 g.

Answer:

(a) 1.35 x 10^6 Joule

(b) 321.45 Kcal

Explanation:

Energy consumed in 1 hour = 270 kJ

So, Energy consumed in 5 hour = 270 x 5 = 1350 kJ

(a) Energy in joules = 1350 x 1000 J = 1.35 x 10^6 J

(b) 4.2 Joule = 1 calorie

So, 1.35 x 10^6 Joule = 1.35 x 10^6 / 4.2 = 0.32 x 10^6 Calorie

                                                                =  0.32 x 1000 Kcal = 321.45 Kcal

The electrical energy consumed by the lightbulb is 1..35 x 10⁶J or 322.65 kcal.

The electrical energy consumed by the lightbulb is the product of power and time of energy consumption.

E = Pt

1 h ------------------- 270 kJ

5 h -------------------- ?

= 5 x 270 kJ

= 1,350 kJ

= 1.35 x 10⁶ J.

1 kJ ---------------- 0.239 kcal

1,350 kJ ------------ ?

= 1,350 x 0.239

= 322.65 kcal.

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Answer:

The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.

Explanation:

Given that,

Mass = 2 kg

Radius = 0.5 m

Angular speed = 3 rad/s

Force = 10 N

(I). We need to calculate the rotational kinetic energy

Using formula of kinetic energy

[tex]K.E =\dfrac{1}{2}\timesI\omega^2[/tex]

[tex]K.E=\dfrac{1}{2}\times mr^2\times\omega^2[/tex]

[tex]K.E=\dfrac{1}{2}\times2\times(0.5)^2\times(3)^2[/tex]

[tex]K.E=2.25\ J[/tex]

(II). We need to calculate the instantaneous change rate of the kinetic energy

Using formula of kinetic energy

[tex]K.E=\dfrac{1}{2}mv^2[/tex]

On differentiating

[tex]\dfrac{K.E}{dt}=\dfrac{1}{2}m\times2v\times\dfrac{dv}{dt}[/tex]

[tex]\dfrac{K.T}{dt}=mva[/tex]....(I)

Using newton's second law

[tex]F = ma[/tex]

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

[tex]a=\dfrac{10}{2}[/tex]

[tex]a=5 m/s^2[/tex]

Put the value of a in equation (I)

[tex]\dfrac{K.E}{dt}=mva[/tex]

[tex]\dfrac{K.E}{dt}=mr\omega a[/tex]

[tex]\dfrac{K.E}{dt}=2\times0.5\times3\times5[/tex]

[tex]\dfrac{K.E}{dt}=15\ J/s[/tex]

Hence, The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.

The rotational kinetic energy of the hoop is 2.25 J, and the instantaneous change rate of the kinetic energy is 15 W.

(a) To determine the rotational kinetic energy (RKE) of the hoop, we use the formula for RKE, which is given by:

[tex]\[ RKE = \frac{1}{2} I \omega^2 \][/tex]

where I is the moment of inertia of the hoop and [tex]\( \omega \)[/tex] is the angular speed. The moment of inertia I for a hoop (or a thin-walled cylinder) rotating about its center is:

[tex]\[ I = m r^2 \][/tex]

where m is the mass of the hoop and r is the radius. Given that  m = 2kg and ( r = 0.5 ) m, we can calculate I:

[tex]\[ I = 2 \times (0.5)^2 = 0.5 \text{ kg} \cdot \text{m}^2 \][/tex]

Now, we know [tex]\( \omega = 3 \)[/tex] rad/s. We can substitute the values into the RKE formula:

[tex]\[ RKE = \frac{1}{2} \times 0.5 \times (3)^2 = \frac{1}{2} \times 0.5 \times 9 = 2.25 \text{ J} \][/tex]

So, the rotational kinetic energy of the hoop is 2.25 J.

(b) The instantaneous change rate of the kinetic energy is equal to the power input to the system. Power P due to a torque [tex]\( \tau \)[/tex] is given by:

[tex]\[ P = \tau \omega \][/tex]

The torque [tex]\( \tau \)[/tex] can be calculated from the force F applied tangentially at the rim of the hoop:

[tex]\[ \tau = F \times r \][/tex]

Given that ( F = 10 ) N and ( r = 0.5 ) m, we find:

[tex]\[ \tau = 10 \times 0.5 = 5 \text{ N} \cdot \text{m} \][/tex]

Now, we can calculate the power:

[tex]\[ P = \tau \omega = 5 \times 3 = 15 \text{ W} \][/tex]

Therefore, the instantaneous change rate of the kinetic energy is 15 W.

Answer:

the lowest frequency is [tex]7.5\times 10^{14} Hz[/tex]

Explanation:

In the question it is given that wavelength(L) in the range of 200μm to 400μm.

let ν be frequency of wave v velocity = 3\times 10^8

velocity v= Lν

therefore ν=[tex]\frac{v}{L}[/tex]

frequency ν be lopwest when L will be heighest

ν(lowest)=[tex]\frac{3\times 10^8}{400\times 10^-9}[/tex]

ν=[tex]7.5\times 10^{14} Hz[/tex]

Answer:

9.4 N

Explanation:

m = mass of the book = 1.83 kg

[tex]\mu _{s}[/tex] = Coefficient of static friction = 0.522

[tex]\mu _{k}[/tex] = Coefficient of kinetic friction = 0.283

[tex]f_{s}[/tex] = Static frictional force

F = force needed to make the book move

force needed to make the book move is same as the magnitude of maximum static frictional force applied by the desk on the book

Static frictional force is given as

[tex]f_{s}[/tex] = [tex]\mu _{s}[/tex] mg

Hence, the force need to move the book is given as

F = [tex]f_{s}[/tex] = [tex]\mu _{s}[/tex] mg

F = [tex]\mu _{s}[/tex] mg

F = (0.522) (1.83 x 9.8)

F = 9.4 N

Answer:

The largest equivalent resistance yu can build using these three resistors is a Serie Resistance with the value of R= 16.74 Ω

Explanation:

Adding Resistances in serie is the way to build de largest equivalent value possible.

Rt= R1+R2+R3

Rt= 6.32 + 8.13 + 2.29

Rt= 16.74Ω

Answer:

The second child must exert a force of magnitude 23.3N to keep the door from moving.

Explanation:

We have to find the moment that the first child exerts and then match it to that exercised by the second child.

F1= 17.5N

d1= 0.6m

F2= ?

d2= 0.45m

M= F * d

M1= 17.5N * 0.6m

M1= 10.5 N.m

M1=M2

M2= F2 * 0.45m

10.5 N.m= F2 * 0.45m

10.5 N.m/0.45m = F2

F2=23.3 N

The force that the second child must exert to keep the door from moving is 23.33 N.

A balanced force occurs when an object subjected to different forces are at equilibrium.

F1r1 = F2r2

(17.5 x 0.6) = F2(0.45)

F2 = 23.33 N

Thus, the force that the second child must exert to keep the door from moving is 23.33 N.

Learn more about balanced force here: https://brainly.com/question/27147344

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Answer:

164.33 N

Explanation:

Given:

The mass of the ball, m = 5.57 kg

Angle made by the sag = 19.4°

it is required to find the tension [tex]F_T[/tex] in the string

Note: Refer the attached figure

The tension in the string will be caused by the weight of the wall

thus,

Weight of the ball, W = 5.57 kg × 9.8 m/s² = 54.586 N

Now the resolving the tension [tex]F_T[/tex] in the string into components as shown in the figure attached, the weight of the ball will be balanced by the sin component of the tension

thus,

W = [tex]F_T[/tex]sinΘ

or

54.586 N = [tex]F_T[/tex] sin 19.4°

or

[tex]F_T[/tex] = 164.33 N

Hence, the tension in the string will be 164.33 N

Answer:[tex]4.802 m/s^2[/tex]

Explanation:

height of slope(h) =90 foot

[tex]\theta =40[/tex]

weight of snowboarder=170 pounds\approx 77.1107 kg

[tex]\mu =0.2[/tex]

as the snowboarder is sliding down the slope therefore

Now net acceleration of snowboarder is

[tex]a_{net}=gsin\theta -\mu \cdot gcos\theta[/tex]

[tex]a_{net }=9.81\times sin\left ( 40\right )-0.2\times 9.81\times cos\left ( 40\right )[/tex]

[tex]a_{net }=4.802 m/s^2[/tex]

Answer:

2.13 h

Explanation:

Simple harmonic motion is:

x = A sin(2π/T t + φ) + B

where A is the amplitude, T is the period, φ is the phase shift, and B is the midline.

This can also be written in terms of cosine:

x = A cos(2π/T t + φ) + B

Here, A = d/2, T = 12.8, φ = 0, and B = d/2.  I'll use cosine so that the highest level is at t=0.

x = d/2 cos(2π/12.8 t) + d/2

When x = d - 0.250 d = 0.750 d:

0.750 d = d/2 cos(2π/12.8 t) + d/2

0.250 d = d/2 cos(2π/12.8 t)

0.500 = cos(2π/12.8 t)

π/3 = 2π/12.8 t

12.8/6 = t

t = 2.13

It takes 2.13 hours to fall 0.250 d from the highest level.

The time it takes for the water to fall a distance 0.250d from its highest level in the tide cycle can be calculated by determining a quarter of the period, which is 3.2 hours.

The time it takes for the water to fall a distance 0.250d from its highest level can be calculated using the concept of simple harmonic motion. Since the period of the tide is 12.8 hours, to find the time for the water to fall 0.250d, we need to determine the fraction of the period corresponding to this distance.

Given that the complete cycle from highest to lowest level is 12.8 hours, the time to fall 0.250d would be one-quarter of that period, which equals 3.2 hours.

Answer:

a)W=20 KJ

b) ΔQ= 220 KJ

Explanation:

Given:

V₁=0.1 m^3,   P₁=200 kPa and heat is added to the system such that system expands under constant pressure.

Therefore V₂= 2V₁= 0.2 m^3

a) Work transfer W= P(V₂-V₁)= [tex]200\times(0.2-0.1)\times10^{5} = 2\times10^4 joules[/tex]

W=20 KJ

b) internal energy change ΔU= 200 KJ

from first law we know that ΔQ(net heat transfer)= ΔU + W

ΔQ= [tex]200\times10^3 +2\times10^4[/tex]

ΔQ=[tex]22\times10^4 J[/tex]

ΔQ= 220 KJ

Answer:

Weight on Venus = 443.75 N

Explanation:

Weight of a body is the product of mass and acceleration due to gravity.

So we have

       Weight =  Mass x Acceleration due to gravity

        W = mg

        Mass, m = 50 kg

        Acceleration due to gravity, g = 8.875 m/s²

        W = 50 x 8.875 = 443.75 N

Weight on Venus = 443.75 N

Answer:

443.75 N

Explanation:

Weight is the force with which a planet can attract anybody towards its centre.

Weight = mass of body × acceleration due to gravity on that planet

Weight = 50 × 8.875 = 443.75 N

Answer:

The work done on the test charge is 4.5 joules

Explanation:

It is given that,

Electric potential difference, [tex]V=1000\ V[/tex]

Charged particle, [tex]q=4.5\times 10^{-3}\ C[/tex]

Work done on the test charge is given by the product of test charge and potential difference i.e.

W = q × V

[tex]W=4.5\times 10^{-3}\ C\times 1000\ V[/tex]

W = 4.5 Joules

So, the work done on the test charge is 4.5 joules. Hence, this is the required solution.