Bill and Janet are pulling on opposite sides of a table. Bill pulls with a force of
450 N to the left, and Janet pulls with a force of 300 N to the right. What is
the net force on the table?
O
A. 450 N to the right
O
B. 450 N to the left
C. 150 N to the left
O
D. 150 N to the right

Answer:

a)

492 kJ

b)

Consistent

Explanation:

Q = Heat stored by woman from food = 600 k J

η = Efficiency of woman = 18% = 0.18

Q' = heat transferred to the environment

heat transferred to the environment is given as

Q' = (1 - η) Q

Inserting the values

Q' = (1 - 0.18) (600)

Q' = 492 kJ

b)

Yes the amount of heat transfer is consistent. The process of sweating produces the heat and keeps the body warm  

A woman climbing the Washington Monument metabolizes food energy with 18% efficiency, meaning 82% of the energy is lost as heat. When we calculate this value, we find that 492 kJ of energy is released as heat, which is consistent with the fact that people quickly warm up when exercising.

The woman climbing the Washington Monument metabolizes 6.00×10² kJ of food energy with an efficiency of 18%. This implies that only 18% of the energy consumed is used for performing work, while the remaining (82%) is lost as heat to the environment.

To calculate the energy lost as heat:

The released heat of 492 kJ is consistent with the fact that a person quickly warms up when exercising, because a significant portion of the body's metabolic energy is lost as heat due to inefficiencies in converting energy from food into work.

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I = 28.6mA.

The magnetic field in the center of a solenoid is given by:

B = μ₀NI/L

Clear I from the equation above, we obtain:

I = BL/μ₀N

With B = 1.00 x 10⁻⁴T, L = 0.42m, μ₀ = 4π x 10⁻⁷T.m/A  and N = 1170turns

I = [(1.00 x 10⁻⁴T)(0.42m)]/[(4π x 10⁻⁷T.m/A)(1170turns)]

I = 0.0286A

I = 28.6mA

Final answer:

Direct current (DC) is the flow of electric current in only one direction. According to the electron flow convention, the current flows from the positive terminal to the negative terminal.

Explanation:

Direct current (DC) is the flow of electric current in only one direction. It refers to systems where the source voltage is constant. The direction of current flow in a direct current circuit is from the positive terminal to the negative terminal. This is according to the electron flow convention, where the movement of negative charges (electrons) is considered as the flow of current.

Answer:

The energy transferred is 2.93 MJ.

Explanation:

Given that,

Time = 1 h = 3600 s

Area[tex]A = 1.6\times4.00[/tex]

Width [tex]\Delta x=0.20 m[/tex]

Temperature [tex] T_{1}=20^{\circ}\ C[/tex]

[tex]T_{2}=5^{\circ}\ C[/tex]

For concrete wall,

The thermal conductivity

k = 1.7 W/mK

Using Fourier's law

[tex]Q=kA\dfrac{dt}{\Delta x}[/tex]

Where, A = area

[tex]\Delta x[/tex] = width

[tex]\Delta t[/tex] = change in temperature

k = thermal conductivity

Put the value into the formula

[tex]Q=1.7\times 1.6\times 4.00\times\dfrac{15}{0.20}[/tex]

[tex]Q=816\ W[/tex]

The energy transferred in 1 hour

[tex]E = Q\times t[/tex]

[tex]E=816\times3600[/tex]

[tex]E=2937600[/tex]

[tex]E = 2.93\times10^{6}\ J[/tex]

[tex]E = 2.93\ MJ[/tex]

Hence, The energy transferred is 2.93 MJ.

Answer:

1.13 Wb

Explanation:

First of all, we need to find the area enclosed by the coil.

The perimeter of the square is 2.0 m, so the length of each side is

[tex]L=\frac{2.0}{4}=0.5 m[/tex]

So the area enclosed by the coil is

[tex]A=L^2 = (0.5 m)^2=0.25 m^2[/tex]

Now we can calculate the magnetic flux through the square, which is given by

[tex]\Phi = B A cos \theta[/tex]

where

B = 9.0 T is the strength of the magnetic field

[tex]A=0.25 m^2[/tex] is the area of the coil

[tex]\theta[/tex] is the angle between the direction of the magnetic field and the normal to the coil; since the field is oriented 30.0° above the horizontal and the coil lies in the horizontal plane, the angle between the direction of the magnetic field and the normal to the coil is

[tex]\theta=90^{\circ}-30^{\circ}=60^{\circ}[/tex]

So the magnetic flux is

[tex]\Phi = (9.0)(0.25)(cos 60^{\circ})=1.13 Wb[/tex]

The magnetic flux through the square is 18.0 T·m²

To find the magnetic flux through the square, we need to calculate the area of the square and the component of the magnetic field perpendicular to the square's plane.

The area of the square is given by A = (side length) = (2.0 m)² = 4.0 m²

The component of the magnetic field perpendicular to the square's plane is B_perpendicular = B × sin(30°) = 9.0 T × sin(30°) = 4.5 T.

Therefore, the magnetic flux through the square is given by the product of the area and the component of the magnetic field perpendicular to the square's plane: flux = B_perpendicular × A = 4.5 T × 4.0 m²= 18.0 T·m²

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(a) The frequency from the stationary whistle as heard by the listener is approximately [tex]\(409.62 \, \text{Hz}\)[/tex].

(b) The frequency from the moving whistle as heard by the listener is approximately [tex]\(454.55 \, \text{Hz}\).[/tex]

(c) The beat frequency detected by the listener is approximately [tex]\(44.93 \, \text{Hz}\).[/tex]

To solve this problem, we can use the Doppler effect formula for sound frequencies.

The formula for the apparent frequency [tex](\(f'\))[/tex] heard by an observer when the source and/or observer is in motion relative to the medium is given by:

[tex]\[ f' = \frac{f \cdot (v + v_o)}{(v + v_s)} \][/tex]

where:

- [tex]\( f \)[/tex] is the frequency emitted by the source,

- [tex]\( v \)[/tex] is the speed of sound in the medium (assumed to be constant),

- [tex]\( v_o \)[/tex] is the speed of the observer relative to the medium,

- [tex]\( v_s \)[/tex] is the speed of the source relative to the medium.

Given:

- [tex]\( f = 392 \, \text{Hz} \)[/tex] (frequency emitted by both whistles),

-[tex]\( v = 343 \, \text{m/s} \)[/tex] (speed of sound in air),

- [tex]\( v_o = 15.0 \, \text{m/s} \)[/tex] (speed of the listener),

- [tex]\( v_s = 35.0 \, \text{m/s} \)[/tex] (speed of the moving whistle).

Let's calculate each part of the problem:

(a) Frequency from the stationary whistle [tex](\(f_1'\))[/tex] as heard by the listener:

[tex]\[ f_1' = \frac{f \cdot (v + v_o)}{(v + v_s)} \]\[ f_1' = \frac{392 \, \text{Hz} \cdot (343 \, \text{m/s} + 15.0 \, \text{m/s})}{(343 \, \text{m/s} + 0 \, \text{m/s})} \]\[ f_1' = \frac{392 \, \text{Hz} \cdot (358.0 \, \text{m/s})}{343 \, \text{m/s}} \]\[ f_1' \approx 409.62 \, \text{Hz} \][/tex]

(b) Frequency from the moving whistle [tex](\(f_2'\))[/tex] as heard by the listener:

[tex]\[ f_2' = \frac{f \cdot (v + v_o)}{(v - v_s)} \]\[ f_2' = \frac{392 \, \text{Hz} \cdot (343 \, \text{m/s} + 15.0 \, \text{m/s})}{(343 \, \text{m/s} - 35.0 \, \text{m/s})} \]\[ f_2' = \frac{392 \, \text{Hz} \cdot (358.0 \, \text{m/s})}{308 \, \text{m/s}} \]\[ f_2' \approx 454.55 \, \text{Hz} \][/tex]

(c) Beat frequency detected by the listener:

The beat frequency is the difference in frequencies between the two whistles as heard by the listener:

[tex]\[ \text{Beat frequency} = |f_1' - f_2'| \]\[ \text{Beat frequency} = |409.62 \, \text{Hz} - 454.55 \, \text{Hz}| \]\[ \text{Beat frequency} \approx 44.93 \, \text{Hz} \][/tex]

So, the answers are:

(a) The frequency from the stationary whistle as heard by the listener is approximately [tex]\(409.62 \, \text{Hz}\)[/tex].

(b) The frequency from the moving whistle as heard by the listener is approximately [tex]\(454.55 \, \text{Hz}\).[/tex]

(c) The beat frequency detected by the listener is approximately [tex]\(44.93 \, \text{Hz}\).[/tex]

Final answer:

The question involves calculating the individual masses of two objects that collectively weigh 5.09 kg and attract each other with a gravitational force of 0.98 x 10^-8 N, separated by 19.5 cm. The gravitational formula F = G×(m1×m2)/r^2 and a system of equations are used to find the solution.

Explanation:

The student's question is about determining the mass of two objects that attract each other with a known gravitational force when separated by a certain distance. The given force is 0.98 x 10-8 N, and the total mass of the two objects is 5.09 kg. The distance between them is 19.5 cm (or 0.195 m in SI units).

To solve this problem, we'll use the formula for the gravitational force between two masses, which is:

F = G×(m1×m2)/r2,

where:

Answer:

The current in wire 3 is 0.25 A.

Explanation:

It is given that, three wires meet at a junction.

Current in wire 1, I₁ = 0.4 A

Current in wire 2, I₂ = -0.65 A  (out of the junction)

We need to find the magnitude of the current in wire 3 (I₃). Applying Kirchhoff's current law which states that the sum of current in the circuit at a junction is equal to zero.

[tex]I_1+I_2+I_3=0[/tex]

[tex]I_3=-(I_2+I_1)[/tex]

[tex]I_3=-(-0.65+0.4)\ A[/tex]

I₃ = 0.25 A

So, the current in wire 3 is 0.25 A. Hence, this is the required solution.

Using the principle of Kirchhoff's junction rule, we calculate the current in Wire 3 as being 0.25 amperes.

The understanding of this question relies on the principle of Kirchhoff's junction rule in Physics, which states the sum of currents entering a junction equals the sum of currents leaving the junction. In this case scenario, Wire 1 has a current of 0.40 A into the junction and wire 2 has a current of 0.65 A out of the junction. The current of Wire 3 can be solved via the equation: I1 + I3 = I2, where I1, I2, and I3 are the currents on Wires 1, 2, and 3 respectively. Plugging in the given values, we get: 0.40 A + I3 = 0.65 A. Solving the equation for I3, we find that I3 = 0.25 A. Hence, the magnitude of the current in Wire 3 is 0.25 A.

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Final answer:

The radius of the circle is approximately 1.392 m (rounded to one decimal place).

Explanation:

To find the radius of the circle, we need to use the formula for the area of a sector. The area of a sector is given by the formula A = (θ/2π) × πr², where θ is the central angle in radians and r is the radius. In this case, we are given that the central angle is 2π/11 radians and the area is 25 m². We can set up the equation as 25 = (2π/11) × πr² and solve for r.

Solution:

25 = (2π/11) × πr²

25 = (2π²/11) × r²

r² = 11/2π

r ≈ √(11/2π)

r ≈ 1.392 m (rounded to one decimal place)

Answer:

here the coil must be oriented in such a way that its plane is perpendicular to the magnetic field

Explanation:

As we know by Faraday's law of electromagnetic induction

Rate of change in magnetic flux will induce EMF in the coil

so here we will have

[tex]EMF = \frac{d\phi}{dt}[/tex]

here we know that

[tex]\phi = NB.A[/tex]

now if the magnetic flux will change with time then it will induce EMF in the coil

[tex]EMF = N\frac{d}{dt}(B.A)[/tex]

so here induced EMF will be zero in the coil if the flux linked with the coil will remain constant

so here the coil must be oriented in such a way that its plane is perpendicular to the magnetic field

In such a way when coil will rotate then the flux linked with the coil will remains constant and there will be no induced EMF in it

To have zero induced emf in a circular loop rotating in a magnetic field, the loop's axis of rotation should be parallel to the magnetic field lines.

The orientation of the loop's axis of rotation must be parallel to the magnetic field lines in a region where a uniform magnetic field points straight down for the induced emf to be zero. When the loop's axis is perpendicular to the magnetic field, the maximum emf is induced. Thus, to minimize the induced emf, the axis should be parallel to the magnetic field.

Answer:

Part a)

[tex]Q = 320 \mu C[/tex]

Part b)

[tex]t = 8.52 \times 10^{-5} s[/tex]

Part c)

Rate of energy = 301.5 J/s

Explanation:

Part a)

Since energy is always conserved in LC oscillating system

So here for maximum charge stored in the capacitor is equal to the magnetic field energy stored in inductor

[tex]\frac{1}{2}Li^2 = \frac{Q^2}{2C}[/tex]

now we have

[tex]Q = \sqrt{LC} i[/tex]

[tex]Q = \sqrt{(3.76 \times 10^{-3})(3.13 \times 10^{-6})} (2.95)[/tex]

[tex]Q = 320 \mu C[/tex]

Part b)

Energy stored in the capacitor is given as

[tex]U = \frac{q^2}{2C}[/tex]

now rate of energy stored is given as

[tex]\frac{dU}{dt} = \frac{q}{C}\frac{dq}{dt}[/tex]

so here we also know that

[tex]q = Q sin(\omega t)[/tex]

[tex]\frac{dq}{dt} = Q\omega cos(\omega t)[/tex]

now from above equation

[tex]\frac{dU}{dt} = \frac{Qsin(\omega t)}{C} (Q\omega cos\omega t)[/tex]

so maximum rate of energy will be given when

[tex]sin\omega t = cos\omega t[/tex]

[tex]\omega t = \frac{\pi}{4}[/tex]

[tex]t = \frac{\pi}{4}\sqrt{LC}[/tex]

[tex]t = 8.52 \times 10^{-5} s[/tex]

Part c)

Greatest rate of energy is given as

[tex]\frac{dU}{dt} = \frac{Q^2\omega}{C}[/tex]

[tex]\frac{dU}{dt} = \frac{(320 \mu C)^2 \sqrt{\frac{1}{(3.76 mH)(3.13 \mu C)}}}{3.13 \mu C}[/tex]

[tex]\frac{dU}{dt} = 301.5 J/s[/tex]

Answer:

157.8 J

Explanation:

m = mass of the cylinder = 7 kg

h = height difference in top and bottom of the incline = 2.3 m

g = acceleration due to gravity = 9.8 m/s²

TE = Total Energy at the bottom

PE = Gravitational potential energy at the top

Using conservation of energy

Total Energy at the bottom = Gravitational potential energy at the top  

TE = PE

TE = m g h

TE = (7) (9.8) (2.3)

TE = 157.8 J

Answer:

Wavelength, [tex]\lambda=3.01\ m[/tex]

Explanation:

It is given that,

Frequency, f = 99.5 MHz = 99.5 × 10⁶ Hz

We need to find the wavelength of the radio waves from an FM station operating at above frequency. The relationship between the frequency and the wavelength is given by :

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

[tex]\lambda=\dfrac{c}{f}{[/tex]

c = speed of light

[tex]\lambda=\dfrac{3\times 10^8\ m/s}{99.5\times 10^6\ Hz}[/tex]

[tex]\lambda=3.01\ m[/tex]

So, the wavelength of the radio waves from an FM station is 3.01 m. Hence, this is the required solution.

Answer:

4.49 x 10^-11 newton

Explanation:

v = 12.3 m/s along north = 12.3 j m/s

B = 4.78 x 10^-5 T downwards = 4.78 x 10^-5 k T

q = 7.64 x 10^-8 C

force on a charged particle when it is moving in a uniform magnetic field is given by

F = q (v x B )

F = 7.64 x 10^-8 {(12.3 i) x (4.78 x 10^-5 k)}

F = 4.49 x 10^-11 (- k) newton

magnitude of force = 4.49 x 10^-11 newton

Answer :

The time is [tex]1\times10^{5}\ yrs[/tex]

(B) is correct option.

Explanation :

Given that,

Diameter = 100000 light year

Velocity = 0.99 c

We need to calculate the time

Using formula of time

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

Put the value into the formula

[tex]t =\dfrac{100000 c}{0.99c}[/tex]

[tex]t =1\times10^{5}\ yrs[/tex]

Hence, The time is [tex]1\times10^{5}\ yrs[/tex]

Answer:

5.6 Tesla

Explanation:

L = 52 cm = 0.52 m

V = 2.75 m/s

e = 8 V

Let B be tha magnitude of magnetic field. Use the formula for the motional emf

e = B × V × L

B = e / V L

B = 8 / (2.75 × 0.52)

B = 5.6 Tesla

Answer:

[tex]N = 8.1 \times 10^{10}[/tex]

Explanation:

Initial charge on the rod is

[tex]Q_i = 14 nC[/tex]

final charge on the rod is

[tex]Q_f = 1 nC[/tex]

now the charge transferred from to the sphere is given as

[tex]\Delta Q = Q_i - Q_f[/tex]

[tex]\Delta Q = 14 - 1 = 13 nC[/tex]

now we also know that

Q = Ne

so number of particles transferred is

[tex]N = \frac{\Delta Q}{e}[/tex]

[tex]N = \frac{13 \times 10^{-9}}{1.6 \times 10^{-19}}[/tex]

[tex]N = 8.1 \times 10^{10}[/tex]

When the charged plastic rod touches the metal sphere, it transfers around 13 nC of charge to the sphere. This amounts to about 8.125 x 10^10 electrons, as each electron carries a charge of approximately -1.6 x 10^-19 C.

When the charged plastic rod touches the metal sphere, charge transfer occurs until both objects have the same charge. Initially, the plastic rod had a charge of -14 nC and after touching it had a charge -1.0 nC, which means that 13 nC was transferred to the metal sphere.

The elementary charge (charge of an electron) is approximately -1.6 x 10^-19 C. Therefore, the number of electrons transferred would be the total transferred charge divided by the charge of one electron, which results in approximately 8.125 x 10^10 electrons.

The charge is negative indicating electrons, which have negative charge, were transferred.

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Answer:

The pilot must keep the tip pointed at 12.1 degrees to the right with respect to the direction of the runway to align the flight path with the runway.

Explanation:

x= -10m/s

y= 47m/s

r= √(x²)+(y²)

r=48.05 m/s

β= tan⁻¹(y/x)

β=102.01°

the runway is at 90 degrees. Considering the wind, the airplane is flying at 102.01 º direction. Must fly at 12.1 degrees to the right with respect to the direction of the runway to contrarest the wind effect.

The reactance of an inductor is given by:

X = 2πfL

X is the inductor's reactance

f is the frequency of the supplied voltage

L is the inductor's inductance

The given values are:

f = 60.0Hz

L = 43.8mH (I'm assuming the value is given in milli Henries because this is within the normal range of inductors)

Plug these values in and solve for X:

X = 2π(60.0)(43.8×10⁻³)

X = 16.512Ω

Round this value to 3 significant figures:

X = 16.5Ω

The relationship between AC voltage and current is given by:

V = IZ

V is the voltage

I is the current

Z is the impedance

For an AC inductor circuit, Z = X = 16.512Ω and V is the rms voltage 120V. Plug these values in to get the rms current:

120 = I×16.512

I = 7.2673A

Round this value to 3 significant figures:

I = 7.27A

Final answer:

The inductive reactance is 16,515 Ohms and the rms current through the inductor is 7.3 mA for an AC source with an rms voltage of 120 V operating at a frequency of 60 Hz.

Explanation:

To determine the inductive reactance and the rms current through the inductor in a purely inductive AC circuit, we use the inductive reactance formula XL = 2πfL, where f is the frequency and L is the inductance of the coil.

In this case, the frequency f is 60 Hz and the inductor has an inductance L of 43.8 H. The inductive reactance, XL, can be calculated as:

XL = 2π × 60 Hz × 43.8 H ≈ 16,515 Ohms (or 16.5 kΩ)

Once we have the inductive reactance, we can calculate the rms current using Ohm's law, I = V/XL, where I is the current and V is the rms voltage of the AC source. With an rms voltage of 120 V, the rms current is:

I = 120 V / 16,515 Ohms ≈ 0.0073 A (or 7.3 mA)

Answer:

a) The 99% confidence interval for the mean noise level = [122.44, 151.96]

b) Sample standard deviation, s = 17.3dB

Explanation:

Noise levels at 5 airports = 147,123,119,161,136

Mean noise level

        [tex]\bar{x} =\frac{ 147+123+119+161+136}{5}=137.2dB[/tex]

Variance of noise level

        [tex]\sigma^2 =\frac{ (137.2-147)^2+(137.2-123)^2+(137.2-119)^2+(137.2-161)^2+(137.2-136)^2}{5}\\\\\sigma^2=164.16[/tex]

Standard deviation,

       [tex]\sigma =\sqrt{164.16}=12.81dB[/tex]

a) Confidence interval  is given by

       [tex]\bar{x}-Z\times \frac{\sigma}{\sqrt{n}}\leq \mu\leq \bar{x}+Z\times \frac{\sigma}{\sqrt{n}}[/tex]

For 99% confidence interval Z = 2.576,

Number of noises, n = 5

Substituting

    [tex]137.2-2.576\times \frac{12.81}{\sqrt{5}}\leq \mu\leq 137.2+2.576\times \frac{12.81}{\sqrt{5}}\\\\122.44\leq \mu\leq 151.96[/tex]

The 99% confidence interval for the mean noise level = [122.44, 151.96]

b) Sample standard deviation

        [tex]s=\sqrt{\frac{ (137.2-147)^2+(137.2-123)^2+(137.2-119)^2+(137.2-161)^2+(137.2-136)^2}{5-1}}\\\\s=17.3dB[/tex]

   Sample standard deviation, s = 17.3dB

Using a formula linking initial and final angular velocities, angular acceleration, and total angle, the final angular velocity can be obtained. Multiplying the final angular velocity and radius gives the final linear velocity. Angular acceleration and linear acceleration on the rim of a wheel are related by their radius.

Given the angular acceleration, a, the number of revolutions, and the radius of the wheel, r, you can determine the angular velocity at the end of 2 revolutions using the formula w_f = sqrt(w_i^2 + 2*a*θ), where w_i is the initial angular velocity, w_f is the final angular velocity, a is angular acceleration, and θ is the total angle swept out in radians. The angular velocity can then be used to find the final linear velocity, v = r*w_f.

Angular acceleration is directly given by a = α*r, where α is angular acceleration and r is the radius. Angular acceleration is also related to the linear (tangential) acceleration, a_t, by a = r*α, where α is angular acceleration and r is the radius. Tangential acceleration involves not only changes in speed (which causes tangential acceleration) but also changes in direction (which causes radial or centripetal acceleration).

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Answer:

speed of boat as

[tex]v_b = 10 mph[/tex]

river speed is given as

[tex]v_r = 2 mph[/tex]

Explanation:

When boat is moving down stream then in that case net resultant speed of the boat is given as

since the boat and river is in same direction so we will have

[tex]v_1 = v_r + v_b[/tex]

Now when boat moves upstream then in that case the net speed of the boat is opposite to the speed of the river

so here we have

[tex]v_2 = v_b - v_r[/tex]

as we know when boat is in downstream then in that case it covers 24 miles in 2 hours

[tex]v_1 = \frac{24}{2} = 12 mph[/tex]

also when it moves in upstream then it covers same distance in 3 hours of time

[tex]v_2 = \frac{24}{3} = 8 mph[/tex]

[tex]v_b + v_r = 12 mph[/tex]

[tex]v_b - v_r = 8 mph[/tex]

so we have speed of boat as

[tex]v_b = 10 mph[/tex]

river speed is given as

[tex]v_r = 2 mph[/tex]

The rate of the boat in still water is 10 mph

The rate of the current is 2 mph

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

Given:

distance covered = d = 24 miles

time for driving down the river = td = 2 hours

time for driving up the river = tu = 3 hours

Unknown:

velocity of the boat in still water = vs = ?

velocity of the current = vc = ?

Solution:

When Marcus drive his boat down the river , the velocity of the boat is in the same direction to the velocity of the current.

[tex]v_s + v_c = \frac{d}{t_d}[/tex]

[tex]v_s + v_c = \frac{24}{2}[/tex]

[tex]v_s + v_c = 12[/tex]

[tex]v_s = 12 - v_c[/tex] → Equation 1

When Marcus drive his boat up the river , the velocity of the boat is in the opposite direction to the velocity of the current.

[tex]v_s - v_c = \frac{d}{t_d}[/tex]

[tex]v_s - v_c = \frac{24}{3}[/tex]

[tex]v_s - v_c = 8[/tex]

[tex]12 - v_c - v_c = 8[/tex] ← Equation 1

[tex]12 - 2v_c = 8[/tex]

[tex]2v_c = 12 - 8[/tex]

[tex]2v_c = 4[/tex]

[tex]v_c = 4 \div 2[/tex]

[tex]\large {\boxed {v_c = 2 ~ mph} }[/tex]

[tex]v_s = 12 - v_c[/tex]

[tex]v_s = 12 - 2[/tex]

[tex]\large {\boxed {v_s = 10 ~ mph} }[/tex]

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate

The launch angle that results in the maximum range of a projectile up an incline depends on the initial speed and the angle of the incline. For conditions neglecting air resistance, the maximum range is obtained at 45 degrees. If air resistance is considered, the maximum angle is around 38 degrees.

The range of a projectile launched up an incline depends on the launch angle. To determine the launch angle that results in the maximum range, we need to consider the initial speed and the angle of the incline. Figure 3.38(b) shows that for a fixed initial speed, the maximum range is obtained at 45 degrees. However, this is only true for conditions neglecting air resistance. If air resistance is considered, the maximum angle is around 38 degrees. It is also interesting to note that for every initial angle except 45 degrees, there are two angles that give the same range, and the sum of those angles is 90 degrees.

Answer:

6593.4

Explanation:

wavelength, λ = 910 nm = 910 x 10^-9 m

Speed of laser = 3 x 10^8 m/s

t = 20 pico seconds

Distance traveled by laser in this time, d = c x t

d = 3 x 10^8 x 20 x 10^-12 = 6 x 10^-3 m

number of wavelengths, n = d / λ

n = (6 x 10^-3) / (910 x 10^-9)

n = 6593.4

Answer:

4.665×10⁻¹¹ N

Explanation:

Intensity of solar radiation=I=1.4×10³ W/m²

Area of panel=A=5.0 m²

speed of sound=c=3×10⁸ m/s

[tex]P_t=\text {Total\ Solar\ Radiation\ Pressure}[/tex]

[tex]P_t=2\frac{I}{c}\\\Rightarrow P_t=2\frac{1.4\times 10^{-3}}{3\times 10^{8}}\\\Rightarrow P_t=0.933\times 10^{-11}[/tex]

∴Total Solar Radiation Pressure=0.933×10⁻¹¹ Ws/m³

Force= Pressure×Area

Force=0.933×10⁻¹¹×5

Force=4.665×10⁻¹¹ Ws/m=4.665×10⁻¹¹ N

∴Force is exerted by solar radiation impinging normally on a perfectly reflecting panel of an artificial satellite orbiting the earth is 4.665×10⁻¹¹ N

The force exerted by solar radiation on the panel is 7.0 × 10³ N.

The force exerted by solar radiation can be calculated using the formula: force = power/area. In this case, the power is 1.4 × 10³ W/m² and the area is 5.0 m². Substituting these values in, the force exerted by solar radiation on the panel is: force = (1.4 × 10³ W/m²) x (5.0 m²) = 7.0 × 10³ N.

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Answer:

Sagittarius

Explanation:

The center of the Milky Way galaxy lies in the direction of the Sagittarius constellation, about 26,000 light-years away.

The center of the Milky Way galaxy is in the direction of the Sagittarius constellation and is about 26,000 light-years away from us.

The center of the Milky Way galaxy lies in the direction of the Sagittarius constellation and is approximately 26,000 light-years away. When we look towards Sagittarius, we are looking towards the densest part of the Milky Way, which is why it appears brighter in the night sky.

A light-year, by definition, is the distance that light can travel in one year. Hence, saying that the center of the Milky Way is 26,000 light-years away means it would take 26,000 years for the light from the center of the galaxy to reach us.

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Answer:

Torque is 93 Nm anticlockwise.

Explanation:

We have value of torque is cross product of position vector and force vector.

A force of 38 N, directed 30° above the x axis in the x-y plane.

        Force, F = 38 cos 30 i + 38 sin 30 j = 32.91 i + 19 j

A particle is located on the x axis 4.9 m and we have to find torque about the origin on the particle.

Position vector, r = 4.9 i

Torque, T = r x F = 4.9 i x (32.91 i + 19 j) = 4.9 x 19 k = 93.1 k Nm

So Torque is 93 Nm anticlockwise.

The orbital velocity of the satellite can be calculated using the formula v = √(GM/r), where v is the orbital velocity, G is the universal gravitational constant, M is the mass of Mars, and r is the radius of the orbit.

To calculate the orbital velocity of the satellite, we can use the formula for orbital velocity:

v = √(GM/r)

where v is the orbital velocity, G is the universal gravitational constant, M is the mass of Mars, and r is the radius of the orbit. Plugging in the known values, we have:

v = √((6.674×10-11 m3/kg/s2)(6.419×1023 kg)/(3396000 m + 300000 m))

Calculating this will give us the orbital velocity of the satellite.

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The orbital velocity of the satellite is approximately 3,404 meters per second.

Sure, here is the solution to the problem:

Given:

Radius of Mars (R) = 3,396 km = 3.396 × 10⁶ m

Mass of Mars (M) = 6.419 × 10²³ kg

Universal gravitational constant (G) = 6.674 × 10⁻¹¹ m³/kg·s²

To find:

Orbital velocity (v)

Formula:

The orbital velocity of a satellite in a circular orbit is given by the following formula:

v = √(GM / r)

where:

G is the universal gravitational constant

M is the mass of the planet

r is the radius of the orbit

Calculation:

First, convert the radius of Mars from kilometers to meters:

r = 3.396 × 10⁶ m

Now, plug in the values into the formula:

v = √((6.674 × 10⁻¹¹ m³/kg·s²) × (6.419 × 10²³ kg) / (3.396 × 10⁶ m))

v ≈ 3,404 m/s

To solve this question, we need to use the concept of the conversion between potential and kinetic energy.

The roller coaster's initial potential energy before it goes down the hill can be calculated using the formula for gravitational potential energy: `PE = m * g * h`,

where `m` is the mass,

           `g` is the acceleration due to gravity, and

           `h` is the height.

Substituting the given values, where `m` is 1200 kg, `g` is 9.81 m/s² and the initial height `h` is 19 m,

`PE_initial = 1200 kg * 9.81 m/s² * 19 m = 223668 J`

This is the energy the roller coaster has due to its position at the top of the 19m high hill before it starts to move.

When the roller coaster reaches the hill that is 13 m high, we can calculate its potential energy at this point the same way we calculated the initial potential energy, with `h` being now the final height of 13 m,

`PE_final = 1200 kg * 9.81 m/s² * 13 m = 153036 J`

This is the energy the coaster has due to its position at the top of the 13m high hill.

The kinetic energy (KE) of the roller coaster at this point is gained by the conversion of some of the initial potential energy into kinetic energy. This conversion is equal to the difference between the initial potential energy and the final potential energy:

`KE = PE_initial - PE_final = 223668 J - 153036 J = 70632 J`

So, when the roller coaster goes over the hill that is 13 m high, its kinetic energy is 70632 J. This is the energy the roller coaster has due to its speed at this point.

Learn more about Conservation of Mechanical Energy here:

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Given:

applied tensile stress, [tex]\sigma[/tex] = 170 MPa

radius of curvature of crack tip,  [tex]r_{t}[/tex] =  [tex]3.5\times 10^{-4}[/tex] mm

crack length = [tex]4.5\times 10^{-2}[/tex] mm

half of internal crack length, a = [tex]\frac{crack length}{2} = \frac{4.5\times 10^{-2}}{2}[/tex]

a =  [tex]2.25\times 10^{-2}[/tex]

Formula Used:

[tex]\sigma _{max} =  2\times\sigma \sqrt{\frac{a}{r_t}}[/tex]

Solution:

Using the given formula:

[tex]\sigma _{max} = 2\times170 \sqrt{\frac{2.25\times 10 ^{-2}}{3.5\times 10^{-4}}}[/tex]

[tex]\sigma _{max}[/tex] = 2726 MPa (395372.9 psi)

The magnitude of the maximum stress at the tip of an internal crack can be determined using the stress concentration factor formula.

The magnitude of the maximum stress at the tip of an internal crack can be determined using the formula for the stress concentration factor, which is the ratio of the maximum stress to the applied stress. The stress concentration factor (Kt) for an internal crack can be calculated using the following equation:

Where Kt is the stress concentration factor, a is the crack length, and r is the radius of curvature of the crack.

Plugging in the given values:

Kt = 1 + 2 * (4.5 x 10-2 mm) / (3.5 x 10-4 mm) = 127

The magnitude of the maximum stress can be calculated by multiplying the stress concentration factor by the applied stress:

Maximum Stress = 127 * 170 MPa = 21,590 MPa