The gauge pressure in your car tires is 2.70 ✕ 105 N/m2 at a temperature of 35.0°C when you drive it onto a ship in Los Angeles to be sent to Alaska. What is their gauge pressure (in atm) later, when their temperature has dropped to −42.0°C? Assume the tires have not gained or lost any air.

Answer:

Option (a), (b) and (c)

Explanation:

The resistance of a conductor depends on the length of the conductor, area of crossection of the conductor and the nature of the conductor.

The formula for the resistance is given by

R = ρ x l / A

Where, ρ is the resistivity of the conductor, l be the length of the conductor and A be the area of crossection of the conductor.

So, It depends on the length, area and the type of material.

Resistance of a material is determined by its length, cross-sectional area, and the type of material; voltage and current do not determine resistance but are influenced by it.

The Quantities Determining the Resistance of a Material

The resistance of a piece of material is determined by several key factors, specifically:

It is important to note that the voltage across the material and the current flowing through the material are not factors that determine resistance. These two quantities are actually influenced by the resistance according to Ohm's law, which states that Voltage (V) equals Current (I) times Resistance (R), or V = IR.

Answer:

Option D is the correct answer.

Explanation:

Since value of angular acceleration is constant, the body has only centripetal acceleration.

Centripetal acceleration

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

We have radius = 7.112 cm = 0.07112 m

Frequency, f = 1975 rpm = 32.92 rps

Angular frequency, ω = 2πf = 2 x π x 32.92 = 206.82 rad/s

Substituting in centripetal acceleration equation,

              [tex]a=r\omega ^2=0.07112\times 206.82^2=3042.17m/s^2[/tex]

Option D is the correct answer.                

Answer:

magnitude : 0.6 m/s²

Direction : Left

Explanation:

m = mass of the cart = 90 kg

Taking force in right direction as positive and force in left direction as negative

F₁ = Force applied by man 1 = 140 N

F₂ = Force applied by other man = - 195 N

a = acceleration of the cart

Force equation for the motion of the cart is given as

F₁ + F₂ = ma

140 + (- 195) = 90 a

a = - 0.6 m/s²

magnitude of acceleration is 0.6 m/s²

The negative sign indicates the direction of acceleration towards left

Final answer:

The magnitude of the cart's acceleration is approximately 0.611 m/s², and the direction is to the left.

Explanation:

When calculating the acceleration of the cart, we need to consider the net force acting on it and its mass. The net force is found by subtracting the smaller force from the larger force, taking into account their directions. With forces of 140 N to the right and 195 N to the left, the net force is the difference, which is 195 N - 140 N = 55 N, directed to the left since the larger force is in that direction. Using Newton's second law, acceleration (a) is the net force (Fnet) divided by the mass (m).

To find the magnitude of the acceleration, calculate a = Fnet / m. So, a = 55 N / 90 kg which equals approximately 0.611 m/s2. Since the larger force was to the left, the direction of the acceleration is also to the left.

Answer:

D. 5.18 x 10⁻¹²

Explanation:

[tex]\frac{dE}{dt}[/tex] = rate at which sun radiates energy = 3.92 x 10²⁶ W

M = mass of sun = 1.99 x 10³⁰ kg

[tex]\frac{dm}{dt}[/tex] = rate at which sun's mass is lost

c = speed of light

Energy is given as

E = m c²

Taking derivative both side relative to "t"

[tex]\frac{dE}{dt}=c^{2}\frac{dm}{dt}[/tex]

[tex]3.92\times 10^{26}=(3\times 10^{8})^{2}\frac{dm}{dt}[/tex]

[tex]\frac{dm}{dt}[/tex] = 4.4 x 10⁹ kg/s

t = time interval = 75 yrs = 75 x 365 days = 75 x 365 x 24 hours = 75 x 365 x 24 x 3600 sec = 2.4 x 10⁹ sec

[tex]m[/tex] = mass lost

mass lost is given as

[tex]m = t\frac{dm}{dt}[/tex]

[tex]m = (2.4\times 10^{9})(4.4\times 10^{9})[/tex]

m = 10.56 x 10¹⁸ kg

fraction is given as

fraction = [tex]\frac{m}{M}[/tex]

fraction = [tex]\frac{10.56\times 10^{18}}{1.99\times 10^{30}}[/tex]

fraction = 5.18 x 10⁻¹²

Answer:

The horizontal distance traveled by the ball before it hits the ground is 48.85 meters.

Explanation:

It is given that,

Speed of golf ball, v = 25 m/s

Angle above horizontal or angle of projection, θ = 65°

We need to find the distance travelled by the ball before it hots the ground or in other words we need to find the range. It is given by R.

[tex]R=\dfrac{v^2\ sin2\theta}{g}[/tex]

[tex]R=\dfrac{(25\ m/s)^2\ sin2(65)}{9.8\ m/s^2}[/tex]

R = 48.85 m

So, the distance travelled by the ball before it hots the ground is 48.85 meters. Hence, this is the required solution.

To calculate the total resistance of three resistors connected in parallel, use the formula 1/R = 1/R1 + 1/R2 + 1/R3. The maximum error in the calculated value of R can be estimated by multiplying the sum of the errors in each resistance by the calculated value of R.

To find the total resistance of three resistors connected in parallel, we use the formula 1/R = 1/R1 + 1/R2 + 1/R3. Given the resistances R1 = 100 Ω, R2 = 25 Ω, and R3 = 10 Ω, we can substitute these values into the formula to calculate the total resistance R. Therefore, 1/R = 1/100 + 1/25 + 1/10 = 0.01 + 0.04 + 0.1 = 0.15. Now, to estimate the maximum error in the calculated value of R, we consider the errors in each resistance. Since each resistance has a possible error of 0.5%, we can calculate the maximum error in R by multiplying the sum of the errors in each resistance by the calculated value of R. Therefore, maximum error in R = 0.005 * 0.15 = 0.00075 Ω.

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The maximum error in the total resistance R of three parallel resistors with a potential error of 0.5% in each resistor is approximately 0.667 ohms.

The question asks to calculate the maximum error in the calculated value of total resistance R when three resistors R1, R2, and R3 are connected in parallel, where R1 = 100 Ω, R2 = 25 Ω, and R3 = 10 Ω, each with a possible error of 0.5%. The resistors in parallel have a total resistance denoted by:

1/R = 1/R1 + 1/R2 + 1/R3

To find the maximum error in the calculated value of R, we will first calculate R and then use derivative rules to estimate the maximum error considering the errors in R1, R2, and R3.

After calculating 1/R using the given resistances:

1/R = 1/100 + 1/25 + 1/10

1/R = 0.01 + 0.04 + 0.1 = 0.15

Therefore, R = 1 / 0.15 = 6.667 Ω

We calculate the maximum possible errors in resistances as:

Using the formula for the propagation of errors for functions of several independent variables, we estimate the maximum error in R (eR) as:

eR ≈ | -R² * eR1/R1² | + | -R² * eR2/R2² | + | -R² * eR3/R3² |

Plugging in the values:

eR ≈ | -6.667² * 0.5/100² | + | -6.667² * 0.125/25² | + | -6.667² * 0.05/10² |

eR ≈ | -0.04446 | + | -0.17784 | + | -0.4446 | = 0.667 Ω (Approximated to three decimal places)

The estimated maximum error in the calculated value of R is therefore approximately 0.667 Ω.

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Answer:

equal

Explanation:

According to the Newton's second law of motion, the force acting on the body is directly proportional to the rate of change of momentum.

If the force is same, then the change in momentum is also same for a given time.

The change of the boulder’s momentum in one second greater is equal to the change of the pebble’s momentum in the same time period.

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration.

The given data in the problem is;

F is the net force=200 N

m is the mass of boulder=100 kg

M is the mass of pebble=100 g

The force acting on the body is directly proportional to the rate of change in momentum, according to Newton's second law of motion.

For a given time, if the force is the same, the change in momentum is likewise the same.

Hence the change of the boulder’s momentum in one second greater is equal to the change of the pebble’s momentum in the same time period.

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Answer:

[tex]F=55146000 N[/tex]

Explanation:

Given:

Pressure on the building = 170 N/m²

Pressure function P(y) = 170 + 2y

Consider a small height of the building as dy

the small force (dF) exerted on the height dy will be given as

dF = P(y)×(Area of the building face)

or

dF = (170 + 2y)(6565)dy

or

dF = 170×6565×dy + 2y×6565×dy

dF = 1116050dy + 13130ydy

integerating the function for the whole height of 40 m

[tex]\int\limits^{40}_0 {F} = \int _0^{40}\:\left(\:1116050dy\:+\:13130ydy\right)[/tex]

[tex]F = \int _0^{40}1116050dy+\int _0^{40}13130ydy[/tex]

[tex]F=\left[1116050y\right]^{40}_0+13130\left[\frac{y^2}{2}\right]^{40}_0[/tex]

[tex]F=44642000+10504000[/tex]

[tex]F=55146000 N[/tex] ; Total force exerted on the face of the building

To calculate the total force on a building due to wind pressure, integrate the pressure function over the building's height and multiply by the building's width. This calculation is vital for determining the building's required resistance to wind forces for safe and stable design.

The question asks to calculate the total force exerted on a large building by a wind blowing against one of its faces. The wind pressure increases with height as described by the function P(y)=170+2y, where y is the height in meters above the ground. To find the total force, we need to integrate this pressure function over the height of the building and multiply by the width of the building's face.

First, we can write the integral of the pressure over the height of the building:

Integrate [P(y) dy] from y=0 to y=40

= Integrate [(170+2y) dy] from y=0 to y=40

Performing this integration will give us the total pressure exerted over the entire height of the building. After integrating, we multiply the result by the width of the building (6565 m) to obtain the total force.

Using the integration results, the total force, F, can be expressed as:

F = (Integral result) × Width of the building

This force calculation is critical for architects and engineers as it provides an estimate of the resistance that the building must be designed to withstand due to wind forces, ensuring structural stability and safety.

Answer:

a) 250 N/m

b) 22.4 rad/s , 3.6 Hz , 0.28 sec

c) 0.3125 J

Explanation:

a)

F = force applied on the spring = 7.50 N

x = stretch of the spring from relaxed length when force "F" is applied = 3 cm = 0.03 m

k = spring constant of the spring

Since the force applied causes the spring to stretch

F = k x

7.50 = k (0.03)

k = 250 N/m

b)

m = mass of the particle attached to the spring = 0.500 kg

Angular frequency of motion is given as

[tex]w = \sqrt{\frac{k}{m}}[/tex]

[tex]w = \sqrt{\frac{250}{0.5}}[/tex]

[tex]w [/tex] = 22.4 rad/s

[tex]f[/tex] = frequency

Angular frequency is also given as

[tex]w [/tex] = 2 π [tex]f[/tex]

22.4 = 2 (3.14) f

[tex]f[/tex]  = 3.6 Hz

[tex]T[/tex] = Time period

Time period is given as

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

[tex]T = \frac{1}{3.6}[/tex]

[tex]T[/tex] = 0.28 sec

c)

A = amplitude of motion = 5 cm = 0.05 m

Total energy of the spring-block system is given as

U = (0.5) k A²

U = (0.5) (250) (0.05)²

U = 0.3125 J

(a) The force constant of the spring is 250 N/m.

(b) The angular frequency of the mass oscillation is 22.36 rad/s, frequency is 3.56 Hz and the period is 0.28 s.

(c)  the total energy of the system is 0.31 J.

The force constant of the spring can be determined by applying Hooke's law as follows;

F = kx

k = F/x

k = (7.5)/0.03)

k = 250 N/m

The angular frequency of the mass oscillation is calculated as follows;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{250}{0.5} }\\\\\omega = 22.36 \ rad/s[/tex]

The angular frequency is calculated as follows;

ω = 2πf

f = ω/2π

f = (22.36)/2π

f = 3.56 Hz

The period of the oscillation is calculated as follows;

T = 1/f

T = 1/3.56

T = 0.28 s

The total energy of the system is calculated as follows;

U = ¹/₂kA²

U = ¹/₂ x 250 x (0.05)²

U = 0.31 J

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Answer:

0.3375

Explanation:

w = angular speed of the DVD drive = 100.0 rpm = [tex]100.0 \frac{rev}{min}\frac{2\pi rad}{1 rev}\frac{1 min}{60 sec}[/tex] = [tex]10.5\frac{rad}{sec}[/tex]

m = mass of the dime = 2 g = 0.002 kg

r = radius = 3 cm = 0.03 m

μ = Coefficient of friction

The frictional force provides the necessary centripetal force to move in circle. hence

frictional force = centripetal force

μ mg = m r w²

μ g =  r w²

μ (9.8) =  (0.03) (10.5)²

μ = 0.3375

The average induced current in the loop is 0.218 A.

The emf induced in the loop is determined by applying Faraday's law as shown below;

emf = dФ/dt

emf = BA/t

where;

A is the area

A = πr² = πd²/4

A = π x (0.025)²/4

A = 4.908 x 10⁻³ m²

emf = (2 x 4.908 x 10⁻³)/(0.45)

emf = 2.18 x  10⁻³ V

The average induced current in the loop is calculated as follows;

I = emf/R

I = 2.18 x  10⁻³/0.01

I = 0.218 A

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Answer:

0.694 m

Explanation:

Case 1 : When only mass of 2.82 kg is hanged from spring

m = mass hanged from the spring = 2.82 kg

x = stretch caused in the spring = 0.331 m

k = spring constant

Using equilibrium of force in vertical direction

Spring force = weight of the mass

k x = m g

k (0.331) = (2.82) (9.8)

k = 83.5 N/m

Case 2 : When both masses are hanged from spring

m = mass hanged from the spring = 3.09 + 2.82 = 5.91 kg

x = stretch caused in the spring = ?

k = spring constant = 83.5 N/m

Using equilibrium of force in vertical direction

Spring force = weight of the mass

k x = m g

(83.5) x = (5.91) (9.8)

x = 0.694 m

Answer:

Force of friction, F = 45.76 N

Explanation:

t is given that,

Mass of the skater, m = 64 kg

Initial velocity of the skater, u = 2.81 m/s

Finally it comes to rest, v = 0

Time, t = 3.93 s

We need to find the force of friction. According to seconds law of motion as :

F = m × a

[tex]F=m\times \dfrac{v-u}{t}[/tex]

[tex]F=64\ kg\times \dfrac{0-2.81\ m/s}{3.93\ s}[/tex]

F = −45.76 N

So, the frictional force exerting on the skater is 45.76 N. Hence, this is the required solution.

Answer:

So the intensity of the light be smaller in second case with the polarizers at 90 degree with the vertical

Explanation:

When two polarizers are arranged at 45 degree and 90 degree in series

so here the intensity of light coming out of the polarizers is given as

[tex]I_1 = I_ocos^45[/tex]

[tex]I_1 = \frac{I_0}{2}[/tex]

now for second polarizer we have

[tex]I_2 = I_1 cos^2(90 - 45)[/tex]

[tex]I_2 = (\frac{I_0}{2})\frac{1}{2}[/tex]

[tex]I_2 = \frac{I_0}{4}[/tex]

now in other case when two polarizers are inclined 90 degree to the vertical

[tex]I = I_ocos^290 = 0[/tex]

so final intensity in second case will be ZERO

So the intensity of the light be smaller in second case with the polarizers at 90 degree with the vertical

Answer:

Depth of well 3.06m

Explanation:

We know that for a pipe closed at one end the frequencies are in ratios if 1:2:3:5:7.... to the fundamental frequency

In our case the given frequencies are in the ratio of

a)[tex]\frac{140}{196}=\frac{5}{7}[/tex]

b) [tex]\frac{196}{252}=\frac{7}{9}[/tex]

Thus the fundamental frequency can be calculated as [tex]140Hz=5n[/tex]

[tex]\therefore n=\frac{140}{5}=28Hz[/tex]

Now we know that

[tex]\lambda_{1}=4l\\\\\frac{v}{f}=4l\\\\l=\frac{v}{4f}[/tex]

Applying values we get

[tex]L=3.06m[/tex]

By identifying 56 Hz as the fundamental frequency and using the formula for the resonance of open-ended tubes, the depth of the well is calculated to be approximately 3.06 meters using the speed of sound as 343 m/s.

The student has observed standing waves at frequencies of 140 Hz, 196 Hz, and 252 Hz in a well. These frequencies represent the natural resonant frequencies or harmonics of the well. To find the depth of the well, we need to consider these frequencies as harmonics of a sound wave corresponding to the lengths of the air column in the well where each harmonic creates a standing wave.

Given the multiple frequencies, there is a constant difference of 56 Hz (196 Hz - 140 Hz, 252 Hz - 196 Hz) between consecutive frequencies. This indicates that 56 Hz is the fundamental frequency of the harmonics. We can represent these frequencies as [tex]F1 = 56 Hz \(n=1\), F2 = 112 Hz \(n=2\)[/tex]ere the observed frequencies correspond to F3, F4, and F5.

To determine the length of the well (L), we use the formula for the resonance of open-ended tubes, [tex]\(L = \frac{v}{2f}\)[/tex]d of sound (343 m/s) and 'f' is the frequency (56 Hz in this case). Therefore, [tex]L = \frac{343 m/s}{2 \times 56 Hz} = 3.0625 m[/tex]ly 3.06 meters.

Answer:This problem has been solved! See the answer. Two charged particles, Q 1 = + 5.10 nC and Q 2 = -3.40 nC, are separated by 40.0 cm. (a) What is the ... has not been graded yet. (b) What is the electric potential at a point midway between the charged particles? (Note: Assume a reference level of potential V = 0 at r = infinity .)

Explanation:

The sign of charge Q2 is positive, and its magnitude can be calculated by setting the gravitational force equal to the electric force due to Q1 and solving for Q2.

To determine the sign and magnitude of the second charge Q2, we use the concept that the net force acting on it must be zero. The forces that act on Q2 include the gravitational force and the electric force due to Q1. Since Q1 is positive and the net force is zero, Q2 must also be positive for the electric force (Coulomb force) to balance the gravitational pull downwards.

The gravitational force can be calculated using Newtons's law of gravitation F = mg, where g is the acceleration due to gravity (9.8 m/s2). The electric force is given by Coulomb's law F = k|Q1Q2|/d2, where k is Coulomb's constant (8.99 x 109 Nm2/C2).

Setting the magnitude of the gravitational force equal to the magnitude of the electric force gives the equation mg = k|Q1Q2|/d2. Solving for Q2, and converting the mass of Q2 from micrograms to kilograms, we calculate the magnitude of charge Q2.

Answer:

Worker's change in gravitational potential energy = 641.69 kJ

Explanation:

Potential energy = Mass x Acceleration due to gravity x Height

PE = mgh

Mass, m = 79 kg

Acceleration due to gravity, g = 9.81 m/s²

Height, h = 828 m

Potential energy, PE = 79 x 9.81 x 828 = 641691.72 J = 641.69 kJ

Worker's change in gravitational potential energy = 641.69 kJ

Answer:

0.25 T

Explanation:

F = Force required to keep the bar moving = 0.080 N

B = magnitude of magnetic field = ?

L = length of the bar = 50 cm = 0.50 m

v = speed of the bar = 0.50 m/s

R = resistance of the bar =0.10 Ω

Force is given as

[tex]F = \frac{B^{2}L^{2}v}{R}[/tex]

[tex]0.08 = \frac{B^{2}(0.50)^{2}(0.50)}{0.10}[/tex]

B = 0.25 T

To find the magnitude of the magnetic field, use the equation F = qvBsinθ and Ohm's Law to solve for q. Then substitute q back into the equation to find the magnetic field.

To find the magnitude of the magnetic field, we can use the equation F = qvBsinθ. In this case, the force (F) is given as 0.080 N, the charge (q) is not given but is not needed to find the magnetic field, the velocity (v) is given as 0.50 m/s, and sinθ is 1 because the angle between the velocity and the magnetic field is 90°. Rearranging the equation to solve for B, we get B = F / (qv sinθ). Substituting the given values, we have B = 0.080 N / (q * 0.50 m/s * 1). The only unknown variable left is the charge (q).

To calculate the charge, we can use the given resistance and Ohm's Law. Ohm's Law states that V = IR, where V is the voltage, I is the current, and R is the resistance. Rearranging the equation to solve for I, we get I = V / R. The voltage across the resistor can be calculated using the formula V = Fd, where d is the distance between the rails, which is given as 50 cm (or 0.5 m). Substituting the given values, we have V = 0.080 N * 0.5 m. Now, we can substitute the calculated current (I) into the equation B = 0.080 N / (q * 0.50 m/s * 1) and solve for q. Finally, we can substitute the calculated charge (q) back into the equation B = 0.080 N / (q * 0.50 m/s * 1) to find the magnitude of the magnetic field.

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Answer:

Moe: 6687.5 J, Larry: -2812.5 J, Curly: 0 J

Explanation:

The work done by each clown is given by:

[tex]W=Fd cos \theta[/tex]

where

F is the magnitude of the force applied

d is the displacement of the box

[tex]\theta[/tex] is the angle between the direction of the force and of the displacement

Let's apply the formula to each clown:

- Moe: F = 535 N, d = 12.5 m, [tex]\theta=0^{\circ}[/tex] (because Moe pushes to the right, and the box also moves to the right)

[tex]W=(535)(12.5)cos 0^{\circ}=6687.5 J[/tex]

- Larry: F = 225 N, d = 12.5 m, [tex]\theta=180^{\circ}[/tex] (because Larry pushes to the left, while the box moves to the right)

[tex]W=(225)(12.5)cos 180^{\circ}=-2812.5 J[/tex]

- Curly: F = 705 N, d = 12.5 m, [tex]\theta=90^{\circ}[/tex] (because Curly pushes downward, while the box moves to the right)

[tex]W=(705)(12.5)cos 90^{\circ}=0[/tex]

The work done by each of the clowns to move 345 kg crate 12.5 m to the right across a smooth floor is,

Work done is the force applied on a body to move it over a distance. Work done to move a body by the application of force can be given as,

[tex]W=Fd\cos\theta[/tex]

Here (F) is the magnitude of force, [tex]\theta[/tex] is the angle of force applied and (d) is the distance traveled.

The mass of the crate is 345 kg. The crate moved by the clowns is 12.5 m to the right across a smooth floor.

As, the Moe applies the force to the crate on the right side and the crate move in the direction of force. Here, the angle made is equal to the 0 degrees.

Therefore, the work done by the Moe is,

[tex]W=535\times12.5\times\cos (0)\\W=6687.5\rm J[/tex]

thus, the work done by the Moe is 6687.5 J.

As, the Larry applies the force to the crate on the left side and the crate move in the opposite direction of force. Here, the angle made is equal to the 180 degrees.

Therefore, the work done by the Larry is,

[tex]W=225\times12.5\times\cos (180)\\W=-2812.5\rm J[/tex]

thus, the work done by the Larry is -2812.5 J.

As, the Curly applies the force to the crate on the straight down and the crate move in the perpendicular direction of force. Here, the angle made is equal to the 90 degrees.

Therefore, the work done by the Curly is,

[tex]W=705\times12.5\times\cos (90)\\W=0\rm J[/tex]

thus, the work done by the Curly is 0 J.

Thus, the work done by each of the clowns to move 345 kg crate 12.5 m to the right across a smooth floor is,

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Answer:

a)  The velocity of the ball upon leaving the foot = 600 m/s

b)  Time to reach the goal posts 40.0 m away = 0.07 seconds

c)  The kick won't e going inside goal post, it is higher by 10.34m.

Explanation:

a) Rate of change of momentum = Force

   [tex]\frac{\texttt{Final momentum - Initial momentum}}{\texttt{Time}}=\texttt{Force}\\\\\frac{0.040v-0.040\times 0}{0.010}=2400\\\\v=600m/s[/tex]

  The velocity of the ball upon leaving the foot = 600 m/s

b) Horizontal velocity = 600 cos20 = 563.82 m/s

   Horizontal displacement = 40 m

   Time

            [tex]t=\frac{40}{563.82}=0.07s[/tex]

   Time to reach the goal posts 40.0 m away = 0.07 seconds

c) Vertical velocity = 600 sin20 = 205.21 m/s

    Time to reach the goal posts 40.0 m away = 0.07 seconds

    Acceleration = -9.81m/s²

    Substituting in s = ut + 0.5at²

             s = 205.21 x 0.07 - 0.5 x 9.81 x 0.07²= 14.34 m

    Height of ball = 14.34 m

    Height of post = 4 m

    Difference in height = 14.34 - 4 = 10.34 m

    The kick won't e going inside goal post, it is higher by 10.34m.

Answer:

104

Explanation:

U = Energy stored in the solenoid = 6.00 μJ = 6.00 x 10⁻⁶ J

i = current flowing through the solenoid = 0.4 A

L = inductance of the solenoid

Energy stored in the solenoid is given as

U = (0.5) L i²

6.00 x 10⁻⁶ = (0.5) L (0.4)²

L = 75 x 10⁻⁶

Inductance is given as

l = length of the solenoid = 0.7 m

N = number of turns

r = radius = 5.00 cm = 0.05 m

Area of cross-section is given as

A = πr²

A = (3.14) (0.05)²

A = 0.00785 m²

Inductance is given as

[tex]L=\frac{\mu _{o}N^{2}A}{l}[/tex]

[tex]75\times 10^{-6}=\frac{(12.56\times 10^{-7})N^{2}(0.00785)}{0.7}[/tex]

N = 73

Winding density is given as

density = n = [tex]\frac{N}{l}[/tex]

n = [tex]\frac{73}{0.7}[/tex]

n = 104

Answer:

4.49 dollars

Explanation:

i = 9 A, V = 240 V, t = 16 h

Energy = V x i x t = 240 x 9 x 16 = 34560 W h = 34.56 kWh

The cost of 1 kWh is 13 cents.

Cost of 34.56 kWh = 13 x 34.56 = 449.28 cents = 449.28 / 100 = 4.49 dollars

The cost of running a water pump drawing 9 A at 240 V for 16 hours, with an energy cost of 13 cents per kWh, is $4.49.

To calculate the cost of running a water pump for 16 hours at an electrical energy cost of 13 cents per kWh, you need to follow these steps:

In dollars, the cost of running the pump for 16 hours is $4.49 (rounded to the nearest cent).

Answer:

The mass of the planet is [tex]2.7\times10^{27}\ kg[/tex].

Explanation:

Given that,

Semi major axis [tex]a= 4.3\times10^{8}[/tex]

Orbital period T=1.516 days

Using Kepler's third law

[tex]T^2=\dfrac{4\pi^2}{GM}a^3[/tex]

[tex]M=\dfrac{4\pi^2}{GT^2}a^3[/tex]

Where, T = days

G = gravitational constant

a = semi major axis

Put the value into the formula

[tex]M=\dfrac{4\times(3.14)^2}{6.67\times10^{-11}(1.516\times24\times60\times60)^2}(4.3\times10^{8})^3[/tex]

[tex]M=2.7\times10^{27}\ kg[/tex]

Hence, The mass of the planet is [tex]2.7\times10^{27}\ kg[/tex].

Answer:

The highest of its trajectory = 0.45 m

Option C is the correct answer.

Explanation:

Considering vertical motion of cat:-

Initial velocity, u =  3.44 sin60 = 2.98 m/s

Acceleration , a = -9.81 m/s²

Final velocity, v = 0 m/s

We have equation of motion v² = u² + 2as

Substituting

   v² = u² + 2as

    0² = 2.98² + 2 x -9.81 x s

    s = 0.45 m

The highest of its trajectory = 0.45 m

Option C is the correct answer.

180N

Using Newton's law of motion;

∑F = m x a       --------------------(i)

Where;

∑F = Resultant force

m = mass of the object (sled in this case)

a = acceleration of the sled

Calculate the resultant force;

Since the direction of motion is horizontal, the horizontal forces acting on the sled are the;

i. Applied force ([tex]F_{A}[/tex]) in one direction and;

ii. Frictional force ([tex]F_{R}[/tex]) in the other direction to oppose motion

Therefore, the resultant force ∑F is the vector sum of the two forces. i.e;

∑F = [tex]F_{A}[/tex] - [tex]F_{R}[/tex]  -----------------------(i)

Frictional force [tex]F_{R}[/tex] is the product of the coefficient of kinetic friction (μ) and weight(W) of the sled. i.e

[tex]F_{R}[/tex] = μ x W

Where;

W = mass(m) x gravity(g)

W = m x g

=> [tex]F_{R}[/tex] = μmg

Substitute [tex]F_{R}[/tex] into equation (ii)

∑F = [tex]F_{A}[/tex] - μmg

Substitute ∑F into equation (i)

[tex]F_{A}[/tex] - μmg = ma  -------------------(iii)

Since the motion is at constant speed, it means acceleration is zero (0)

Substitute a = 0 into equation (iii) to give;

[tex]F_{A}[/tex] - μmg = 0

=> [tex]F_{A}[/tex] = μmg

Substitute the values of μ = 0.3, m = 60kg and g = 10m/s² into the above equation to give;

=> [tex]F_{A}[/tex]  = 0.3 x 60 x 10

=> [tex]F_{A}[/tex] = 180N

This means that the applied force should be 180N

The amount of force the football player must apply to the sled is 176.4 Newton.

Given the following data:

We know that acceleration due to gravity (g) on Earth is equal to 9.8 [tex]m/s^2[/tex]

To find how much force the football player must apply to the sled:

Mathematically, the force of kinetic friction is given by the formula;

[tex]Fk = umg[/tex]

Where;

Substituting the given parameters into the formula, we have;

[tex]Fk = 0.30\times60\times9.8[/tex]

Force, Fk = 176.4 Newton.

Read more: https://brainly.com/question/13754413

Final answer:

To find the nozzle's inside diameter, use the principle of continuity. Set up the equation A1V1 = A2V2 and solve for r2. Finally, calculate the nozzle's inside diameter.

Explanation:

To find the nozzle's inside diameter, we can use the principle of continuity, which states that the flow rate of a fluid is constant throughout a pipe or nozzle.

The formula for continuity is A1V1 = A2V2, where A is the cross-sectional area and V is the fluid velocity.

In this case, we can set up the equation as (pi * r12) * 2.17 = (pi * r22) * 14.5, where r1 and r2 are the radii of the hose and nozzle respectively.

By rearranging the equation and solving for r2, we find that r2 = sqrt((r12 * 2.17) / 14.5).

Since the radius is half of the diameter, the nozzle's inside diameter can be calculated as 2 * r2 in cm.

Answer:

(a) 0.816 s

(b) 0.816 m

(c) 5.656 m

Explanation:

u = 8 m/s, theta = 30 degree,

(a) Use the formula for time of flight

T = 2 u Sin theta / g

T = ( 2 x 8 x Sin 30 ) / 9.8 = 0.816 s

(b) Use the formula for maximum height

H = u^2 Sin^2theta / 2 g

H = ( 8 x 8 x Sin^2 30) / ( 2 x 9.8)

H = 0.816 m

(c) Use the formula for horizontal range

R = u^2 Sin 2 theta / g

R = ( 8 x 8 x Sin 60) / 9.8

R = 5.656 m

Answer:

75 m/s is faster

Explanation:

MKS units stands for meter kilogram seconds

75 miles per hour = 75 mph

1 mile = 1609.34 meters

1 hour = 60×60 = 3600 seconds

1 mph = 1609.34/3600 = 0.44704 m/s

75 mph = 75×0.44707 = 33.52792 m/s

Comparing 75 mph = 33.52792 m/s with 75 m/s it can be seen that 75 m/s is faster. Even without calculating the values you can know the answer. 75 mph means that in 1 hour the object will move 75 miles. 75 m/s means that in one second the object will cover 75 meters multiply by 3600 and you will get 270000 m/h that is 270 km/h divide it by 1.6 and you can approximately get the value in mph that will be around 168 mph which is faster than 75 mph.

I have seen this question before and the correct answer would be B

Hope this helped!!

Answer:

The correct answer would be B

Explanation:

Answer:

Rise in temperature is given as

[tex]\Delta T = 8.4 \times 10^{-3} ^0C[/tex]

Explanation:

Initial kinetic energy of the block is given as

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

here we will have

m = 3.3 kg

v = 3.2 m/s

now we will have

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

now we will have

[tex]KE = \frac{1}{2}(3.3)(3.2)^2[/tex]

[tex]KE = 17 J[/tex]

now we know that 74% of initial kinetic energy is absorbed as internal energy of the block

so the rise in temperature of the block is given as

[tex]KE = ms\Delta T[/tex]

[tex]0.74 \times 17 J = (3.3)(452)\Delta T[/tex]

[tex]12.5 = 1491.6 \Delta T[/tex]

[tex]\Delta T = 8.4 \times 10^{-3} ^0C[/tex]

Answer:

The temperature increases 0.0084ºC

Explanation:

Please look at the solution in the attached Word file.