Consider a satellite in a circular low Mars orbit, 300 km above the planetary surface. Use Newton's Law of Universal Gravitation and the concepts introduced in this section to answer the questions below. Use the following quantities in your calculations and pay close attention to unit conversions.Radius of Mars: R=3396km Mass of Mars: M=6.419×1023kg Universal gravitational constant: G=6.674×10−11m3/kg/s2 What is the orbital velocity of the satellite? g

Answers

Answer 1
Final answer:

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.

Explanation:

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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Answer 2

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


Related Questions

The primary coil of a transformer contains100 turns; the secondary has 200 turns. The primary coil isconnected to a size-AA battery that supplies a constant voltage of1.5 volts. What voltage would be measured across the secondarycoil? zero

Answers

Answer:

There  will be no voltage across  secondary coil because DC voltage source is used in primary coil so there is no electromotive force induced in secondary coil.

Explanation:

In this question we have given

AA battery is used to supplies a constant voltage of 1.5 volts to primary coil of  transformer. In this case, voltage source is DC source which is providing constant voltage and for a transformer to work it is necessary to use an AC source.

Therefore, no EMF will induce in the secondary coil

and we know that in a transformer,

[tex]\frac{E_{2} }{E_{1}} =\frac{V_{2} }{V_{1} }=\frac{N_{2} }{N_{1} }[/tex]...........(1)

hence from above equation it is clear that,

[tex]\frac{E_{2} }{E_{1}} =\frac{V_{2} }{V_{1}}[/tex]............(2)

Here ,

[tex]E_{2}=0[/tex]

Put [tex]E_{2}=0[/tex] in equation (2)

We got

[tex]V_{2}=0[/tex]

There  will be no voltage across  secondary coil because there is no electromotive force induced in secondary coil.

A gun with a muzzle velocity of 100 m/s is fired horizontally from a tower. Neglecting air resistance, how far downrange will the bullet be 8 seconds later? g

Answers

Answer:

313.6 m downward

Explanation:

The distance covered by the bullet along the vertical direction can be calculated by using the equation of motion of a projectile along the y-axis.

In fact, we have:

[tex]y(t) = h +u_y t + \frac{1}{2}at^2[/tex]

where

y(t) is the vertical position of the projectile at time t

h is the initial height of the projectile

[tex]u_y = 0[/tex] is the initial vertical velocity of the projectile, which is zero since the bullet is fired horizontally

t is the time

a = g = -9.8 m/s^2 is the acceleration due to gravity

We can rewrite the equation as

[tex]y(t)-h = \frac{1}{2}gt^2[/tex]

where the term on the left, [tex]y(t)-h[/tex], represents the vertical displacement of the bullet. Substituting numbers and t = 8 s, we  find

[tex]y(t)-h= \frac{1}{2}(-9.8)(8)^2 = -313.6 m[/tex]

So the bullet has travelled 313.6 m downward.

A windmill is turning at 12.5 rad/s in a clockwise direction. The wind suddenly shifts, producing an angular acceleration of 0.75 rad/s2 in a counterclockwise direction. After 12.0 s, how many revolutions has the windmill blades turned?

Answers

Answer:

Windmill will complete 0.27 revolutions.

Explanation:

We have equation of motion s = ut + 0.5at²

Initial speed, u = 12.5 rad/s

Time, t = 12 seconds.

Angular acceleration, a = -0.75 rad/s²

Substituting

               s = 12.5 x 12 - 0.5 x 0.75 x 12² = 96 rad

1 revolution = 360 rad

[tex]96rad=\frac{96}{360}=0.27revolution[/tex]

So, windmill will complete 0.27 revolutions.

The momentum of an isolated system is conserved A) only in inelastic collisions. B) only in elastic collisions. C) in both elastic and inelastic collisions

Answers

Answer:

It is actually in both.

Explanation:

The momentum of an isolated system is conserved for both elastic and inelastic collision. OPtion C is correct

Collision is the process by which two bodies come into contact with the release of energy.

Collision can be elastic or inelastic.

Foe elastic collision, both momentum, and energy is conserved while for inelastic collision only momentum is conserved.

From the explanation above, we can see that momentum of an isolated system is conserved for both elastic and inelastic collision

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In getting ready to slum-dunk the ball, a basketball player starts from rest and sprints to a speed of 6.0 m/s in 1.5s. Assuming that the player accelerates uniformly, determine the distance he runs.

Answers

Answer:

Distance ran by basketball player = 4.5 m

Explanation:

We have equation of motion, v = u+at

Substituting

           6 = 0 + a x 1.5

           a = 4m/s²

Now we need to find distance traveled, we have the equation of motion

           v² = u² + 2as

Substituting

           6² = 0² + 2 x 4 x s

           8 s = 36

              s = 4.5 m

Distance ran by basketball player = 4.5 m

A wire carrying a current is shaped in the form of a circular loop of radius 3.0mm If the magnetic field strength that this current produces at the center of the loop is 1.1mT, what is the magnitude of the current that flows through the wire? (μo = 4π x10^-7 T. m/A) A) 5.3A B) 16A C) 23 A D) 9.1A

Answers

Final answer:

The magnitude of the current that flows through the wire is approximately 16A.

Explanation:

To calculate the magnitude of the current that flows through the wire, we can use the formula for the magnetic field produced by a current-carrying circular loop: B = μ0 * I / (2 * R), where B is the magnetic field strength, μ0 is the permeability of free space, I is the current, and R is the radius of the loop.

Given that the radius of the loop is 3.0 mm and the magnetic field strength at the center of the loop is 1.1 mT, we can rearrange the formula to solve for I:

I = (2 * B * R) / μ0

Substituting the given values, we get:

I = (2 * 1.1 x 10^-3 T * 3.0 x 10^-3 m) / (4π x 10^-7 T.m/A)

Simplifying the expression, we find that the magnitude of the current is approximately 16 A. Therefore, the correct answer is B) 16A.

A proton moves through a region containing a uniform electric field given by E with arrow = 54.0 ĵ V/m and a uniform magnetic field B with arrow = (0.200 î + 0.300 ĵ + 0.400 k) T. Determine the acceleration of the proton when it has a velocity v with arrow = 170 î m/s.

Answers

Answer:

acceleration is 9.58 × [tex]10^{7}[/tex] (- 14 ĵ + 51 k  ) m/s

Explanation:

given data

uniform electric field E  = 54.0 ĵ V/m

uniform magnetic field B = (0.200 î + 0.300 ĵ + 0.400 k) T

velocity v  = 170 î m/s.

to find out

acceleration

solution

we know magnetic force for proton is

i.e = e (velocity × uniform magnetic field)

magnetic force = e (170 î × (0.200 î + 0.300 ĵ + 0.400 k) )

magnetic force = e (- 68 ĵ + 51 k  )   ..................1

and now for electric force for proton i.e

= uniform electric field × e ĵ

electric force = e (54 ĵ )   ............2

so net force will be  add magnetic force + electric force

from equation 1 and 2

e (- 68 ĵ + 51 k  )  + e (54 ĵ )

e (- 14 ĵ + 51 k  )

so the acceleration (a)  for proton will be

net force = mass × acceleration

a = e (- 14 ĵ + 51 k  )  / 1.6 [tex]10^{-19}[/tex] / 1.67 × [tex]10^{-27}[/tex]

acceleration = 9.58 × [tex]10^{7}[/tex] (- 14 ĵ + 51 k  ) m/s

Final answer:

The acceleration of the proton can be determined using the equation F = qvBsinθ. Plug in the given values to calculate the acceleration.

Explanation:

The acceleration of the proton can be determined using the equation F = qvBsinθ, where F is the force, q is the charge of the proton, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. Since the question does not provide the angle, we assume that θ = 0. Therefore, the acceleration of the proton is given by:

a = (q * v * B) / m

where m is the mass of the proton. Plugging in the given values, we can calculate the acceleration.

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When a charged particle moves at an angle of 25° with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90°) with respect to this field will this particle, moving at the same speed, experience a magnetic force of magnitude 2F?

Answers

Answer:57.7

Explanation:

Force on a moving charge in  a magnetic field is given by

[tex]F=q\times V\times Bsin\theta [/tex]

Where F=Force experienced by charge

q=charge of particle

V=velocity of particle

B=magnetic field

F=qVBsin(25)-------1

[tex]2F=qVBsin(\theta )------2[/tex]

Divide (1)&(2)

[tex]\frac{1}{2}[/tex]=[tex]\frac{sin(25)}{sin(\theta)}[/tex]

[tex]Sin(\theta )=2\times sin(25)[/tex]

[tex]\theta =57.7 ^{\circ}[/tex]

Given the constraints of the question and the formula for magnetic force on a moving charge, the situation described isn't possible.

The magnetic force experienced by a moving charged particle is given by the formula F = qvBsin(θ), where q is the charge, v is the speed, B is the magnetic field strength, and θ is the angle between the velocity of the charge and the magnetic field. When the force is F at 25°, to experience a force of 2F, the sin of the new angle must be double that of sin(25°) because all other factors (charge, speed, magnetic field strength) are constant.

In the case given, sin(θ) would need to be 2sin(25°), but since the maximimum value sin(θ) can have is 1 (at 90°), this situation isn't possible within the constraints of the question (angle less than 90°). Hence, the student might have misunderstood the problem or there might be an error the question itself. However, if it is assumed that the question intends to find when the component of the force perpendicular to the field is double, the answer would be when θ is 90 degrees because the sin(90°) = 1. This is because when a charged particle moves perpendicular to a magnetic field, the sin(θ) factor in the force equation is maximized.

How much charge is on each plate of a 3.00-μF capacitor when it is connected toa 15.0-V battery? b) If this same capacitor is connected to a 2.00-V battery, what charge is stored?

Answers

Answer:

(a) 45 micro coulomb

(b) 6 micro Coulomb

Explanation:

C = 3 micro Farad = 3 x 10^-6 Farad

V = 15 V

(a) q = C x V

where, q be the charge.

q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb

(b)

V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad

q = C x V

where, q be the charge.

q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb

40-turn circular coil (radius = 4.0 cm, total resistance = 0.20 ) is placed in a uniform magnetic field directed perpendicular to the plane of the coil. The magnitude of the magnetic field varies with time as given by B = 50 sin(10 t) mT where t is measured in s. What is the magnitude of the induced current in the coil at 0.10 s?

Answers

Final answer:

To find the magnitude of the induced current in the coil at 0.10 s, we need to calculate the magnitude of the magnetic flux through the coil and use Faraday's law of electromagnetic induction.

Explanation:

To find the magnitude of the induced current in the coil at 0.10 s, we need to first calculate the magnitude of the magnetic flux through the coil. The magnetic flux is given by Φ = B * A, where B is the magnetic field and A is the area of the coil. The area of the coil is calculated as A = π * r^2, where r is the radius of the coil.

Next, we can use Faraday's law of electromagnetic induction, which states that the induced emf is equal to the rate of change of magnetic flux. Mathematically, this can be written as emf = -dΦ/dt. Since the coil has resistance, we can use Ohm's Law to find the magnitude of the induced current, which is given by I = emf/R. Plugging in the values and calculating, we can find the magnitude of the induced current at 0.10 s.

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Final answer:

To find the magnitude of the induced current in the coil, we need to use Faraday's Law of Electromagnetic Induction. The magnetic field B is given as B = 50sin(10πt) mT. At t = 0.1s, the magnetic field is B = 50sin(10π(0.1)) mT.

Explanation:

To find the magnitude of the induced current in the coil, we need to use Faraday's Law of Electromagnetic Induction, which states that the induced emf (voltage) is equal to the negative rate of change of magnetic flux through the coil.

The magnetic field B is given as B = 50sin(10πt) mT, where t is measured in seconds.

The magnetic flux through the coil is given by Φ = B.A, where A is the area of the coil.

Substituting the given values, we have A = πr^2 = π(0.04^2) = 0.005 cm^2.

At t = 0.1s, the magnetic field is B = 50sin(10π(0.1)) mT.

We can now calculate the magnetic flux through the coil using Φ = B.A.

Finally, we can use Faraday's Law to find the induced emf and divide it by the resistance of the coil to find the magnitude of the induced current.

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The angle of elevation to the top of a very tall Building is found to be 6° from the ground at a distance of 1 mi from the base of the building. Using this information, find the height of the building. (Round your answer to the nearest foot.)

Answers

Answer:

555

Explanation:

The scenario is shown in the image attached below. A Right Angled Triangle is formed. We have an angle of 6 degrees, the side adjacent to angle which measures 1 mile and we need to find the side opposite to the angle.

Since 1 mile = 5280 feet, the side adjacent to the angle has a measure of 5280 feet.

Tangent ratio relates the opposite and adjacent side by following formula:

[tex]tan(\theta)=\frac{Opposite}{Adjacent}[/tex]

Using the given values, we get:

[tex]tan(6)=\frac{x}{5280}\\\\ x=tan(6) \times 5280\\\\ x = 555[/tex]

Thus, rounded to nearest foot, the height of the building is 555 feet

Final answer:

To find the height of the building, we can use trigonometry and the given angle of elevation. The height of the building is approximately 555 feet.

Explanation:

To find the height of the building, we can use trigonometry and the given angle of elevation. Let x be the height of the building. From the given information, we have the opposite side (x) and the adjacent side (1 mile = 5280 feet). The tangent function is used to relate these sides: tan(6°) = x/5280 feet. Solving for x, we get x = 5280 feet × tan(6°).

Using a calculator, we find that tan(6°) is approximately 0.1051. Multiplying this by 5280 feet gives us the height of the building in feet: 0.1051 × 5280 = 554.928 feet.

Rounding to the nearest foot, the height of the building is approximately 555 feet.

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A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calculate (a) the current in the circuit, (b) the terminal voltage of the battery, Vab, and (c) the power dissipated in the resistor R and in the battery’s internal resistance r.

Answers

Answer:

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

[tex]V_{ab}[/tex] = i R

[tex]V_{ab}[/tex] = (0.1832) (65)

[tex]V_{ab}[/tex] = 11.91 Volts

(c)

Power dissipated in the resister R is given as

[tex]P_{R}[/tex] = i²R

[tex]P_{R}[/tex] = (0.1832)²(65)

[tex]P_{R}[/tex] = 2.18 Watt

Power dissipated in the internal resistance is given as

[tex]P_{r}[/tex] = i²r

[tex]P_{r}[/tex] = (0.1832)²(0.5)

[tex]P_{r}[/tex] = 0.0168 Watt

Final answer:

To compute the values in the question, one can utilize Ohm’s law and the power dissipation formula. The current in the circuit is 0.183 A, the terminal voltage of the battery is 11.908 V, the power dissipated in the 65.0 Ω resistor is 2.18 W, and the energy wasted in the battery's internal resistance is 0.0167 W.

Explanation:

To calculate the current in the circuit, we can use Ohm's law which states that the current is equal to the voltage divided by the resistance. This gives us: I = emf/(R+r) where I is the current, emf is the electromotive force of the battery, R is the load resistance and r is the internal resistance of the battery.

Substituting the values given in the question, we get: I = 12.0 V/(65.0 Ω + 0.5 Ω) = 0.183 A. Therefore, the current flowing through the circuit is 0.183 Amperes.

Now that we have the current, we can calculate the terminal voltage of the battery. The terminal voltage (Vab) is given by the equation: Vab = emf - Ir. Substituting the values, we get: Vab = 12.0V - (0.183A * 0.5Ω) = 11.908 V. Therefore, the terminal voltage of the battery is 11.908 Volts.

Lastly, we can calculate the power dissipated in the resistor R and in the battery’s internal resistance r using the formula: Power = I²R. For the resistor, power dissipation = (0.183A)² * 65.0Ω = 2.18 Watts, and for the battery's internal resistance, power dissipation = (0.183A)² * 0.5Ω = 0.0167 Watts.

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Suppose a 50 turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.325 m^2. It is squished to have no area in 0.225 s. What is the magnitude of the average induced emf in volts if the uniform magnetic field has a strength of 1.5 T?

Answers

Answer:

EMF = 108.3 Volts

Explanation:

As per Faraday's law of electromagnetic induction we know that rate of change in magnetic flux will induce EMF in closed loop

So it is given as

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

[tex]EMF = N\frac{BA - 0}{\Delta t}[/tex]

now we know that

N = 50 turns

B = 1.5 T

A = 0.325 m^2

[tex]\Delta t = 0.225 s[/tex]

now we have

[tex]EMF = (50)(\frac{1.5\times 0.325}{0.225})[/tex]

[tex]EMF = 108.3 Volts[/tex]

The magnitude of the average induced emf is 110.12 volts.

The magnitude of the average induced emf can be calculated using Faraday's law. The formula is given as:

emf = -N * A * (dB/dt)

Where:

N is the number of turns in the coilA is the area of the coildB/dt is the rate of change of the magnetic field

In this case, the number of turns is 50, the area is 0.325 m², and

the rate of change of the magnetic field is (0 T - 1.5 T) / 0.225 s = -6.67 T/s.

Substituting these values into the formula, we get:

emf = -(50)(0.325)(-6.67) = 110.12 V

The magnitude of the average induced emf is 110.12 volts.

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A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.9 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

Answers

Final answer:

By using similar triangles concept and rates of change in Mathematics, we can find the man's shadow length is decreasing at a rate of 0.5 m/s when he is 4 meters away from the building.

Explanation:

The problem you're describing is typically solved using similar triangles concept in Mathematics and rates of change. As the man walks towards the building, his shadow becomes shorter. We can set up a ratio between the man's height and his distance from the building, and the length of the shadow and the wall. When the man is 4 meters away from the building, a right-triangle is formed with the man's height (2m), his distance from the building (4m), and his shadow's length.

Let s gives the length of the shadow, then the similar triangles imply 12/s = 2/4 which gives us s = 6 meters.

To find the rate at which the man's shadow is decreasing, we can differentiate the similar triangles ratio with respect to time to get -12/s^2 ds/dt =-1/2, sub in s = 6 into the equation to find ds/dt = -0.5 m/s.

So the length of the man's shadow on the building is decreasing at a rate of 0.5 m/s when he is 4 m from the building.

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Final answer:

To find how fast the man's shadow is decreasing, we use similar triangles and related rates in calculus. When the man is 4 m from the building, his 2 m tall shadow is shortening at 3.8 m/s, twice his walking speed of 1.9 m/s.

Explanation:

The question involves determining how fast the length of a man's shadow on a building is decreasing when he walks toward the building and away from the spotlight. This requires the application of similar triangles and the concept of related rates in calculus to solve. If we let the distance between the man and the building be x, and the length of the shadow be y, as the man walks towards the building, both x and y are changing with respect to time.

Given that the man is 2 m tall and the spotlight is 12 m away from the building, we can set up a proportion between the heights of the man and his shadow, and their respective distances from the spotlight. When he is 4 m from the building, the similar triangles can be represented as 2/y = (12-x)/(12). Differentiating both sides with respect to time t yields the related rate of change for the length of the shadow y when x is 4 m and the speed of the man is known, which is 1.9 m/s.

By solving this proportion and applying the derivative, we can find that the rate at which the length of his shadow decreases is exactly double the speed at which the man is walking, because in this scenario, the man and the tip of his shadow form a straight line with the light source. Thus, the length of the shadow decreases at a rate of 3.8 m/s when the man is 4 m from the building.

The differential equation below models the temperature of a 91°C cup of coffee in a 21°C room, where it is known that the coffee cools at a rate of 1°C per minute when its temperature is 71°C. Solve the differential equation to find an expression for the temperature of the coffee at time t. (Let y be the temperature of the cup of coffee in °C, and let t be the time in minutes, with t = 0 corresponding to the time when the temperature was 91°C.) dy/dt =(− 1/50)(y − 21)

Answers

Answer:

[tex]T=21+70e^{\frac{-t}{50} }[/tex]

Explanation:

We start from:

[tex]\frac{dT}{dt}=\frac{-1}{50}(T-21)[/tex]

Separating variables:

[tex]-50dT=(T-21)dt[/tex]

[tex]-50\frac{dT}{T-21}=dt[/tex]

Integrating with initial conditions:

[tex]-50\int\limits^{T}_{91} {\frac{1}{T-21} } \, dT= \int\limits^t_{0} {} \, dt[/tex]

[tex]-50ln(\frac{T-21}{91-21})=t[/tex]

[tex]ln(\frac{T-21}{71})=\frac{-t}{50}[/tex]

Isolating T:

[tex]\frac{T-21}{70} =e^\frac{-t}{50} }[/tex]

[tex]T=21+70e^{\frac{-t}{50} }[/tex]

You may note that when t is zero the temperature is 91 ºC, as is specified by the problem. As well, when t is bigger (close to infinite), the temperature tends to be the room temperature (21 ºC)

Final answer:

The given differential equation can be solved to yield the function y(t) = 70 e^(-t/50) + 21, which describes the temperature of the coffee as a function of time.

Explanation:

In your given differential equation, dy/dt = (− 1/50)(y − 21), y represents the temperature of the coffee at time t, and the equation describes how the temperature changes over time. This is a type of first-order linear differential equation, which can be solved using an integrating factor. The general solution of such equation is given by y(t) = [integral(t, e^(-t/50)*(-1/50)dt] + C e^(t/50), where C is a constant. To solve for C, we use the initial condition: at t = 0, y = 91°C, which yields C = (91 - 21), or C = 70. Substituting back into the original equation provides the final formula for the temperature of the coffee at given time t: y(t) = 70 e−t/50 + 21. It states that, the temperature of coffee decreases over time from its initial temperature, and it will eventually cool to the same temperature as the room (21°C).

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Acharged particle moving through a magnetic field at right angles to the field with a speed of 36.2 m/s experiences a magnetic force of 7.38x104 N. Determine the magnetic force on an identical particle when it travels through the same magnetic field with a speed of 12.1 m/s at an angle of 17° relative to the magnetic field. Express your answer in microNewtons.

Answers

Answer:

7212.3 N

Explanation:

F = 7.38 x 10^4 N, v = 36.2 m/s

Let b be the strength of magnetic field and charge on the particle is q.

F = q v B Sin theta

Here theta = 90 degree

7.38 x 10^4 = q x 36.2 x B x Sin 90

q B = 2038.7 .....(1)

Now, theta = 17 degree, v = 12.1 m/s

F = q v B Sin theta

F = 2038.7 x 12.1 x Sin 17     ( q v = 2038.7 from equation (1)

F = 7212.3 N

A machine gun fires 33 g bullets at a speed of 342 m/s. The gun fires 115 bullets/min.What is the average force the shooter must exert to keep the gun from moving?

Answers

Final answer:

The required force is approximately 21.63 newtons.

Explanation:

The question involves determining the average force that a shooter must exert to keep a gun from moving while it fires bullets. To find this force, we can use the concept of impulse, which relates force, time, and change in momentum.

The gun fires 115 bullets per minute, which is 115/60 bullets per second. Multiplying the number of bullets per second by the mass and speed of each bullet gives the total momentum per second, which is the same as the force when mass is in kg and speed is in m/s.

First, we find the momentum of each bullet:

Bullet speed (v) = 342 m/s

F ≈ 21.63 N

The shooter must exert an average force of approximately 21.63 newtons to keep the gun from moving.

A 8.0 N force acts on a 0.30-kg object for 0.20 seconds. By how much does the object's momentum change (in kg-m/s)? (NEVER include units in the answer to a numerical question.)

Answers

Answer:

The change in momentum of an object is 1.6 kg-m/s.  

Explanation:

It is given that,

Mass of the object, m = 0.3 kg

Force acting on it, F = 8 N

Time taken, t = 0.2 s

We need to find the change in momentum of an object. The change in momentum is equal to the impulse imparted on an object i.e.

[tex]\Delta p=J=F.\Delta t[/tex]

[tex]\Delta p=8\ N\times 0.2\ s[/tex]

[tex]\Delta p=1.6\ kg-m/s[/tex]

So, the change in momentum of an object is 1.6 kg-m/s. Hence, this is the required solution.

The change in the momentum of the object is 1.6 kg-m/s.

What is change in momentum?

Change in momentum is the product of force acting on a body and time

To calculate the change in momentum of the object, we use the formula below

Formula:

ΔM = Ft...............Equation 1

Where:

ΔM = change in momentum of the objectm = Force acting on the objectt = time.

From the question,

Given:

F = 8 Nt = 0.20 seconds

Substitute these values into equation 1

ΔM = (8×0.2)ΔM = 1.6 kg-m/s

Hence, The change in the momentum of the object is 1.6 kg-m/s.

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A vertical block-spring system on earth has a period of 6.0 s. What is the period of this same system on the moon where the acceleration due to gravity is roughly 1/6 that of earth? A)1.0s B)2.4s C)6.0s D)15s E)36s

Answers

Final answer:

The period of the vertical block-spring system on the moon is approximately 2.45 seconds.

Explanation:

The period of a vertical block-spring system is determined by the square root of the ratio of the mass of the block to the spring constant. Since the mass and spring constant remain the same, the period will only be affected by the acceleration due to gravity.

On the moon, where the acceleration due to gravity is about 1/6 that of Earth, the period of the system will be sqrt(6), or approximately 2.45 seconds.

Learn more about Vertical block-spring system here:

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A light ray is traveling in a diamond (n = 2.419). If the ray approaches the diamond-air interface, what is the minimum angle of incidence that will result in all of the light reflected back into the diamond? The index of refraction for air is 1.000.

Answers

Answer:

24.2 degree

Explanation:

n = 2.419

Let the minimum angle of incidence is i.

The value of minimum angle of incidence so that all the light reflects back into the diamond is the critical angle for air diamond interface.

The relation for the critical angle and the refractive index of diamond is given by

Sin ic = 1 / n

where, ic is the critical angle for air diamond interface and n be the refractive index for diamond.

Sin ic = 1 / 2.419 = 0.4133

ic = 24.4 degree

Thus, the value of minimum value of angle of incidence for which all the light reflects back into diamond is 24.4 degree.

Explanation:

A light ray is traveling in a diamond (n = 2.419). If the ray approaches the diamond-air interface, what is the minimum angle of incidence that will result ...

Which of the following is a nonferromagnetic material? (a) aluminum (b) iron (c) cobalt (d) gadolinium

Answers

Answer:

A

Explanation:

Iron and gadlinium are both very easily made into magnetic substances.  Cobalt is also capable of being magnetized. Aluminum, put in an alloy, can make a magnetic substance, but

Aluminum by itself is not able to be magnetized.

A metal wheel 50 cm in radius rotates on a desk anti-clockwise at a constant rate of 800 rev/min about its central axis. Determine (a) its angular velocity (not speed!!!),
(b) its linear speed at a point 30 cm from its center of rotation,
(c) the radial acceleration of a point on the rim, and
(d) the total distance a point on the rim moves in 120 seconds.
Note: You must draw a diagram and pick appropriate axes to aid your calculation.
Write down the detail process.

Answers

Answer:

(a) 83.73 rad/s

(b) 251.2 m/s

(c) 3505.4 rad/s^2

(d) 5024 m

Explanation:

R = 50 cm = 0.5 m, f = 800 rpm = 800 / 60 rps

(a) Angular velocity, w = 2 x 3.14 x f = 2 x 3.14 x 800 / 60 = 83.73 rad / s

(b) The relation between linear speed and the angular speed is

V = r w

Here, r = 30 cm = 0.3 m

V = 0.3 x 83.73 = 25.12 m/s

(c) Radial acceleration = R w^2 = 0.5 x 83.73 x 83.73 = 3505.4 rad/s^2

(d) Time period T = 2 x 3.14 / w

T = 2 x 3.14 / 83.73 = 0.075 sec

In 0.075 second, angle turn = 360 degree

In 120 second, the angle turn = 360 x 120 / 0.075 = 576000 degree

In 360 degree, the distance traveled = 2 x pi x R

In 576000 degree, the distance traveled = 2 x 3.14 x 0.5 x 576000 / 360

                                                                     = 5024 m

Ball 1 is launched with an initial vertical velocity v1 = 146 ft/sec. Ball 2 is launched 2.3 seconds later with an initial vertical velocity v2. Determine v2 if the balls are to collide at an altitude of 234 ft. At the instant of collision, is ball 1 ascending or descending?

Answers

Answer:

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s    

At the time of collision velocity of ball one is descending.

Explanation:

Velocity of ball 1 = 146 ft/sec = 44.50m/s

The balls are to collide at an altitude of 234 ft

H = 234 ft = 71.32 m

We have equation of motion

         v² = u² + 2as

         v² = 44.50² + 2 x (-9.81) x 71.32

         v = ±24.10 m/s.

Time for each velocity can be calculated using equation of motion

        v = u + at

         24.10 = 44.50 - 9.81 t , t = 2.07 s

         -24.10 = 44.50 - 9.81 t , t = 6.99 s      

Since the second ball throws after 2.3 seconds  we can avoid case with t = 2.07 s.

So at the time of collision velocity of ball one is descending.

The collision occurs at t = 6.99 s.

Time of flight of ball 2 = 6.99 - 2.3 = 4.69 seconds.

Height traveled by ball 2 = 71.32 m

We need to find velocity

We have

           s = ut + 0.5 at²

           71.32 = u x 4.69 - 0.5 x 9.81 x 4.69²

           u = 38.21 m/s = 125.36 ft/s

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s    

Answer:

v2=139 ft

Explanation:

First we just look at the motion of the first particle. It is moving vertically in a gravitational field so is decelerating with rate g = 9.81 m/s^2 = 32.18 ft/s^2. We can write it's vertical position as a function of time.  

h_1=v_1*t-(a*t/2)

We set this equal to 234 ft to find when the body is passing that point, a solve the quadratic equation for t.  

t_1,2=v_1±(√v_1^2-4*a/2*h_1)/a=2.57 s, 7.44 s

Since we know the second ball was launched after 2.3 seconds, we know that the time we are looking for is the second one, when the first ball is descending. The second ball will have 2.3 seconds less so the time we further use is t_c = 7.44 - 2.3 = 5.14 s. With this the speed of the second ball needed for collision at given height, can be found.

Solving a similar equation, but this time for v2 to obtain the result.

h_2=234 ft=v2*t_c-(a*t_c^2/2)--->v2=139 ft

f a vector has magnitude V_1 = 46 N, and direction theta_1 = 30 degree Another vector V_2 = 63 N, and direction theta_2 = 151 degree. Calculate the reference angle of the vector V_1 + V_2. Round your answer such that it has no decimal.

Answers

Answer:

105.8 degree from + X axis

Explanation:

V1 = 46 N at 30 degree

V2 = 63 N at 151 degree

Write the vector form

V1 = 46 (Cos 30 i + Sin 30 j) = 39.84 i + 23 j

V2 = 63 (Cos 151 i + Sin 151 j) = - 55 i + 30.54 j

The resultant of V1 and V2 is given by

V = V1 + V2 = 39.84 i + 23 j - 55 i + 30.54 j = - 15.16 i + 53.54 j

The angle made by resultant of V1 and V2 with X axis is

tan∅ = 53.54 / (- 15.16) = - 3.53

∅ = 105.8 degree from + X axis.

What is the value of (8 104)2, written with the correct mumber of significant figures? A) 206.324 B) 206.323 C)2063 D) 206 E) 200

Answers

Answer:

B) 206.323

Explanation:

The value of (8 104)2, written with the correct mumber of significant figures is 206.323.

Answer:

B. 206.323

Explanation:

A 3.00-kg object undergoes an acceleration given by a = (2.00 i + 5.00 j) m/s^2. Find (a) the resultant force acting on the object and (b) the magnitude of the resultant force.

Answers

Answer:

(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).

(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .

Explanation:

m= 3kg

a= 2 i + 5 j = 5 .38 < 68.19 º

F= m * a

F= 3* ( 5.38 < 68.19º )

F= 16.4 N < 68.19º

Fx= F * cos(68.19º)

Fx= 5.99

Fy= F* sin(68.19º)

Fy= 14.98

The resultant force acting on the object is (6.00 i + 15.00 j) N, and the magnitude of the resultant force is 16.16 N.

To find the resultant force acting on a 3.00-kg object with an acceleration given by a = (2.00 i + 5.00 j) m/s2, we use Newton's second law of motion (Force = mass  imes acceleration).

Calculate the resultant force: F = m  imes a where m is the mass and a is the acceleration vector.Substitute the given values into the equation: F = 3.00 kg  imes (2.00 i + 5.00 j) m/s2.This gives F = (6.00 i + 15.00 j) N.

For the magnitude of the resultant force, we calculate it using the Pythagorean theorem:

|F| = \\(6.002 + 15.002)\sqrt N.Which simplifies to |F| = \\(36 + 225)\sqrt N = \\(261)\sqrt N.Thus, the magnitude of the resultant force is |F| = 16.16 N.

We are given a piece of copper of mass m=0.82 g to form a cylindrical wire of resistance R=0.87.2. What should be the length / of this wire?: 2.16 m. What would be the diameter d of this wire?: d 0.232 mm. To make things certain, let us use copper's mass density of 8.96 g/cm3 and resistivity p= 1.7 x 10-8 2.m.

Answers

Answer:

Length of wire=2.16 m

Diameter of wire=0.232 mm

Explanation:

m= mass of copper wire= 0.82 g

R= Resistance of copper wire= 0.87 ohms

D= Density of copper= 8.96 g/cm^3

ρ= Resistivity= 1.7×10^-8 Ωm

[tex]Density=\frac{mass}{volume}\\\Rightarrow volume=\frac{mass}{density}\\\Rightarrow volume=\frac {0.82}{8.96}\\\Rightarrow volume=0.091\ cm^3\\ volume = \pi r^2 l\\\Rightarrow \pi r^2=\frac{volume}{l}\\ \Rightarrow \pi r^2=\frac {0.091}{l}\\[/tex]

[tex]\rho=R\frac{A}{l}\\\Rightarrow 1.7\times 10^{-8}=0.87 \frac{\pi r^2}{l}\\\Rightarrow 1.7\times 10^{-8}=0.87 \frac{0.091\times 10^{-6}}{l^2}\\\Rightarrow l^2=\frac {0.87\times 0.091\times 10^{-6}}{1.7\times 10^{-8}}\\\Rightarrow l^2=0.046\times 10^2\\\Rightarrow length=2.16\ m[/tex]

[tex]\pi r^2=\frac {0.091}{l}\\\Rightarrow r^2=\frac {0.091\times 10^{-6}}{2.16 \pi}\\\Rightarrow r^2=1.34\times 10^{-8}\\\Rightarrow r=0.00011\ m\\\Rightarrow d=0.000232\ m\\\therefore diameter=0.232\ mm[/tex]

A ball thrown straight upwards with an initial velocity of 15 m/s. Calculate how long the ball will remain in the air and the maximum height the ball reaches?

Answers

Answer:

The ball remain in air for 3.06 seconds.

Maximum height reached = 22.94 m

Explanation:

We have equation of motion v = u + at

At maximum height, final velocity, v =0 m/s

       Initial velocity = 15 m/s

       acceleration = -9.81 m/s²

Substituting

       0 = 15 -9.81 t

        t = 1.53 s

Time of flight = 2 x 1.53 = 3.06 s

The ball remain in air for 3.06 seconds.

We also have equation of motion v² = u² + 2as

At maximum height, final velocity, v =0 m/s

       Initial velocity = 15 m/s

       acceleration = -9.81 m/s²

Substituting

       0² = 15² - 2 x 9.81 x s

        s = 22.94 m

Maximum height reached = 22.94 m

Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20 kW of shaft power. The free surface of the upper reservoir is 45 m higher than that of the lower reservoir. If the flow rate of water is measured to be 0.03 m3/s, determine mechanical power that is converted to thermal energy during this process due to frictional effects.

Answers

Final answer:

To calculate the power converted to thermal energy due to frictional effects, subtract the ideal mechanical power required to lift water to the higher reservoir (13.2435 kW) from the actual pump power (20 kW), yielding 6.7565 kW.

Explanation:

The student is asking to determine the mechanical power that is converted to thermal energy due to frictional effects when water is pumped from a lower reservoir to a higher reservoir. Given the shaft power of the pump is 20 kW, height difference is 45 m, and flow rate is 0.03 m³/s, we can first calculate the ideal mechanical power required to lift the water to that height.

The gravitational potential energy given to the water per second (which is the power) by pumping it to the height is calculated using the formula P = ρghQ, where ρ (rho) is the density of water (1000 kg/m³), g is the acceleration due to gravity (9.81 m/s²), h is the height difference (45 m), and Q is the flow rate (0.03 m³/s). This results in P = 1000 kg/m³ * 9.81 m/s² * 45 m * 0.03 m³/s = 13243.5 W or 13.2435 kW.

The difference between the actual power supplied by the pump (20 kW) and the ideal power required (13.2435 kW) is the power lost to thermal energy due to friction. Therefore, the power converted to thermal energy is 20 kW - 13.2435 kW = 6.7565 kW.

Determine the inductive reactance for a 50 mH inductor that is across a 15 volt, 400 Hz source.

Answers

Answer:

Inductive reactance is 125.7 Ω

Explanation:

It is given that,

Inductance, [tex]L=50\ mH=0.05\ H[/tex]

Voltage source, V = 15 volt

Frequency, f = 400 Hz

The inductive reactance of the circuit is equivalent to the impedance. It opposes the flow of electric current throughout the circuit. It is given by :

[tex]X_L=2\pi fL[/tex]

[tex]X_L=2\pi \times 400\ Hz\times 0.05\ H[/tex]

[tex]X_L=125.66\ \Omega[/tex]

[tex]X_L=125.7\ \Omega[/tex]

So, the inductive reactance is 125.7 Ω. Hence, this is the required solution.

To calculate the inductive reactance for a 50 mH inductor with a 400 Hz source, apply the formula XL = 2πfL, resulting in an inductive reactance of 125.6 ohms.

To determine the inductive reactance of a 50 mH inductor across a 15 volt, 400 Hz source, we use the formula for inductive reactance, which is:

L = 2πfL

where:

XL is the inductive reactance

f is the frequency of the AC source (400 Hz in this case)

L is the inductance of the coil (50 mH or 0.050 H)

Substituting the given values:

XL = 2π x 400 Hz x 0.050 H

XL = 125.6 Ω

Therefore, the inductive reactance is 125.6 ohms (Ω).

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