The IKAROS spacecraft, launched in 2010, was designed to test the feasibility of solar sails for spacecraft propulsion. These large, ultralight sails are pushed on by the force of light from the sun, so the spacecraft doesn't need to carry any fuel. The force on IKAROS's sails was measured to be 1.12 mN. If this were the only force acting on the 290 kg spacecraft, by how much would its speed increase after 3.0 months of flight? Assume there are 30 days in each month.

Explanation:

The X- rays are not emitted by the black hole itself but by the  from hot gas orbiting around the black hole forming a disc known as accretion disk. When a black hole rips apart a star it attracts gases towards it. The pressure that is generated in the gas heats it up to millions of degrees thus emitting radiation among which x rays are also present which are detected by radio telescopes on the earth.

Answer:

The speed of the cylinder has decreased

Explanation:

Given data:

Distance between the center of the turn table and the cylinder = R

Distance between the center of the turn table and the cylinder when it is moved to a new point, R' = R/2

the tangential velocity in circular motion is given as:

[tex]V=\frac{2\pi R}{T}[/tex]

where T is the time

now for R = R' = R/2

the new velocity V' comes as:

[tex]V'=\frac{2\pi \frac{R}{2}}{T}[/tex]

or

[tex]V'=\frac{\pi R}{T}[/tex]

thus the velocity becomes half or we can say the velocity decreases

Now, the acceleration (a) in circular motion is given as:

a = V²/R

now for

R = R' = R/2

the velocity V' = V/2

thus,

new acceleration a' comes as:

[tex]a = \frac{(\frac{V}{2})^2}{R/2}[/tex]

or

[tex]a'=\frac{V^2}{2R}[/tex]

thus,

the acceleration decreases by 2 times

The statement that accurately describes the motion of the cylinder at the new location is : ( 1 )  and ( 3 )

Given data :

Distance between cylinder and center of the turn table = R

Distance between cylinder and turntable when moved = R/2

where ;

Tangential Velocity/speed =  [tex]\frac{2\pi R}{T}[/tex]  

R = distance between objects

T = time

∴ The initial Tangential velocity of the cylinder at rest ( V ) = [tex]\frac{2\pi R}{T}[/tex]  ----- ( 1 )

 while the

Tangential velocity/speed of the cylinder when it moves ( V' ) = [tex]\frac{2\pi R/2}{T}[/tex]

                                                                                                       = [tex]\frac{\pi R}{T}[/tex] ----- ( 2 )

Therefore comparing equations ( 1 ) and ( 2 )  the speed of the cylinder has decreased as the cylinder moves from R to R/2 .

Also Given that Tangential acceleration = v² / R .

Since velocity decreases after the cylinder moves, the magnitude of the acceleration of the cylinder will decrease as well.

Hence we can conclude that the acceleration and speed of the cylinder has decreased with the movement of the cylinder.

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Answer:

d

velocity multiplied to mass

Answer:

Momentum of the person, p = 75 kg-m/s

Explanation:

It is given that,

Mass of the person, m = 15 kg

The person is moving with a velocity of, v = 5 m/s

We need to find the walker's momentum. It is equal to the product of mass and velocity with which it is moving. Mathematically, it is given by :

p = mv

[tex]p=15\ kg\times 5\ m/s[/tex]

p = 75 kg-m/s

So, the momentum of the person is 75 kg-m/s. Hence, this is the required solution.

Answer: the photograph will likely show only one star.

Explanation:

Since their angular separation is smaller than the telescope's angular resolution, the picture will apparently show only one star rather than two.

The photograph will likely show only one star.

And their mathematical relationship is tan θ = a / d. θ ≈ a / d. Remember that the perfect circle is 2π radians = 360 degrees. The magnitude or distance of the angle can also be measured in arc minutes (60 arcs = 1 degree) or arc seconds (60 arcs = 1 arc).

RA takes values ​​from 0 to 360 degrees, and declination takes values ​​from -90 to +90 degrees). Next, the angular distance A between the two stars 1 and 2 in degrees is determined by the following relationship: cos (A) = sin (devil.

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Answer: [tex]3.97(10)^{-19}J[/tex]

Explanation:

The energy [tex]E[/tex] of a photon is given by:

[tex]E=h\nu[/tex]   (1)

Where:

[tex]h=6.626(10)^{-34}\frac{m^{2}kg}{s}[/tex] is the Planck constant

[tex]\nu=6(10)^{14}s^{-1}[/tex] is the frequency of green light

Solving (1):

[tex]E=(6.626(10)^{-34}\frac{m^{2}kg}{s})(6(10)^{14}s^{-1})[/tex]   (2)

[tex]E=3.9756(10)^{-19}\frac{m^{2}kg}{s^{2}}=3.9756(10)^{-19}J[/tex]   (3)

Answer:

a. The net force acting on it is zero.

An object can remain at rest if the net force acting on it is zero (a) OR there are no forces at all acting on it (c).

The magnitude of the electric field is found by using the components of tension in the string, related to the horizontal electric force and the vertical gravitational force, and applying trigonometry to solve for E.

To find the magnitude and direction of the electric field, we must recognize that there are two forces acting on the plastic ball: the force of gravity and the electric force. The force of gravity acts downward and has a magnitude of mg, where m is the mass of the ball and g is the acceleration due to gravity. The electric force acts horizontally and has a magnitude of Eq, where E is the electric field strength and q is the charge on the ball.

These two forces result in a tension in the string that holds the ball in equilibrium. The electric force provides the horizontal component of this tension, and the force of gravity provides the vertical component. As the string makes an angle of 17.4° with the wall, we can use trigonometry to relate the forces. From the vertical component, we have:

Tcos(θ) = mg

And from the horizontal component, we have:

Tsin(θ) = Eq

Dividing these two equations gives us:

tan(θ) = Eq/mg

Substituting the values provides:

tan(17.4°) = E(1.10 × 10-3 C)/(0.013 kg × 9.81 m/s2)

Solving for E yields:

E = mg(tan(θ))/q

By calculating this, we can find the electric field's magnitude.

Answer:

Electric field, [tex]E=5.4\times 10^6\ N/C[/tex]

Explanation:

It is given that,

Electromagnetic force acting on the proton when it is at rest, [tex]F=8.7\times 10^{-13}\ N[/tex] (in +x direction)

Speed of proton, [tex]v=1.5\times 10^6\ m/s[/tex]

We need to find the magnitude of the electric field. We know that when the charged particle is at rest it experiences electric force which is given by :

F = q E

[tex]E=\dfrac{F}{q}[/tex]

q is charge on proton

[tex]E=\dfrac{8.7\times 10^{-13}\ N}{1.6\times 10^{-19}\ C}[/tex]

E = 5437500 N/C

or

[tex]E=5.4\times 10^6\ N/C[/tex]

So, the magnitude of electric field is [tex]E=5.4\times 10^6\ N/C[/tex]. hence, this is the required solution.

Answer:

option 'b' and 'c'

Explanation:

when we throw ball in upward direction.

acceleration on the body will never be zero because there will always be acceleration due to gravity i.e. 'g' will be acting on it.

and force  of 'mg' will also be acting on the body, where m is the mass of the body.

so,

at the top the parameters which will be zero will be velocity and speed.

at the top most point the ball will change it's direction for that velocity will have to be zero at that point.

At the highest point of the ball's arc, both the velocity and acceleration are zero, with gravity being the only force acting on the ball.

The velocity of the ball at the highest point is zero. At the top of its arc, the acceleration of the ball is also zero because at that moment, the ball changes its direction from going upwards to downwards. The only force acting on the ball at the highest point is due to gravity.

Doubling the battery voltage applied to a light bulb will result in nearly quadrupling its power, assuming the bulb's resistance remains constant, due to the power being proportional to the square of the voltage (P = V²/R).

The luminosity of a light bulb, which can be thought of as its power output, is directly proportional to the electric potential energy converted into light and heat. The relationship between power (P), voltage (V), and current (I) is given by P = ΔVI, where ΔV represents the voltage across the bulb, and I represents the current flowing through it. When we discuss this relationship in terms of resistance (R), we can also express power as P = V²/R. According to this formula, doubling the voltage while keeping the resistance constant will result in a near quadrupling of power because the voltage is squared in the expression.

For example, a 25-W bulb designed to operate at a certain voltage would have its power increased nearly to 100 W if the voltage is doubled, assuming the resistance remains constant. However, in actual practice, the resistance of a bulb increases with temperature; thus, although the power increase is substantial, it is not exactly quadrupled.

Answer:

d. The waves must have the same frequency.

Explanation:

We can observe two waves that have interference between them and they create a pattern that can be destructive or constructive, in order for us to see them the waves have to meet three conditions: They need to have the same wavelenght, they must have a constant phase difference, and the same amplitude, otherwise they would interfere with eachother, the only that is not necessary is that they must have the same frequency.

Explanation:

1. The entire span of possible sound waves is called the acoustic spectrum. It is subdivided into infrasonic sounds, audible sounds, and ultrasonic sounds.

2. The difference between a musical note and another note at twice the frequency is called an octave.

3. Sound intensity varies with the inverse square of distance.

The possibility of sound frequencies is called the sound spectrum. Notes at double the frequency are an octave higher, and the perception of sound varies with distance from the source.

The total possibility of sound frequencies is referred to as the sound spectrum. The difference between a single musical note and another note at twice the frequency of the first is known as an octave in music. The note that is twice the frequency appears higher in pitch.

The relationship of sound to the distance between the source and receiver affects the perceived loudness of the sound. This is due to the way sound waves spread out and weaken as they travel away from the source. As a result, the further somebody is from the source of the sound, the quieter it will appear.

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Answer:

b. the vertical component of the velocity is zero

Explanation:

Tossing a basketball towards the basket is an example of projectile motion. When it comes to projectiles, the horizontal velocity is always constant. This is because in projectile motion, only gravity is acting upon the object. So this would mean that the vertical component is what is changes.

At the highest point of the arc, the vertical component of the velocity is zero because this is the point where it starts to descend. Notice that when you throw something in the air, when it reaches its maximum height it stops but it contines to move forward then goes down. At that point, the object is not moving and thus the velocity is zero.

1. 8.11 m/s

In order to find the speed of the bird, we need to combine both the horizontal and the vertical component of its velocity.

The horizontal component is equal to the tangential velocity of the circular motion of the bird. We have:

r = 6.00 m radius of the circle

T = 5.00 s period of the circular motion

So the tangential velocity (=horizontal component of the velocity) is

[tex]v_x=\frac{2\pi r}{T}=\frac{2\pi (6.00 m)}{5.00 s}=7.54 m/s[/tex]

The vertical component of the velocity is given by the problem

[tex]v_y = 3.00 m/s[/tex]

So the speed of the bird is the magnitude of its velocity:

[tex]v=\sqrt{v_x^2 + v_y^2}=\sqrt{(7.54 m/s)^2+(3.00 m/s)^2}=8.11 m/s[/tex]

2. 9.48 m/s^2

The only component of the acceleration of the bird is the one that determine the circular motion in the horizontal plane - so it is the centripetal acceleration.

The centripetal acceleration of the bird is given by:

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

where

[tex]v_x = 7.54 m/s[/tex] is the tangential velocity of the bird

r = 6.00 m is the radius of the circular trajectory

Substituting,

[tex]a=\frac{(7.54 m/s)^2}{6.00 m}=9.48 m/s^2[/tex]

3. Towards the centre of the circle

The bird is moving upward at constant velocity and in a straight line, so there is no component of the acceleration in the vertical direction.

In the horizontal plane, however, the bird is moving in a uniform circular motion: this means that the tangential acceleration is zero, while there is a centripetal acceleration (calculated in the previous part of the exercise). The direction of the centripetal acceleration is always towards the centre of the circular trajectory: so, the acceleration of the bird has exactly the same direction as the centripetal acceleration, i.e. towards the centre of the circle.

4. [tex]21.7^{\circ}[/tex]

We said that the components of the velocity of the bird are:

[tex]v_x = 7.54 m/s[/tex] in the horizontal direction

[tex]v_y = 3.00 m/s[/tex] in the vertical direction

So we can find the angle of the velocity with the horizontal by using the equation:

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

Substituting the values in, we find

[tex]\theta = tan^{-1} (\frac{3.00 m/s}{7.54 m/s})=21.7^{\circ}[/tex]

The tension in each rope supporting the child on the swing can be calculated by summing the forces (weight of the child and the pulling force) and dividing by 2, because the force is being equally distributed among the ropes due to the symmetry of the situation. This gives a tension of approximately 130N in each rope. The closest answer to this is B: 120N.

This problem relates to forces and tension. As the child is at rest, the total net force on the child is zero. This means that the sum of the upward forces (tension in the ropes) is equal to the downward forces (the weight of the child and the horizontal force pulling him backwards).

The weight of the child can be calculated using the formula Weight = m*g, where m is the mass and g is the acceleration due to gravity. In this case, the weight of the child is 160 N. As for the horizontal pull, it is given as 100N.

The combination of these two forces is then equally distributed across the two ropes holding the swing, due to the symmetry of the situation. Therefore, we sum the forces (160N for weight and 100N for pull) and divide by 2 to find the tension in each rope.

The calculation is, therefore, (160N + 100N) / 2 = 130N.

The answer closest to this value is B: 120N. Without more precise numbers, we choose the closest value.

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Acceleration is the rate of the change in velocity.

The answer would be the rate at which velocity changes.

Answer:

The speed is 20 m/s, and the direction is "downward". Acceleration is the rate of change of velocity. Usually, acceleration means the speed is changing, but not always. When an object moves in a circular path at a constant speed, it is still accelerating, because the direction of its velocity is changing.

Explanation:

Answer:

7.0 m

Explanation:

We can solve the problem by using the work-energy theorem, which states that the work done on the block is equal to the block's change in kinetic energy:

[tex]W=K_f - K_i[/tex]

where:

W is the work done (by gravity, since it is the only force acting on the block)

[tex]K_i[/tex] is the initial kinetic energy

[tex]K_f = 0[/tex] is the final kinetic energy, which is zero since the block comes to a stop

The work done by gravity against the block is:

[tex]W=-(mgsin \theta) d[/tex]

where

m is the mass of the block

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

[tex]\theta=20.5^{\circ}[/tex] is the angle of the incline

[tex](mg sin \theta)[/tex] is the component of the force of gravity that acts along the plane parallel to the incline, and the negative sign is due to the fact that this force acts opposite to the displacement of the block, d

d is the displacement of the block

Instead, the initial kinetic energy is

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

where

v = 4.90 m/s is the initial speed of the block

Substituting everything into the first equation, we can solve to find d, the displacement covered by the block up the incline:

[tex]-mgsin \theta d = -\frac{1}{2}mv^2\\d = \frac{v^2}{g sin \theta}=\frac{(4.90)^2}{(9.8)(sin 20.5^{\circ})}=7.0 m[/tex]

Final answer:

To find how far a block slides up an incline given an initial velocity and angle, we use energy conservation and kinematic equations, considering the acceleration due to gravity and the incline's angle.

Explanation:

The question involves calculating the distance a block slides up a frictionless incline given an initial velocity, making use of concepts from kinematics and energy conservation in physics. Using the energy conservation principle, the initial kinetic energy of the block is converted into potential energy at the highest point of its trajectory on the incline. Given an initial velocity of 4.90 m/s and an incline angle of 20.5°, we can calculate the distance using the formula v^2 = u^2 + 2as, where v is the final velocity (0 m/s at the highest point), u is the initial velocity, a is the acceleration (negative due to gravity acting opposite to the motion), and s is the distance travelled. The acceleration component along the incline can be calculated as a = -g*sin(θ), where g is the acceleration due to gravity (9.81 m/s²) and θ is the incline angle. Hence, the distance s can be found by rearranging the formula. This approach demonstrates the application of kinematic equations and the concept of work-energy theorem in solving problems related to motion on an incline.

Answer:

B) 12.1 m/s

Explanation:

Sum of the forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of the forces in the radial direction:

∑F = ma

F = m v² / r

Nμ = m v² / r

Substituting and solving for v:

mgμ = m v² / r

gμ = v² / r

v = √(gμr)

Given that μ = 0.200 and r = 75.0 m:

v = √(9.81 m/s² × 0.200 × 75.0 m)

v = 12.1 m/s

The car can turn through the circular track with a maximum speed of 12.1 m/s. Hence, option (B) is correct.

Given data:

The radius of circular track is, r = 75.0 m.

The coefficient of friction between the tire and road is, [tex]\mu = 0.200[/tex].

When the car moves around a circular track, then for safe turn through the track it is necessary to have the value of frictional force and centripetal force in a balanced amount. Therefore,

Fc = Ff

[tex]\dfrac{m \times v^{2}}{r} = \mu \times m \times g\\\\\\\dfrac{v^{2}}{r} = \mu \times g\\\\v =\sqrt{\mu \times r \times g}[/tex]

Solving as,

[tex]v =\sqrt{0.200 \times 75.0 \times 9.8}\\\\v =12.1 \;\rm m/s[/tex]

Thus, we can conclude that the car can turn through the circular track with a maximum speed of 12.1 m/s. Hence, option (B) is correct.

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Potential energy can be found using this formula:

PE= m * g * h

where:

PE= potential energy

m=mass

g=gravitational acceleration constant (9.8 m/s^2)

h= height

So your answer is height because you also use the gravitational constant.

Answer:

We know, U = m * g * h

So, seema needs the value of g & h

In short, Your Answer would be Option C) acceleration due to gravity and the height the ball reaches

Explanation:

Answer:

Total distance covered equals [tex]\frac{48}{7}feet[/tex]

Explanation:

The situation is represented in the attached figure

Distance in first bounce = [tex]d_{1}=2\times 3ft[/tex]

Distance in second bounce =[tex]d_{2}=2\times \frac{3}{8}ft[/tex]

Distance in third bounce=[tex]d_{3}=2\times \frac{3}{8^{2}}ft[/tex]

Thus the total distance covered = [tex]d_{1}+d_{2}+d_{3}+...[/tex]

Applying values we get

Total distance covered = [tex]2\times 3+2\times \frac{3}{8}+2\times \frac{3}{8^{2}}+2\times \frac{3}{8^{3}}+....\\\\=6(1+\frac{1}{8}+\frac{1}{8^{2}}+\frac{1}{8^{3}}+...)[/tex]

Summing the infinite geometric series we get total distance covered as[tex]S_{\infty }=\frac{a}{1-r}[/tex]

[tex]D=6(\frac{1}{1-\frac{1}{8}})\\\\D=\frac{48}{7}feet[/tex]

Answer:

a). same as

b). less than

Explanation:

a). When a bicycle is moving, the linear speed at the top of the rear wheel is same as the linear speed at the top of the front wheel. Since the clown's bicycle is a rigid body, both the wheels that is the front wheel and the rear wheel will move with the same linear speed.

b). Since we know that angular speed varies inversely to the radius of the wheel.

That is ω = 1 / r

Since the rear wheel has twice the radius of that of the front wheel, therefore real wheel will have less angular speed than the front wheel.

Therefore, the angular speed of the rear wheel is less than the angular speed of the front wheel.

Answer:

Maximum distance of image from mirror is equal to focal length of the mirror

Explanation:

As we know by the equation of mirror we have

[tex]\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}[/tex]

here we know for convex mirror

object position is always negative as it will be placed behind the mirror always

while the focal length of the convex mirror is always taken positive

So here we have

[tex]\frac{1}{d_i} + \frac{1}{-d_o} = \frac{1}{f}[/tex]

[tex]\frac{1}{d_i} = \frac{1}{d_o} + \frac{1}{f}[/tex]

so here maximum value of image distance is equal to focal length of the mirror

Answer:

Decreasing rate of thickness when x=3 in is [tex]\dfrac{dx}{dt}= -7.28\times 10^{-3} in/min[/tex]

Explanation:

Lets assume that thickness of ice over the spherical iron ball =x

So radius diameter of Sphere=5+x in

    Inner radius=5 in

So volume V=[tex]\dfrac{4}{3}\pi [(5+x )^3-5^3][/tex]

         V=[tex]\dfrac{4}{3}\pi [x^3+75x^2+15x][/tex]

Now given that ice melts rate=[tex] 15in^3/min[/tex]

[tex]\dfrac{dV}{dt}= -15in^3/min[/tex]

[tex]\dfrac{dV}{dt}=\dfrac{4}{3}\pi [3x^2+150x+15]\dfrac{dx}{dt}[/tex]

When x=3 in [tex]\dfrac{dV}{dt}= -15in^3/min[/tex]

[tex]-15=\dfrac{4}{3}\pi [3\times 3^2+150\times 3+15]\dfrac{dx}{dt}[/tex]

[tex]\dfrac{dx}{dt}= -7.28\times 10^{-3} in/min[/tex]

So decreasing rate of thickness when x=3 in is [tex]\dfrac{dx}{dt}= -7.28\times 10^{-3} in/min[/tex]

Answer:

W = 506.75 N

Explanation:

tension = 2300 N

Rider is towed at a constant speed means there no net force acting on the rider.

hence taking all the horizontal force and vertical force in consideration.

net horizontal  force:

F cos 30° - T cos 19° = 0

F cos 30° = 2300 × cos 19°

F = 2511.12 N

net vertical force:

F sin 30° - T sin 19°- W = 0

W = F sin 30° - T sin 19°

W =  2511.12 sin 30° - 2300 sin 19°

W = 506.75 N

The weight of the rider in parasailing can be found by calculating the vertical component of the tension in the rope. This is achieved by multiplying the total tension (2300 N) by the cosine of the angle from the horizontal (cos(19 degrees)), resulting in approximately 2167 N.

To determine the weight of the rider in parasailing, we need to look at the forces acting on the rider. If the rider is towed at a constant speed by a rope with a tension of 2300 N at an angle of 19 degrees from the horizontal, the vertical component of this tension supports the rider's weight. The vertical component of the tension can be found using trigonometry, specifically by multiplying the tension by the cosine of the angle from the horizontal (T*cos(θ)). Thus, the weight (W) is W = T * cos(19°), where T is the tension in the rope.

Mathematically, this is W = 2300 N * cos(19°). Using a calculator, we find the vertical component of the tension to be approximately 2167 N. Since the rider is at a constant speed and there is no vertical acceleration, this vertical component of the tension must equal the weight of the rider. Therefore, the weight of the rider is approximately 2167 N.

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Answer:

1. 1800 lb ft, 5.27 J

2. 905

Explanation:

1. Work is the change in energy.

W = mgh

W = (120 lb) (15 ft)

W = 1800 lb ft

W = mgh

W = (0.250 kg) (9.8 m/s²) (2.15 m)

W = 5.27 J

2. The IMA (ideal mechanical advantage) of a screw is the circumference of the screw divided by the pitch.

IMA = 2πr / p

The IMA of a handle (lever) is the circumference of the handle divided by the circumference of the screw.

IMA = 2πR / 2πr

The total IMA is the product:

IMA = (2πR / 2πr) (2πr / p)

IMA = 2πR / p

IMA = 2π (36) / (0.25)

IMA = 905

Answer:

a). F = 3.376 N, θ = 59.18°

b). W = 1.3x [tex]10^{9}[/tex] J      

Explanation:

We know

Gravitational constant, G = 6.673 x [tex]10^{-11}[/tex] N-[tex]m^{2}[/tex]/[tex]kg^{-2}[/tex]

Mass of the earth, M = 5.97 x [tex]10^{24}[/tex] kg

mass of the moon, m = 7.35 x [tex]10^{22}[/tex] kg  

Mass of the satellite, [tex]m_{s}[/tex] = 1250 kg

Distance between the objects, r = 3.84 x [tex]10^{5}[/tex] km

                                                      = 3.84 x [tex]10^{8}[/tex] m

Now

The force on the satellite due to moon

[tex]F_{m}= \frac{G\times m\times m_{s}}{r^{2}}[/tex]

[tex]F_{m}= \frac{6.673\times 10^{-11}\times 7.35\times 10^{22}\times 1250}{(3.84\times 10^{8})^{2}}[/tex]

[tex]F_{m}[/tex] = 0.0415 N ( in the positive x direction )

The force on the space craft due to the earth

[tex]F_{m}= \frac{G\times M\times m_{s}}{r^{2}}[/tex]

[tex]F_{m}= \frac{6.673\times 10^{-11}\times 5.97\times 10^{24}\times 1250}{(3.84\times 10^{8})^{2}}[/tex]

[tex]F_{m}[/tex] = 3.377 N ( at 60° to x axis )

Now component of force of earth along x axis

[tex]F_{e_{x}} = F_{e}\times cos 60[/tex]

                     = 3.377 x 0.5

                     = 1.6885 N

Now component of force of earth along y axis

[tex]F_{e_{y}} = F_{e}\times sin 60[/tex]

                      = 3.377 x 0.86

                      = 2.90 N

∴ Net force on the space craft due to earth and moon along x axis

[tex]F_{x}[/tex] = [tex]F_{e}[/tex] cos 60+[tex]F_{m}[/tex]

                       = 1.3885+0.0415

                        = 1.73 N

Net force on the space craft due to earth and moon along y axis  

[tex]F_{x}[/tex] = [tex]F_{e_{y}}[/tex]

                         = 2.90 N

Therefore, total force F = [tex]\sqrt{(F_{x}^{2})+(F_{y}^{2})}[/tex]

                                    F = [tex]\sqrt{(1.73^{2})+(2.90^{2})}[/tex]

                                    F = 3.376 N

∴ Magnitude of the net gravitational force on the space craft is 3.376 N

Direction of net force on the space craft is given by

[tex]\Theta = \arctan \left (\frac{F_{y}}{F_{x}}\right )[/tex]

[tex]\Theta = \arctan \left (\frac{2.90}{1.73}\right )[/tex]

[tex]\Theta = 59.18[/tex]°

Therefore this direction is 59.18° from the line joining earth and the space craft.

b).

∴ Gravitational potential energy of the space craft is given by

[tex]E = \frac{G.M.m_{s}}{r}+\frac{G.m.m_{s}}{r}[/tex]

[tex]E = \frac{G\times m_{s}\left ( M+m \right )}{r}[/tex]

[tex]E = \frac{6.673\times 10^{-11}\times 1250\left ( 5.97\times 10^{24}+7.35\times 10^{22} \right )}{3.84\times 10^{8}}[/tex]

E = 1312769385 J

E = 1.3 x [tex]10^{9}[/tex] J

Therefore minimum work done is 1.3x [tex]10^{9}[/tex] J

The net gravitational force exerted on the spacecraft by the earth and moon is 1.7923 * 10^19 N in the direction of 60 degrees measured from a line connecting the earth and the spacecraft. The minimum amount of work required to move the spacecraft to a point far from the earth and moon is 1.32 * 10^10 J.

(a) To find the magnitude and direction of the net gravitational force exerted on the spacecraft by the earth and moon, we need to first calculate the individual gravitational forces exerted by each object on the spacecraft and then find their vector sum. The gravitational force between two objects can be calculated using the equation:

F = G * (m1 * m2) / r^2

where F is the force, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between them.

Given that the mass of the spacecraft is 1250 kg, the mass of the earth is 5.97 * 10^24 kg, the mass of the moon is 7.35 * 10^22 kg, and the distance between the earth and the spacecraft (and the moon and the spacecraft) is 3.84 * 10^5 km, we can substitute these values into the equation to find the gravitational force exerted by each object. Since the triangle is equilateral, the angle between the line connecting the earth and the spacecraft and the line connecting the moon and the spacecraft is 60 degrees.

Using the equation for gravitational force, we can calculate the force exerted by the earth:

Fearth = (6.674 * 10^-11 N*m^2/kg^2) * ((5.97 * 10^24 kg) * (1250 kg)) / (3.84 * 10^8 m)^2

= 1.79 * 10^19 N

Similarly, we can calculate the force exerted by the moon:

Fmoon = (6.674 * 10^-11 N*m^2/kg^2) * ((7.35 * 10^22 kg) * (1250 kg)) / (3.84 * 10^8 m)^2

= 3.23 * 10^14 N

To find the net gravitational force, we need to find the vector sum of these forces:

Fnet = Fearth + Fmoon

= (1.79 * 10^19 N) + (3.23 * 10^14 N)

= 1.79 * 10^19 N + 3.23 * 10^14 N

= 1.7923 * 10^19 N

To find the angle, we can use trigonometry. Since the triangle is equilateral, the angle between the line connecting the earth and the spacecraft and the line connecting the moon and the spacecraft is 60 degrees.

Therefore, the magnitude of the net gravitational force exerted on the spacecraft by the earth and moon is 1.7923 * 10^19 N, and the direction is 60 degrees measured from a line connecting the earth and the spacecraft.

(b) The minimum amount of work required to move the spacecraft to a point far from the earth and moon can be calculated using the formula for gravitational potential energy:

PE = -G * (m1 * m2) / r

where PE is the potential energy, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between them.

Since the gravitational potential energy is defined as zero at an infinite distance, we need to calculate the potential energy at the starting point and subtract the potential energy at the final point:

PEinitial = (-6.674 * 10^-11 N*m^2/kg^2) * ((5.97 * 10^24 kg) * (1250 kg)) / (3.84 * 10^8 m)

= -1.32 * 10^10 J

Assuming we move the spacecraft to a point far from the earth and moon, the potential energy becomes zero:

PEfinal = 0 J

The minimum amount of work done is equal to the change in potential energy:

Work = PEfinal - PEinitial

= 0 J - (-1.32 * 10^10 J)

= 1.32 * 10^10 J

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Answer:

B). to the right

Explanation:

Since the direction of magnetic field is into the page

So here we know that

[tex]B = B_o(-\hat k)[/tex]

now the velocity is from bottom to top

so we have

[tex]v = v_o \hat j[/tex]

now the force on the moving charge is given as

[tex]\vec F = q(\vec v \times \vec B)[/tex]

now we have

[tex]\vec F = (-e)(v_o \hat j \times B_o(-\hat k))[/tex]

[tex]\vec F = e v_o B \hat i[/tex]

so force will be towards Right

The force on an electron moving upward in a magnetic field directed into the page will be to the left due to the negative charge of the electron and the application of the right-hand rule.

Since the magnetic field is directed into the page and the electron is moving from the bottom edge to the top edge of the page, we can use the right-hand rule for charged particles in a magnetic field to determine the direction of the force. However, because electrons are negatively charged, we need to reverse the direction given by the right-hand rule. Therefore, if a positively charged particle moving upward would experience a force to the right due to the magnetic field into the page, an electron would experience a force in the opposite direction, which is to the left. Hence, the correct answer to the question is B. to the left.

Answer:

The speed of the wave is 2 m/s

Explanation:

It is given that,

Distance covered by water waves or the wavelength of water wave, [tex]d=2\ m[/tex]

The water waves on a lake goes by a piece of floating cork that bobs up and down one complete cycle each second, t = 1 s

The speed of the wave is equal to the distance covered divided by total time taken i.e.

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

v = 2 m/s

So, the speed of the wave is 2 m/s. Hence, this is the required solution.  

The speed of the 2-m long water waves is determined using the formula Speed = Wavelength × Frequency. With a wavelength of 2 m and a frequency of 1 Hz, the speed of the waves is 2 m/s.

The speed of a wave is calculated by multiplying its wavelength by its frequency. In the case of a 2-m long water wave and a cork bobbing up and down once every second, the frequency of the wave is 1 cycle per second (this is also known as 1 hertz).

To find the speed of the wave, we use the formula:

Speed = Wavelength × Frequency

Speed = 2 m × 1 Hz = 2 m/s

Therefore, the speed of the water waves on the lake is 2 meters per second.

Final answer:

Assuming an AC source for illustration, a transformer with a primary coil of 100 turns connected to a 1.5 V source and a secondary coil of 200 turns would yield a voltage of 3.0 V across the secondary coil, calculated by the turns ratio and input voltage.

Explanation:

The question pertains to the operation of a transformer, which is a device that uses electromagnetic induction to change the voltage levels between circuits. A transformer consists of a primary coil and a secondary coil wound around a common core. When an alternating current (AC) flows through the primary coil, it creates a changing magnetic field, which induces a voltage across the secondary coil. The voltage induced in the secondary coil is related to the voltage in the primary coil by the ratio of the number of turns in each coil.

For a step-up transformer, the voltage is increased from the primary to the secondary coil. The relationship can be denoted as Vp/Vs = Np/Ns, where Vp and Vs are the voltages in the primary and secondary coils, respectively, and Np and Ns are the number of turns in the primary and secondary coils, respectively. Since this question refers to a direct current (DC) source (a size AA battery) which would not work effectively with a transformer, let's consider it as an AC source for the purpose of explanation.

If we apply the formula Vs = (Ns/Np) * Vp, where Vp is 1.5 volts, Np is 100 turns, and Ns is 200 turns, we get:

Vs = (200/100) * 1.5 V

Vs = 2 * 1.5 V

Vs = 3.0 V

The voltage across the secondary coil would theoretically measure 3.0 volts if the system operated with an AC input and perfect efficiency.

Answer:

1. 610,000 lb ft

2. 490 J

Explanation:

1. First, convert mi/hr to ft/s:

100 mi/hr × (5280 ft / mi) × (1 hr / 3600 s) = 146.67 ft/s

Now find the kinetic energy:

KE = ½ mv²

KE = ½ (1825 lb / 32.2 ft/s²) (146.67 ft/s)²

KE = 610,000 lb ft

2. KE = ½ mv²

KE = ½ (5 kg) (14 m/s)²

KE = 490 J