A rectangular field is 300 meters long and 300 meters wide. What is the area of the field in square kilometers? Do not round your answer. Be sure to include the correct unit in your answer.

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.

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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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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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:

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:

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.

Answer:

C. Nodes

Explanation:

There are some points on a standing  wave that never move. These points are called Nodes.

Final answer:

Nodes are points on a standing wave with no motion due to destructive interference, while antinodes are where motion is maximized due to constructive interference.

Explanation:

The points on a standing wave that appear to remain flat and do not move are called nodes. Nodes occur due to complete destructive interference, where the amplitude of the wave at these points is always zero. In contrast, antinodes are points on a standing wave that experience the maximum motion as a result of constructive interference, resulting in the wave amplitude reaching its highest values at these points.

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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Explanation:

It is given that,

Mass of the stone, m = 20 kg

Frictional force, F = -2 N

A player releases a stone that travels 27.9 m before coming to rest, s = 27.9 m

We need to find the initial velocity of the stone. It can be calculated using third equation of motion as :

[tex]v^2-u^2=2as[/tex]

v = 0

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

[tex]a=\dfrac{-2\ N}{20\ kg}=-0.1\ m/s^2[/tex]

[tex]0-u^2=2\times -0.1\ m/s^2\times 27.9\ m[/tex]

[tex]u=2.36\ m/s[/tex]

So, the speed of the stone when it was released is 2.36 m/s. Hence, this is the required solution.

The speed of the stone when the player released it was about 2.4 m/s

[tex]\texttt{ }[/tex]

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

[tex]\large {\boxed {F = ma }[/tex]

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

mass of stone = m = 20 kg

distance traveled = d = 27.9 m

magnitude of friction force = f = 2.0 N

final speed of stone = v = 0 m/s

Asked:

initial speed of stone = u = ?

Solution:

Firstly, we will use Newton's Second Law of Motion to calculate the deceleration of the stone:

[tex]\Sigma F = ma[/tex]

[tex]-f = m a[/tex]

[tex]-2.0 = 20 a[/tex]

[tex]a = -2.0 \div 20[/tex]

[tex]\boxed{a = -0.1 \texttt{ m/s}^2}[/tex]

[tex]\texttt{ }[/tex]

Next, we could calculate the initial speed of stone as follows:

[tex]v^2 = u^2 + 2ad[/tex]

[tex]0^2 = u^2 + 2( -0.1) (27.9)[/tex]

[tex]u^2 = 5.58[/tex]

[tex]u = \sqrt{5.58}[/tex]

[tex]\boxed{u \approx 2.4 \texttt{ m/s}}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Dynamics

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). 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:

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:

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

Explanation:

Chemical weathering is the set of destructive processes through which rocky materials go trhough. These processes cause changes in the color, texture, composition, firmness and shape of the material.

It should be noted that this happens when the rocks come into contact with atmospheric agents such as oxygen and carbon dioxide.

Another important aspect is that rocks are able to break up more easily thanks to this type of weathering, since the mineral grains within the rock lose adherence and dissolve better under the action of some physical agents, such as humidity (rainfall included) and temperature.

Therefore:

Chemical weathering is greatest under conditions of higher mean annual rainfall and temperatures.

Chemical weathering is greatest under circumstances of hot temperatures and abundant moisture, such as in tropical rainforests. These conditions accelerate weathering reactions such as hydrolysis, oxidation and dissolution, impacting soil characteristics and promoting biological activity.

Chemical weathering occurs most under conditions of high temperature and abundant moisture. These conditions accelerate the reactions that lead to weathering. For example, processes like hydrolysis, oxidation, and dissolution, which are facilitated by climate conditions of high temperature and moisture, lead to a more extensive break down of rocks. Therefore, regions with these climate conditions, such as tropical rainforests, experience the greatest degree of chemical weathering. This intensifies the development of soil characteristics and also promotes biological activity, a vital element of soil quality.

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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:

The wheel's rotational kinetic energy is 57.6 J.

Explanation:

Given that,

Moment of inertia = 5.00 kg.m²

Torque = 3.00 N.m

Time = 8.00 s

We need to calculate the angular acceleration

Using formula of the torque act on the wheel

[tex]\tau=I\alpha[/tex]

[tex]\alpha=\dfrac{\tau}{I}[/tex]

Where, I = moment of inertia

[tex]\alpha[/tex] = angular acceleration

[tex]\tau[/tex] = torque

Put the value into the formula

[tex]\alpha=\dfrac{3.00}{5.00}[/tex]

[tex]\alpha=0.6\ rad/s^2[/tex]

We need to calculate the final angular velocity

Initially wheel at rest so initial velocity  is zero.

Using formula of angular velocity

[tex]\alpha=\dfrac{\omega_{f}-\omega_{i}}{t}[/tex]

[tex]\omega_{f}=\omega_{i}+\alpha t[/tex]

Put the value into the formula

[tex]\omega_{f}=0+0.6\times8.00[/tex]

[tex]\omega_{f}=4.8\ rad/s[/tex]

We need to calculate the rotational kinetic energy of the wheel

Using formula of the rotational kinetic energy

[tex]K.E_{rot}=\dfrac{1}{2}I\omega^2[/tex]

[tex]K.E_{rot}=\dfrac{1}{2}\times5.00\times(4.8)^2[/tex]

[tex]K.E_{rot}=57.6\ J[/tex]

Hence, The wheel's rotational kinetic energy is 57.6 J.

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).

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.

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.

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:

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.

The Mercalli seismological scale is a scale of 12 degrees, where 1 is very weak and 12 catastrophic.

This scale owes its name to the Italian physicist Giuseppe Mercalli and was developed to assess the intensity of earthquakes through the effects and damages caused to different structures.

This means the Mercalli scale measures how strong a earthquake has been by its consequences and not by its magnitude, therefore it is based on empirical observations.

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.

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]

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.

The correct choice is b) ΔSh < 0, ΔSc > 0, ΔStotal > 0. The hot object loses heat decreasing its entropy, the cold object gains heat increasing its entropy, and the total entropy of the system increases as it achieves equilibrium as per the second law of thermodynamics.

In thermodynamics, the principle of entropy states that energy distribution will always proceed towards the most probable distribution or a state of equilibrium. When a hot and cold object are brought into thermal contact within an isolated system, heat flows from the hot object to the cold object until they reach a common temperature. In the process, the entropy or disorder of the system increases.

So, the correct choice is b) ΔSh < 0, ΔSc > 0, ΔStotal > 0. This indicates that the entropy of the hot object decreases (as it loses heat), the entropy of the cold object increases (as it gains heat), while the total entropy of the system increases (headed towards equilibrium). The inequality reflects the second law of thermodynamics which states that the total entropy of an isolated system can only increase over time.

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Answer:

4 years

Explanation:

5,000 times .03= 150

600/150=4

Answer:

All the three quantities will have non zero joules.

Explanation:

At the initial position of rest the system will have only gravitational potential energy while the other 2 quantities will be zero.

when the system reaches a height (y-h) only kinetic energy will be zero while the other 2 quantities will be non zero

At the position of (y-h/2) all the 3 quantities will be non zero.

Answer:

Translational, elastic, and gravitational energy

Explanation:

The location is neither a maximum or minimum, so there is both kinetic and potential energy on the ball.