A lens collects light and focuses it into a small spot. This increases the ________ of the light wave.A lens collects light and focuses it into a small spot. This increases the ________ of the light wave.

Answers

Answer 1

Answer:

intensity.

Explanation:

when the light collected by the lens is focused into a small spot it tends to increase the intensity of the light.

as different path of light with different intensity combines from passing through the lens it tends to make the light path and intensity coherent and after being coherent there intensity increases.


Related Questions

A wheel of moment of inertia of 5.00 kg∙m2 starts from rest and accelerates under a constant torque of 3.00 N∙m for 8.00 s. What is the wheel's rotational kinetic energy at the end of 8.00 s?

Answers

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.

A hot object and a cold object are placed in thermal contact and the combination is isolated. They transfer energy until they reach a common temperature. The change ΔSh in the entropy of the hot object, the change ΔSc in the entropy of the cold object, and the change ΔStotal in the entropy of the combination are:
a) ΔSh > 0, ΔSc > 0, ΔStotal > 0 b) ΔSh < 0, ΔSc > 0, ΔStotal > 0 c) ΔSh < 0, ΔSc > 0, ΔStotal < 0 d) ΔSh > 0, ΔSc < 0, ΔStotal > 0 e) ΔSh > 0, ΔSc < 0, ΔStotal < 0

Answers

Final answer:

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.

Explanation:

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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Which of the following conditions is not necessary to observe an interference pattern between two sets of waves?
a. The waves must have the same wavelength.
b. The waves must have a constant phase difference.
c. The waves must have the same amplitude.
d. The waves must have the same frequency.

Answers

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.

The density of lead is greater than iron, and both metals are denser than water. Is the buoyant force on a solid lead object (a) greater than, (b) equal to, or (c) less than the buoyant force acting on a solid iron object of the same dimensions?

Answers

Answer:

b) equal

Explanation:

in question we it is given that the density of the iron is greater than the density of lead and the density of the water is less than both  the metal.

As both the body is submerged in the water so for the buoyant force calculation  we will take the density of water.

i.e. for buoyant force calculation both the metal will have equal buoyant force acting on it.

Final answer:

The buoyant forces on a solid lead object and a solid iron object of the same dimensions are equal because they displace the same volume of water. The correct option is b) equal to.

Explanation:

The question concerns the buoyant force acting on objects submerged in water. According to Archimedes' principle, the buoyant force on an object is equal to the weight of the fluid it displaces.

When two objects have the same dimensions but different densities, they will displace the same volume of water when fully submerged, therefore the buoyant force on each object will be the same. It does not matter whether the object is made of lead or iron; the buoyant force depends on the volume of water displaced, not on the object's density.

An object cannot remain at rest unless which of the following holds?

a. The net force acting on it is zero.
b. The net force acting on it is constant and nonzero.
c. There are no forces at all acting on it.
d. There is only one force acting on it.

Answers

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).

In the winter sport of curling, two teams alternate sliding 20 kg stones on an icy surface in an attempt to end up with the stone closest to the center of a target painted on the ice. During one turn, a player releases a stone that travels 27.9 m before coming to rest. The friction force acting on the stone is 2.0 N. What was the speed of the stone when the player released it?

Answers

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]

Further explanation

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]

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[tex]\texttt{ }[/tex]

Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

There are some points on a standing
wave that never move. What are
these points called?
A. Harmonics
B. Normal Modes
C. Nodes
D. Anti-nodes
E. Interference

Answers

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.

1. A racing car with the driver weighs 1825 lb. Find the kinetic energy in ft*lb when traveling with a speed of 100 mi/hr.


3. A mass of 5 kg initially at rest is given a speed of 14 m/s. What is its kinetic energy?

PLEASE HELP ME

Answers

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

Suppose that the angular separation of two stars is 0.1 arcseconds, and you photograph them with a telescope that has an angular resolution of 1 arcsecond. How will the stars appear in the photograph?

Answers

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.

How do you find the angular distance?

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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The mercalli scale is a scale from ________.

Answers

Answer: 1 to 12.

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.

Which statements describe intensity? Check all that apply.



Intensity is the amount of energy that flows in an area per unit of time.


Intensity is the number of wavelengths that flow in an area per unit of time.


Intensity that is lower than or equal to 85 units can harm human hearing.


Intensity that is greater than or equal to 85 units can harm human hearing.


Intensity is measured in decibels.


Intensity is measured in hertz.


(A, D, E) Your Welcome)

Answers

Answer:

it's A, D, E

Explanation:

The statements that best describe intensity are as follows:

Intensity is the amount of energy that flows in an area per unit of time.An intensity that is greater than or equal to 85 units can harm human hearing.Intensity is measured in decibels.

Thus, the correct options for this question are A, D, and E.

How is intensity measured?

The intensity measure is a non-random measure and is defined as the expectation value of the random measure of a set, hence it corresponds to the average volume the random measure assigns to a set. The intensity measure contains important information about the properties of the random measure.

It is generally represented as the amount of energy that flows in an area per unit of time. It is greater than or equal to 85 units and can harm human hearing in an ideal range. The significance of intensity is always measured in decibels (dB).

Therefore, the correct options for this question are A, D, and E.

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At a certain instant, the earth, the moon, and a station- ary 1250-kg spacecraft lie at the vertices of an equilateral triangle whose sides are 3.84 * 105 km in length. (a) Find the magnitude and direction of the net gravitational force exerted on the spacecraft by the earth and moon. State the direction as an angle measured from a line connecting the earth and the spacecraft. In a sketch, show the earth, the moon, the spacecraft, and the force vector. (b) What is the minimum amount of work that you would have to do to move the spacecraft to a point far from the earth and moon? Ignore any gravitational effects due to the other planets or the sun.

Answers

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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What is the maximum speed at which a car can safely travel around a circular track of radius 75.0 m if the coefficient of friction between the tire and road is 0.200?

A) 3.87 m/s
B) 12.1 m/s
C) 15.0 m/s
D) 147 m/s

Answers

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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The primary coil of a transformer contains 100 turns; the secondary has 200 turns. The primary coil is connected to a size aa battery that supplies a constant voltage of 1.5 volts. What voltage would be measured across the secondary coil?

Answers

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.

A ball is attached to a vertical spring. The ball is initially supported at a height y so that the spring is neither stretched nor compressed. The ball is then released from rest and it falls to a height y - h before moving upward. Consider the following quantities: translational kinetic energy, gravitational potential energy, elastic potential energy. When the ball was at a height y - (h/2), which of the listed quantities has (have) values other than zero joules

Answers

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.

In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A rider is towed at a constant speed by a rope that is at an angle of 19 ∘ from horizontal. The tension in the rope is 2300 N. The force of the sail on the rider is 30∘ from horizontal.What is the weight of the rider?

Answers

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

Final answer:

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.

Explanation:

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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Review When at rest, a proton experiences a net electromagnetic force of magnitude 8.7×10−13 N pointing in the positive x direction. When the proton moves with a speed of 1.5×106 m/s in the positive y direction, the net electromagnetic force on it decreases in magnitude to 7.5×10−13 N , still pointing in the positive x direction. You may want to review (Pages 773 - 777) . Part A Find the magnitude of the electric field. Express your answer using two significant figures.

Answers

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.

Sand is falling onto a cone-shaped pile at 10 cubic feet per minute. The diameter of the base of this cone is always 3 times the height of the cone. At what rate is the height of the sand pile increasing when the pile is 5 feet high?

Answers

Answer:

[tex]\frac{dh}{dt} = 0.056 ft/min[/tex]

Explanation:

rate of falling of cone[tex] \frac {dv}{dt} = 10 ft^3/min[/tex]

height of pile is 5 feet

diameter is 3 times the height of cone

d = 3h

2r =3h

[tex]r \frac{3}{2} h[/tex]

volume of cone is given as

[tex]v = \frac{1}{3} \pi [\frac{2}{3} h]^{2} h[/tex]

[tex]v =\frac{3}{4} \pi* h^{3}[/tex]

[tex]\frac{dv}{dt} =\frac{3}{4}*\pi*3*h^{2}\frac{dh}{dt}[/tex]

[tex]\frac {dv}{dt} =\frac{9}{4}*\pi h^{2}}\frac{dh}{dt}[/tex]

[tex]\frac{dh}{dt} = \frac{4}{9} \pi*5^{2}*10}[/tex]

[tex]\frac{dh}{dt} = \frac{40}{706.5}[/tex]

[tex]\frac{dh}{dt} = 0.056 ft/min[/tex]

Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21⁰ and 61⁰, respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 60.0 w/cm² after it passes through the stack.

If the incident intensity is kept constant:

(a) What is the intensity of the light after it has passed through the stack if the second polarizer is removed?
(b) What is the intensity of the light after it has passed through the stack if the third polarizer is removed?

Answers

Apply Malus' law for light passing between two polarizing filters:

I = I₀cos²(θ)

I is the intensity after passing through, I₀ is the original intensity, and θ is the relative angle between the two filters.

For light passing through more than two filters, keep multiplying I₀ by cos²(θ) for each filter.

First let's solve for I₀ so we can answer parts a and b. We have three filters, so:

I = I₀cos²(θ₁)cos²(θ₂)

Given values:

I = 60.0W/cm²

θ₁ = 21°

θ₂ = 40° (we care about the relative angle between polarizers 2 & 3, not 1 & 3)

Plug in the values and solve for I₀:

60.0 = I₀cos²(21°)cos²(40°)

I₀ = 117W/cm²

a) Remove the second filter. Now the light passing through filter 1 only passes through filter 3. To find the resulting intensity:

I = I₀cos²(θ)

Where θ = 61° (relative angle between filter 1 & 3)

I = 117cos²(61°)

I = 27.5W/cm²

b) Remove the third filter. Now the light passing through filter 1 only passes through filter 2. To find the resulting intensity:

I = I₀cos²(θ)

Where θ = 21° (relative angle between filter 1 & 2)

I = 117cos²(21°)

I = 102W/cm²

A succession of 2-m long water waves on a lake goes by a piece of floating cork that bobs up and down one complete cycle each second. What is the speed of the wave?

Answers

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.

What is the resistance of a 7 W light bulb connected to a 7 V battery source. The answer key says .39 but not sure how they got that?

Answers

Explanation:

Power is current times voltage.

P = IV

And from Ohm's law, voltage is current times resistance.

V = IR

So if we solve for I in Ohm's law and substitute into the power equation:

I = V/R

P = (V/R) V

P = V² / R

Given that P = 7 W and V = 7 V:

7 = 7² / R

R = 7 Ω

Make sure you copied the problem correctly.  If you did, the answer key may be wrong.

A ball is dropped and begins bouncing. On the first bounce, the ball travels 3 feet. Each consecutive bounce is 1/8 the distance of the previous bounce. What is the total distance that the ball travels? Round to the nearest hundredth.

Answers

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]

If a black hole is truly black and has an escape velocity greater than the speed of light such that no light can escape it, where do the X-rays come from in the black hole candidates so far identified?

Answers

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.

Chemical weathering is greatest under conditions of

Answers

Answer: higher mean annual rainfall and temperatures.

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.

Final answer:

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.

Explanation:

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.

Learn more about Chemical Weathering here:

https://brainly.com/question/31833236

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You toss a basketball toward the basket. At the highest point of its arc,

a. The horizontal component of the velocity is zero.
b. The vertical component of the velocity is zero.
c. Both the horizontal and vertical components of the velocity are zero.

Answers

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.

Assume that a uniform magnetic field is directed into thispage. If an electron is released with an initial velocity directedfrom the bottom edge to the top edge of the page, which of thefollowing describes the direction of the resultant force acting onthe electron?

A. out of the page
B. to the right
C. to the left
D. into the page

Answers

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.

When a ball is tossed in the air and reaches the top of its arc, which of the following quantities are zero? Choose one or more: A. the acceleration of the ball B. the speed of the ball C. the velocity of the ball D. the force on the ball

Answers

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.

1. Calculate the work done in raising 120 lb to a height of 15.0 ft. Find the work done in raising 250 g a distance of 215 cm.

2. A screw jack has a handle of radius 36 in and thread pitch of .25in. Calculate the IMA of the jack


PLEASE HELP ME

Answers

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

Mary borrows $5,000 dollars from her mother at a 3% simple interest rate and pays her $600 in interest after (Math
Processing Error) years. What is the value of (Math Processing Error)?
2
3
5
4

Answers

Answer:

4 years

Explanation:

5,000 times .03= 150

600/150=4

In Part H, you discovered that the luminosity of a light bulb increases if the current increases. The rate at which electric potential energy is converted into heat depends on the current flowing through the bulb and the voltage across the bulb. This energy is supplied by the battery. Mathematically, the luminosity P of the light bulb is given by P=ΔVI, where ΔV is the voltage across the bulb and I is the current.What happens to the luminosity of the light bulb if the voltage of the battery is doubled? (Note that the PhET simulation does not display a numerical value for the luminosity, so you should use the relationship between the luminosity, the voltage across the bulb, and the current.)

Answers

Final answer:

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).

Explanation:

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.

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