Which one of the following types of electromagnetic wave travels through space the fastest?

a.) infrared
b.) radio waves
c.) ultraviolet
d.) microwaves
e.) They all travel through space at the same speed.

Answers

Answer 1

Answer:

e.) They all travel through space at the same speed.

Answer 2

All electromagnetic waves travel through space at the same speed. therefore the correct answer is option E

What is an electromagnetic wave?

The oscillation of an electric field and a magnetic field produces electromagnetic waves, which are waves. In other words, electromagnetic waves (EM waves) are made up of vibrating magnetic and electric fields that are orthogonal to one another. Transverse waves are another name for electromagnetic waves since they move in a transverse direction.

These waves are used to transfer light & heat as a form of electromagnetic radiation, these electromagnetic waves are of various kinds such as radio waves, visible light, ultraviolet waves, x-rays, infrared waves, microwaves, gamma rays, etc.

All different kinds of electromagnetic waves have different wavelengths and frequencies but they all travel through space at the same speed of light.

Thus, All electromagnetic waves travel through space at the same speed. therefore the correct answer is option E

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Related Questions

Which type of energy is thermal energy a form of?

Answers

Answer:

Kinetic Energy

Explanation:

Heat energy is another name for thermal energy. Kinetic energy is the energy of a moving object. As thermal energy comes from moving particles, it is a form of kinetic energy.


kinetic or thermal but can i get brainliest please

A loop of radius r = 3.0 cm is placed parallel to the xy-plane in a uniform magnetic field = 0.75 T . The resistance of the loop is 18 Ω. Starting at t = 0, the magnitude of the field decreases uniformly to zero in 0.15 seconds. What is the magnitude of the electric current produced in the loop during that time?

Answers

Answer:

i = 7.777 × 10⁻⁴ A = 0.77 mA

Explanation:

Given:

loop radius, r = 3.0 cm = 0.03 m

Area, A = π x r² = π x 0.03² = 0.0028 m²

Magnetic Field, B = 0.75 T

Loop resistance, R = 18 Ω

time, t = 0.15 seconds

Now,

the induced emf is given as:

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

also

EMF = i x R

Where, i is the current flowing

equating both the formulas for EMF, we get

[tex]{i}{R}=-\frac{BA}{t}[/tex]

or

[tex]{i}=-\frac{BA}{tR}[/tex]

substituting the values in the above equation we get

[tex]{i}=-\frac{0.75\times 0.0028}{0.15\times 18}[/tex]

or

the magnitude of the current, i = 7.777 × 10⁻⁴ A = 0.77 mA

Final answer:

The electric current produced in a loop due to a changing magnetic field can be calculated using Faraday's law of electromagnetic induction. Based on the given data, the magnitude of the induced current in the loop is approximately 1.18 mA.

Explanation:

The scenario described in the question depicts a situation where a magnetic field decreases uniformly in magnitude, thus inducing an electric current within a circular loop according to Faraday's law of electromagnetic induction.

The electric current produced in the loop as a result of the change in the magnetic field can be calculated using the expression, I = ΔΦ/ Δt * R, where ΔΦ is the change in magnetic flux, Δt is the time interval, and R is the resistance of the loop.

In this case, the initial magnetic flux, Φ1 = B1 * A, where B1 is the initial magnetic field and A is the area of the loop. Since the magnetic field decreases to zero, the final magnetic flux, Φ2, is zero. So, ΔΦ = Φ2 - Φ1 = - B1 * A. Here, A = πr², where r is the radius of the loop.

So, I = - [ (B1* πr²) / ( Δt * R ) ]. Substituting the given values into the equation, we get I = (0.75 T * π * (0.03 m)²) / ( 0.15 s * 18 Ω ) = 0.00118 A or 1.18 mA.

So, the magnitude of the induced current is approximately 1.18 mA.

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When the temperature goes up 3^\circ on the Cantor scale, it goes up 8^\circ on the Frobenius scale. On both scales, 18^\circ is the same temperature. How many Frobenius degrees are equal to 30^\circ Cantor?

Answers

Answer:

50°

Explanation:

It is given that the both scales are the same at 18˚

Now,

at 30˚ on the Cantor scale, the temperature is 30˚ − 18˚ = 12˚ above

the 18˚ mark.

also, it is given that with every 3˚ increase in the temperature on the Cantor scale is there is an 8˚ increase on the Frobenius scale,

mathematically, we can write it as (by unitary method)

3˚ increase in the temperature on the Cantor =  8˚ increase on the Frobenius scale

or

1˚ increase in the temperature on the Cantor =  (8/3)˚ increase on the Frobenius scale

thus, for x˚ increase in the temperature on the Cantor =  ((8/3)˚ × x) increase on the Frobenius scale

hence, for 12° increase we have

12˚ increase in the temperature on the Cantor =  ((8/3)˚ × 12) increase on the Frobenius scale

or

12˚ increase in the temperature on the Cantor =  32° increase on the Frobenius scale

hence, the final reading on the Frobenius scale will be, 18˚ + 32˚ = 50˚.

A circular loop of flexible iron wire has an initial circumference of 167 cm, but its circumference is decreasing at a constant rate of 15.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.500 T, which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.A) Find the magnitude of the emf E induced in the loop after exactly 8.00 s has passed since the circumference of the loop started to decrease.B) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field. (Clockwise or Counterclockwise?)

Answers

Answer:

Part a)

[tex]EMF = 5.6 \times 10^{-3} V[/tex]

Part b)

Since the radius is decreasing so induced current will increase the flux through the coil

So it would be clockwise in direction

Explanation:

As we know that magnetic flux linked with the coil is given as

[tex]\phi = \pi r^2 B[/tex]

now the rate of change in flux is given as

[tex]\frac{d\phi}{dt} = 2\pi r \frac{dr}{dt} B[/tex]

now we know that circumference is decreasing at rate of 15 cm/s

so here we know the length of circumference as

[tex]C = 2\pi r[/tex]

So rate of change in circumference is

[tex]\frac{dC}{dt} = 2\pi \frac{dr}{dt}[/tex]

[tex]\frac{1}{2\pi}(15 cm) = \frac{dr}{dt}[/tex]

final length of circumference at t = 8 s

[tex]C = 167 - (15)(8) = 47[/tex]

Part a)

Now the induced EMF is given as

[tex]EMF = (2\pi r)(\frac{1}{2\pi})(0.15)(0.5)[/tex]

[tex]EMF = (0.47)(\frac{1}{2\pi})(0.15)(0.5)[/tex]

[tex]EMF = 5.6 \times 10^{-3} V[/tex]

Part b)

Since the radius is decreasing so induced current will increase the flux through the coil

So it would be clockwise in direction

Final answer:

The problem is solved using Faraday's Law of Electromagnetic Induction: the induced emf equals to the change in magnetic flux divided by change in time. Then, Lenz's law is used to determine the direction of the induced current.

Explanation:

The problem you're dealing with is an application of Faraday's Law of Electromagnetic Induction. This law states that the electromotive force (or emf) induced in a circuit is equivalent to the rate of change of magnetic flux through that circuit.

For Part A, you know that the rate of change of the circumference is constant and you can convert that to a rate of change of radius (since circumference = 2πr). Using these formulas, you can determine the rate of change of area as πr(dr/dt), where dr/dt is the rate of change of radius. The induced emf is proportional to the rate of change of magnetic flux, which is the product of magnetic field strength and area. Thus, induced emf equals to (change in flux/change in time), or -Bπr(dr/dt).

For part B, you apply Lenz's law, which states that the direction of induced current is such that the magnetic field due to it opposes the change in the initial magnetic field. Since the field is getting smaller (since the loop is contracting), then the current will act in such a way as to try to keep the size of the field the same. This means going in the counterclockwise direction.

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Two airplanes leave an airport at the same time and travel in opposite directions. One plane travels 87 km/h faster than the other. If the two planes are 11,865 km apart after 7 hours, what is the rate of each plane?

Answers

Answer:

Speed of A = 891 km/h

Speed of B = 804 km/h

Explanation:

Let the speed of aeroplane B is v, the speed of aeroplane is A is 87 km/h faster than B.

So, the speed of aeroplane A is 87 + v.

Distance traveled by A after 7 hours, d1 = (87 + v) x 7

Distance traveled by B after 7 hours, d2 = v x 7

Total distance traveled = 11865 km

So, d1 + d2 = 11865

(87 + v) x 7 + v x 7 = 11865

609 + 14 v = 11865

14 v = 11256

v = 804 km/h

So, the speed of A = 87 + 804 = 891 km/h

Speed of B = 804 km/h

Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice was so loud in the astronaut’s space helmet that it was picked up by the astronaut’s microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light (3.00×108 m/s). Give your answer in thousands of km.

Answers

Answer: 768000km

Explanation:

Velocity is given by the relation between the distance [tex]d[/tex] and the time it takes to travel that distance [tex]t[/tex]:

[tex]V=\frac{d}{t}[/tex]   (1)

In this problem we are told the time it takes for radio wave to travel from the Earth to the Moon and back is the "echo":

[tex]t=2.56s[/tex]  (2)

In addition, we know radio waves are electromagnetic waves (light), and its velocity is:

[tex]V=3(10)^{8}m/s[/tex]   (3)

Substituting (2) and (3) in (1):

[tex]3(10)^{8}m/s=\frac{d}{2.56s}[/tex]   (4)

And finding [tex]d[/tex]:

[tex]d=(3(10)^{8}m/s)(2.56s)[/tex]   (5)

Finally we can obtain the distance:

[tex]d=768000000m=768000km[/tex]  

Answer:

384,000 km

Explanation:

Given

Velocity of radio wave [tex]v = 3.00 \times 10^{8}m/s[/tex]

Duration of echo T = 2.56 s

Solution

Time taken for the radio wave to travel to moon and to travel back to earth as it was picked up by the astronaut's microphone is 2.56 s

Since any time delays in the electronic equipment can be ignored

time taken for the radio wave to reach moon

[tex]t = \frac{T}{2}\\\\t = \frac{2.56}{2} \\\\t = 1.28 s[/tex]

[tex]v = \frac{d}{t}\\\\d = vt\\\\d = 3 \times 10^8 \times 1.28\\\\d = 3.84 \times 10^8 m\\\\d = 384,000 km[/tex]

A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object moving to the left at 1.0 m/s. What is the total kinetic energy after the collision?

Answers

Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

[tex]E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2[/tex]

[tex]E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2[/tex]

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

Displacement is the slope of a velocity v. time graph.

A. True
B. False

Answers

Answer:

B. False

Explanation:

Acceleration is the slope of a velocity vs. time graph.

Displacement is the area under a velocity vs. time graph.

Which of the following best explains why snow predictions by meteorologists are sometimes incorrect?
A. Weather data are misinterpreted.

B. Weather instruments are extremely precise.

C. Interference from the sun causes data to be collected inaccurately.

D. Local variations in weather are too small for weather instruments to replicate.

Answers

Answer:

C. Interference from the sun causes data to be collected inaccurately.

Explanation:

Snow predictions by meteorologists are sometimes incorrect because from the sun causes data to be collected inaccurately.

A small charged ball lies within the hollow of a metallic spherical shell of radius R. For three situations, the net charges on the ball and shell, respectively, are (1) +4q, 0; (2) –6q, +10q; (3) +16q, –12q. Rank the situations according to their charge on (a) the inner surface of the shell and (b) the outer surface, most positive first

Answers

Answer:

Part a)

2)  >  1)  >  3)

Part b)

1)  =  2)  =  3)

Explanation:

Due to charge induction the magnitude of charge on the inner surface of the outer shell is having same charge as that of the small sphere inside but the sign of charge must be opposite.

So here we can say

1)+ 4q, 0

so inner surface has charge - 4q and outer surface charge is +4q

2) -6q , +10q

so inner surface charge is +6q, outer surface charge is +4q

3) +16q , -12q

so inner surface charge is -16q, outer surface charge is +4q

Part a)

situations in which inner surface charge is arranged in decreasing order is given as

2)  >  1)  >  3)

Part b)

Situations in which outer surface charge is arranged in decreasing order is given as

1)  =  2)  =  3)

Final answer:

Situation (1) has the most positive charge on the inner surface of the shell, while situation (2) has the most positive charge on the outer surface.

Explanation:

To rank the situations according to their charge on the inner and outer surfaces of the metallic spherical shell, we need to consider the net charges on the small charged ball and the shell. Let's analyze each situation:

For situation (1) with a net charge of +4q on the ball and 0 charge on the shell: (a) The inner surface of the shell has a charge of +4q, and (b) the outer surface has a charge of 0.For situation (2) with a net charge of -6q on the ball and +10q on the shell: (a) The inner surface of the shell has a charge of -6q, and (b) the outer surface has a charge of +10q.For situation (3) with a net charge of +16q on the ball and -12q on the shell: (a) The inner surface of the shell has a charge of +16q, and (b) the outer surface has a charge of -12q.

Therefore, ranking the situations according to the charge on the inner surface would be: (1), (3), (2) - from most positive to least positive. For the outer surface, the ranking would be: (2), (1), (3) - from most positive to least positive.

You have a pick-up truck that weighed 4,000 pounds when it was new. you are modifying it to increase its ground clearance. when you are finished

Answers

Based on the provided information, the correct statement is:

b. The bottom of your front bumper must not be more than 28 inches above the pavement.

The modification to increase ground clearance in the pick-up truck involves specific regulations for the front bumper height.

The statement "The bottom of your front bumper must not be more than 28 inches above the pavement" indicates a legal restriction to maintain a certain elevation for safety and compliance reasons.

This limitation aims to prevent potential hazards, such as underride collisions, and ensures that the modified truck adheres to regulatory standards. By specifying the maximum height of the front bumper, authorities aim to strike a balance between vehicle customization and road safety.

This regulation emphasizes the importance of maintaining a reasonable height for front bumpers to mitigate risks and uphold public safety standards, reflecting the broader considerations in vehicular modifications within the legal framework.

The probable question may be:

You have a pick-up truck that weighed 4,000 pounds when it was new. You are modifying it to increase its ground clearance. When you are finished

a. The bottom of your front bumper can be up to 32 inches above the pavement.

b. The bottom of your front bumper must not be more than 28 inches above the pavement.

c. You will no longer be required to have a rear bumper.

Question Part Points Submissions Used If an object with mass m is dropped from rest, one model for its speed v after t seconds, taking air resistance into account, is v = mg c (1 − e−ct/m) where g is the acceleration due to gravity and c is a positive constant describing air resistance. (a) Calculate lim t→∞ v.

Answers

Answer:

[tex]\lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}[/tex]

Explanation:

the velocity as a function of time is

[tex]v(t)=\frac{mg}{c}(1-e^{\frac{-ct}{m}})[/tex]

[tex]\therefore v(t)=\frac{mg}{c}(1-\frac{1}{e^{\frac{ct}{m}}})[/tex]

[tex]\therefore v(t)=\frac{mg}{c}(1-\frac{1}{e^{\frac{ct}{m}}})\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}(1-\frac{1}{\infty })\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}(1-0)\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}[/tex]

Final answer:

The limit as t approaches infinity of the speed v for an object with mass m dropped from rest with air resistance considered, is the terminal velocity mg/c. The exponential term in the equation approaches zero as time becomes very large, leading to the object reaching a constant terminal velocity.

Explanation:

To calculate the limit as t approaches infinity of the speed v for an object with mass m dropped from rest with air resistance taken into account, we use the provided equation v = mg/c (1 − e^{-ct/m}), where g is the acceleration due to gravity, which averages 9.80 m/s², and c is a positive constant representing air resistance. The limit represents the object's terminal velocity, which is the constant speed an object reaches when the force of gravity is balanced by the drag force of air resistance.

As t → ∞ (increases towards infinity), the exponential term e^{-ct/m} approaches zero. Hence, the limit of the speed v becomes:

∑ lim t→∞ v = lim t→∞ mg/c (1 − e^{-ct/m}) = mg/c.

The value mg/c is known as the terminal velocity, which is the maximum speed that the object will reach as it continues to fall.

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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:

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.

Substances A and B, initially at different temperatures, come in contact with each other and reach thermal equilibrium. The mass of substance A is twice the mass of substance B. The specific heat capacity of substance B is twice the specific heat capacity of substance A.
Which statement is true about the final temperature of the two substances once thermal equilibrium is reached?1) The final temperature will be closer to the initial temperature of substance B than substance A.2) The final temperature will be exactly midway between the initial temperatures of substances A and B.3) The final temperature will be closer to the initial temperature of substance A than substance B.

Answers

Answer: .2) The final temperature will be exactly midway between the initial temperatures of substances A and B.

Explanation:

[tex]heat_{absorbed}=heat_{released}[/tex]

As we know that,  

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]m_A\times c_A\times (T_{final}-T_A)=-[m_B\times c_B\times (T_{final}-T_2)][/tex]         .................(1)

where,  

q = heat absorbed or released

[tex]m_A[/tex] = mass of A = 2x

[tex]m_B[/tex] = mass of B = x

[tex]T_{final}[/tex] = final temperature = z

[tex]T_A[/tex] = temperature of A

[tex]T_2[/tex] = temperature of B

[tex]c_A[/tex] = specific heat capacity of A = y

[tex]c_B[/tex] = specific heat capacity of B = 2y

Now put all the given values in equation (1), we get

[tex]2x\times y\times (z-T_A)=-[x\times 2y\times (z-T_B)][/tex]

[tex]2z=T_B+T_A[/tex]

[tex]z=\frac{T_B+T_A}{2}[/tex]

Therefore, the final temperature of the mixture will be exactly midway between the initial temperatures of substances A and B.

Final answer:

When two substances with different temperatures come into contact and reach thermal equilibrium, heat flows until they reach the same final temperature. In this case, the final temperature will be closer to the initial temperature of substance B because substance B requires more heat to raise its temperature and has a smaller mass. The specific heat capacity ratio determines the proportion of heat absorbed or released by the substances.

Explanation:

When two substances at different temperatures come into contact with each other and reach thermal equilibrium, heat flows from the hotter substance to the cooler substance until they reach the same final temperature. In this case, substance A has twice the mass of substance B but substance B has twice the specific heat capacity of substance A. The specific heat capacity determines how much heat is needed to raise the temperature of a substance. Since substance B requires more heat to raise its temperature and has a smaller mass, it will be able to absorb more heat from substance A than substance A can absorb from substance B.

This means that the final temperature will be closer to the initial temperature of substance B than substance A. The specific heat capacity ratio determines the proportion of heat absorbed or released by the substances.

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A 55.6-kg skateboarder starts out with a speed of 2.44 m/s. He does 80.4 J of work on himself by pushing with his feet against the ground. In addition, friction does -244 J of work on him. In both cases, the forces doing the work are non-conservative. The final speed of the skateboarder is 7.24 m/s. (a) Calculate the change (PEf - PE0) in the gravitational potential energy. (b) How much has the vertical height of the skater changed? Give the absolute value.

Answers

Final answer:

Applying the work-energy theorem, which states that the work done by nonconservative forces equates to the change in kinetic plus potential energy, the skateboarder's change in potential energy results in -18.4J, and the absolute change in height becomes 0.032 m.

Explanation:

The student is asking about the concept of conservation of energy in physics, particularly in the context of the work-energy theorem. This theorem states that work done by nonconservative forces (like friction) is equal to the change in the kinetic and potential energy of the system. According to the given problem, the initial kinetic energy of a 55.6-kg skateboarder is KE0 = 0.5 * m * v², and after performing 80.4J of work on himself, while friction does -244J of work on him, the final kinetic energy becomes KEf = m * v² / 2 where 'v' is the final speed of 7.24 m/s.

(a) Considering that there are no conservative forces, the work-energy theorem states that Wnc = ΔKE + ΔPE. From here, we can calculate that ΔPE or PEf - PE0 results in -18.4J.

(b)The change in vertical height can be obtained by Δh = ΔPE / (m*g), where 'g' is the gravitational constant. Thus, Δh =-18.4J divided by (55.6 * 9.8), which results in -0.032 m, but since we are asked for the absolute value, the height change is 0.032 m.

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Suppose that on a hot (33.0°C) and sticky (80% humidity) afternoon in the spring, a tornado passes over the high school. If the air pressure in the lab (volume of 180.0 m³) was 1.00 atm before the storm and 0.800 atm after the storm, to what volume would the laboratory try to expand in order to make up for the large pressure difference outside?
1. 211 m3
2. 134 m3
3. 1800 m3
4. 7,150 m3

Answers

Answer:

   V₂  =  225 m^{3}

so no exact option is match with the calculated answer.

Explanation:

According to Boyle's Law, we have

                                     P_1 V_1  =  P_2 V_2    ----------- (1)

Data Given;

                  P_1  =  1.0 atm

                  V_1  =  180 m³

                  P_2  =  0.80 atm

                  V_2  =  ?

Solving equation 1 for V₂,

                  [tex]V_2= \frac{P_1 V_1}{P_1}[/tex]

Putting values,

                   [tex]V_2 = \frac{1.0*180}{0.80}[/tex]

                  [tex]V_2 = 225 m^{3}[/tex]

so no exact option is match with the calculated answer.

                 

Answer:

E

Explanation:

The force of gravity on an object varies directly with its mass. The constant of variation due to gravity is 32.2 feet per second squared. Which equation represents F, the force on an object due to gravity according to m, the object’s mass?F = 16.1mF = F = 32.2mF =

Answers

Answer:

F = 32.2m

Explanation:

The force of gravity on an object is given by:

[tex]F=mg[/tex]

where

m is the mass of the object

g is the acceleration due to gravity

Here we have:

- An object of mass m

- The acceleration of gravity is expressed as [tex]g=32.2 ft/s^2[/tex]

Therefore, substituting into the formula above, we find that the force of gravity on the object is

[tex]F=m\cdot 32.2 = 32.2m[/tex]

Answer:

F = 32.2m

Explanation:

Julie and Eric row their boat (at a constant speed) 63 miles downstream for 7 hours helped by the current. Rowing at the same rate, the trip back against the current takes 9 hours. Find the rate of the boat in still water.

Answers

Answer:

Boat speed = 8 miles/hr

Explanation:

Let the speed of the boat be U

Let the speed of the current be V.

Therefore, the downstream speed is U+V

and the upstream speed is U-V

Now we know that Speed = distance / time

Therefore, downstream speed, U+V = 63 / 7

                                                    U+V = 9 miles/hr   ------(1)

                  Upstream speed, U-V = 63 / 9

                                                U-V = 7 miles/hr      --------(2)

Therefore subtracting (2) from (1), we get

( U+V) - ( U-V ) = 9-7

2V = 2

V = 1 miles/hr

Therefore the speed of the current is V = 1 mile/hr

Now from (1) we get

U+V = 9

U+1 = 9

U = 8

Therefore, the speed of the boat is U = 8 miles/hr

A rock is suspended by a light string. When the rock is in air, the tension in the string is 43.8 N . When the rock is totally immersed in water, the tension is 31.3 N . When the rock is totally immersed in an unknown liquid, the tension is 18.3 N .

Answers

Answer:

Density of unknown liquid is [tex]2047 kg/m^3[/tex]

Explanation:

When rock is suspended in air then the weight of the rock is counter balanced by the tension force in the string

So here we have

[tex]T = mg = 43.8 N[/tex]

now when the rock is immersed in water then the tension in the string is and buoyancy force due to water is counter balanced by the weight of the object

so here we have

[tex]T_1 + F_b = mg[/tex]

[tex]31.3 + F_b = 43.8[/tex]

[tex]F_b = 43.8 - 31.3 = 12.5 N[/tex]

now we have

[tex](1000)V(9.81) = 12.5[/tex]

[tex]V = 1.27 \times 10^{-3} m^3[/tex]

now when the rock is immersed into other liquid then we have

[tex]T_2 + F_b' = mg[/tex]

[tex]18.3 + F_b' = 43.8[/tex]

[tex]F_b' = 25.5 N[/tex]

now we have

[tex]\rho(1.27 \times 10^{-3})(9.81) = 25.5[/tex]

[tex]\rho = 2047 kg/m^3[/tex]

If an object is propelled upward from a height of 128 feet at an initial velocity of 112 feet per​ second, then its height h after t seconds is given by the equation h equals negative 16 t squared plus 112 t plus 128. After how many seconds does the object hit the​ ground? Round to the nearest tenth of a second.

Answers

Explanation:

The equation of motion of an object is given by :

[tex]h(t)=-16t^2+112t+128[/tex]

Where

t is the time in seconds

We need to find the time when the object hits the ground. When the object hits the ground, h(t) = 0

So,

[tex]-16t^2+112t+128=0[/tex]

[tex]-t^2+7t+8=0[/tex]

On solving above equation using online calculator, t = 8 seconds. So, the object hit the ground after 8 seconds. Hence, this is the required solution.

Energy transformations that involve one transformation from one type of energy to another are called...............transformations.

Answers

Answer:

It is called single transformation

Answer:

Single transformation

Explanation:

Energy can neither be created nor be destroyed, it can only covert from one form to another. Sometimes, to get a work done one form of energy only require to be transformed into another form then it is known as single transformation.

For Example, cell phone transforming electrical energy into electromagnetic energy that makes it's working feasible.

Moving vehicle converting chemical energy to kinetic energy.

A rock is thrown into a still pond. The circular ripples move outward from the point of impact of the rock so that the radius of the circle formed by a ripple increases at the rate of 3 feet per minute. Find the rate at which the area is changing at the instant the radius is 4 feet. When the radius is 4 ​feet, the area is changing at approximately nothing square feet per minute. ​(Type an integer or a decimal. Round to the nearest thousandth as​ needed.)

Answers

Answer:

Rate of change of area is [tex]75.398ft^{2}/sec[/tex]

Explanation:

[tex]Area=\pi r^{2}\\\\\frac{d(Area)}{dt}=\frac{\pi r^{2}}{dt}=2\pi r\frac{dr}{dt}[/tex]

Applying values we get [tex]\frac{d(Area)}{dt}=2\pi 4\times 3=75.398ft^{2}/sec[/tex]

A gas is compressed by an adiabatic process that decreases its volume by a factor of 2.In this process, the pressurea) increases by a factor of more than 2.b) increases by a factor of 2.c) does not change.d) increases by a factor of less than 2.

Answers

Answer:

a) Increase by a factor of more than 2

Explanation:

The compression process is an adiabatic one. The opposite of this kind of process is the isothermal process, where the heat flow makes the temperature to be constant.

Let's consider this kind of process by means of an equation of state; for example, the ideal gas law:

[tex]P=\frac{RT}{v}[/tex]

Be [tex]v_{0}[/tex] the initial volume. And [tex]v_{f}=2v_{0}[/tex] the final volume.

If we consider the ideal gas law, it is evident that if the temperature remains constant (isothermal process), the pressure increases by a factor of 2; but in an adiabatic process the temperature of a gas tends to increase its temperature, so the pressure will be a higher than the resultant for the isothermal process.

A box at rest on a ramp is in equilibrium, as shown.

What is the force of static friction acting on the box? Round your answer to the nearest whole number.
_______N

What is the normal force acting on the box? Round your answer to the nearest whole number.
_______ N

Answers

Answer:

Ffs = 251 N

Fn = 691 N

Explanation:

Take the y direction to be normal to the ramp and the x direction to be parallel to the ramp.

The angle of the ramp is 20°, so the angle that the weight vector makes with the normal is also 20°.  Therefore:

Fgx = Fg sin 20°

Fgy = Fg cos 20°

Sum of the forces in the x direction (parallel to the ramp):

∑F = ma

Ffs − Fgx = 0

Ffs = Fgx

Ffs = Fg sin 20°

Ffs = 735 sin 20°

Ffs ≈ 251

Sum of the forces in the y direction (normal to the ramp):

∑F = ma

Fn − Fgy = 0

Fn = Fgy

Fn = Fg cos 20°

Fn = 735 cos 20°

Fn ≈ 691

To answer that question we need to apply equations of movement ( from Newton´s laws )

In equilibrium:

∑F  = 0         or    ∑Fx = 0     ;  ∑Fy = 0

Solution is:

a) F(sf) = 251 [N]

b) Fn = 691 [N]

From the attached drawings we can see:  ( Body free diagram)

∑ Fₓ  = F(sf)  - Pₓ  = 0           where P = m×g  = 735 [N] ( the weigth)

and Pₓ = P× cos20°

Then     F(sf)  = Pₓ × sin 20°

F(sf) = m×g×cos20°  =  735× 0.34202 [N]

F(sf) = 251.3847

F(sf) = 251 [N]

rounding to the nearest number

F(sf) = 691 [N]

∑ Fy = 0

∑ Fy = Fn - Py  = 0                     Py = P×cos20°      Py = m×g×cos20°

Py = 735×0.939693 [N]     Py =   [N]

Fn = Py = 690.674 [N]

rounding to the nearest whole number

Fn = 691 [N]

Related question  Brainly.com/question/ 11544804

easy bio A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 438 N and have a center of gravity that is 1.28 m above the floor. His upper legs weigh 144 N and have a center of gravity that is 0.760 m above the floor. Finally, his lower legs and feet together weigh 87 N and have a center of gravity that is 0.250 m above the floor. Relative to the floor, find the location of the center of gravity for his entire body.

Answers

Answer:

1.034 m above the floor

Explanation:

The location of center of body for a compound body, when the weights are given is calculated as:

[tex]\bar x = \frac{W_1x_1+W_2x_2+W_3x_3+W_4x_4+W_5x_5+.......+Wnx_n}{W_1+W_2W_3W_4W_5+.....+W_n}[/tex]

where,

[tex]\bar x[/tex] is the center of gravity of the entire body

W = weight of the individual body

x = center of gravity of the individual body

Thus on substituting the values we get,

[tex]\bar x = \frac{438\times 1.28+144\times 0.760+87\times 0.250}{438+144+87}[/tex]

or

[tex]\bar x = \frac{691.83}{669}[/tex]

or

[tex]\bar x =1.034m[/tex]

Hence, the center of gravity of the entire body lies 1.034 m above the floor

A box of negligible mass rests at the left end of a 2.00-m, 25.0-kg plank (Fig. P11.43). The width of the box is 75.0 cm, and sand is to be distributed uniformly throughout it. The center of gravity of the nonuniform plank is 50.0 cm from the right end. What mass of sand should be put into the box so that the plank balances horizontally on a fulcrum placed just below its midpoint?

Answers

Answer:

Required mass of sand is 20 kg

Explanation:

Given:

Mass of the plank = 25 kg

Distance of the Center of gravity of the Plank from the fulcrum = [tex]\frac{2}{2}-0.50 = 0.5m[/tex]

Distance of the Center of gravity of the sand box from the fulcrum = [tex]\frac{2}{2}-\frac{0.75}{2}= 0.625m[/tex]

Balancing the torque due to the plank and the sand box with respect to the fulcrum

Torque = Force × perpendicular distance

thus, we get

(25 × g) × 0.5 = weight of sand × 0.625

where, g is the acceleration due to gravity

or

(25 × g) × 0.5 = (mass of sand × g) × 0.625

or

mass of sand = 20 kg

Hence, the required mass of the sand is 20 kg

Answer:

20 Kg mass of sand should be put into the box so that the plank balances horizontally on a fulcrum placed horizontally on a fulcrum placed just below its midpoint.

Explanation:

Use the second condition of equilibrium.

[tex]$\sum} \tau=0$[/tex]

[tex]MgL-$M g x_{c m}=0$[/tex]

[tex]$M=\frac{m x_{c m}}{L}[/tex]

[tex]=\frac{25(0.50)}{0.625}[/tex]

[tex]=20 \mathrm{~kg}$[/tex]

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A sensor on a traffic light is most likely to produce electromagnetic waves at which of these frequencies?

Answers

Answer:

Option-(D): 10¹¹ waves per second.

Explanation:

Electromagnetic waves:

The electromagnetic waves is such form of a energy transfer or wave propagation through any space with or without having any medium(particles).As, the medium or particles inside a space are able to transfer the amount of energy from the origin towards the receiver, which makes it very easy for the wave propagation through a medium.Now, the electromagnetic waves are generated from the sensor on a traffic light when the frequency,f level of the wave generation is about 10¹¹ Hertz(Hz).

Answer:

answer is A

Explanation:

How to tell if something is an electrolyte

Answers

Answer: If it has ions, it is an electrolyte

Explanation:

Let's start by explaining that electrolytes are compounds that contain charged particles or ions, which can be cations (positive ions) or anions (negative ions).

So, it is this composition that makes an electrolytic material conduct electricity.

In this sense, the way to identify if a material is an electrolyte or not, is knowing whether it is composed of ions or not.

Model rocket engines are rated by their thrust force and by the impulse they provide. You can use this information to determine the time interval at which the engines fire. Two rocket engines provide the same impulse. The first engine provides 6 N of thrust for 2 s; the second provides 4 N of thrust.

For how long does this second engine fire?

Answers

Answer: 3 seconds

Explanation:

Since they provide the same impulse,

Impulse= Ft

F1 = 6N

t1 = 2s

F2= 4N

t2= ?

F1= Force of first engine

t1= time elapsed by first engine

F2= force of second engine

t2= time elapsed by second engine

F1t1 = F2t2

6 × 2 = 4t2

t2= 12/4

t2 = 3 seconds

This second engine fires for 3 s

[tex]\texttt{ }[/tex]

Further explanation

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

thrust of first engine = F₁ = 6 N

elapsed time of first engine = t₁ = 2 s

thrust of second engine = F₂ = 4 N

Asked:

elapsed time of second engine = t₂ = 2 s

Solution:

We will use Newton's Law of Motion to solve this problem as follows:

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

[tex]\Sigma F = m \Delta v \div t[/tex]

[tex]\Sigma F = I \div t[/tex]

[tex]\boxed {I = \Sigma F \times t}[/tex] → Impulse Formula

[tex]\texttt{ }[/tex]

Two rocket engines provide the same impulse :

[tex]I_1 = I_2[/tex]

[tex]\Sigma F_1 \times t_1 = \Sigma F_2 \times t_2[/tex]

[tex]6 \times 2 = 4 \times t_2[/tex]

[tex]12 = 4 \times t_2[/tex]

[tex]t_2 = 12 \div 4[/tex]

[tex]\boxed {t_2 = 3 \texttt{ s}}[/tex]

[tex]\texttt{ }[/tex]

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[tex]\texttt{ }[/tex]

Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

A small frictionless cart is attached to a wall by a spring. It is pulled 14 cm from its rest​ position, released at time tequals​0, and allowed to roll back and forth for 5 seconds. Its position at time t is s equals 14 cosine left parenthesis pi t right parenthesis. a. What is the​ cart's maximum​ speed? When is the cart moving that​ fast? Where is it​ then? What is the magnitude of the acceleration​ then? b. Where is the cart when the magnitude of the acceleration is​ greatest? What is the​ cart's speed​ then?

Answers

Answer:given below

Explanation:

Cart is pulled 14 cm from mean position

and its position is given by

[tex]x=14cos\left ( \pi t\right )[/tex]

therefore its velocity is acceleration is given by

[tex]v=-14\pi sin\left ( \pi t\right )[/tex]

[tex]a=-14\pi ^2cos\left ( \pi t\right )[/tex]

[tex]\left ( a\right ) cart\ max.\ speed\ is[/tex]

[tex]v_{max}=14\pi at\ t=0.5sec[/tex]

and its position is x=0

acceleration at t=0.5sec

a=0

[tex]\left ( b\right )[/tex]

[tex]a_{max}=14\pi ^2 cm/s^2[/tex]

at t=0,1,2 sec

at t=0

x=14 cm

v at t=0

v=0 cm/s

for t=1 sec

x=-14 cm i.e. 14 cm behind mean position

v=0 m/s

Final answer:

The cart's maximum speed is 14π m/s and occurs at the equilibrium point in its oscillation at integer multiples of the half period. At these moments, the magnitude of the acceleration is 14π² m/s². The magnitude of the acceleration is greatest when the cart is at the extreme points of its oscillation, where its speed is zero.

Explanation:

The cart, undergoing simple harmonic motion, has a maximum speed when it passes through equilibrium - the midpoint of its oscillating path. From the equation for the straightforward harmonic motion, we know that the speed v of the cart is given by the derivative of s(t) = 14cos(πt). The derivative of this function is v(t) = -14πsin(πt), and the maximum speed occurs when sin(πt) = ±1, which gives a maximum speed of ±14π m/s.

The maximum speed is attained at integer multiples of the half period (1/2, 3/2, 5/2 seconds, and so on). At this moment, the cart is at the equilibrium position, s=0. Acceleration a(t) is given by the second derivative of the position function s(t), which is a(t) = -14π²cos(πt). The magnitude of the acceleration at the time of maximum speed is 14π² m/s².

In the case of the maximum magnitude acceleration, this is attained at the turning points of the oscillation when cos(πt) = ±1. At these moments, the speed of the cart is zero.

Learn more about Simple Harmonic Motion here:

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