A car has a mass of 850 kg. By pushing on the car, Evan increases its speed
from 3.5 m/s to 5 m/s. What impulse did Evan apply to the car?

A. 4250 kg•m/s
B. 1275 kg•m/s
C. 850 kg•m/s
D. 2975 kg•m/s

Answers

Answer 1

Answer:

B. 1275 kg*m/s

Explanation:

I = F(deltaT) = (deltaP) = mv2- mv1

Therefore,

I = mv2-mv1

m = 850 kg

v2 = 5 m/s

v1 = 3.5 m/s

I = (850)(5)-(850)(3.5)

I = 1275 kg* m/s


Related Questions

A uniform electric field has a magnitude of 10 N/C and is directed upward. A charge brought into the field experiences a force of 50 N downward. The charge must be:_______.

Answers

Answer:

q = 5 C

Explanation:

The electric field is defined as the force experienced by a unit charge when it is brought into the field. Hence, the formula used to find the electrical field is given as follows:

E = F/q

where,

E = Electric Field Magnitude = 10 N/C

F = Force Experienced by the test charge = 50 N

q = Magnitude of the Charge = ?

Therefore,

10 N/C = 50 N/q

q = 50 N/(10 N/C)

Therefore,

q = 5 C


Find the mass of an object on planet F if its weight is 650 N (g = 13m/s^2)

Answers

Answer:

the object's mass is 50 kg

Explanation:

We use Newton's second law to solve for the mass:

F = m * a , then   m = F / a

In our case, the acceleration is the gravitational acceleration on the planet, and the force is the weight of the object on the planet. So we get:

m = w / a = 650 N / 13 m/s^2 = 50 kg

Then, the object's mass is 50 kg.

A current of 3.75 A in a long, straight wire produces a magnetic field of 2.61 μT at a certain distance from the wire. Find this distance.

Answers

Given :

Current, I = 3.75 A .

Magnetic Field, [tex]B = 2.61\times 10^{-4}\ T[/tex]

To Find :

The distance from the wire.

Solution :

We know,

[tex]B = K\dfrac{2i}{d}\\\\d = 10^{-7}\times \dfrac{2\times 3.75}{2.61\times 10^{-4}}\\\\d = 0.00287\ m \\\\d = 2.87\times 10^{-3}\ m[/tex]

Hence, this is the required solution.

A 36.3 kg cart has a velocity of 3 m/s. How much kinetic energy does the object have?

Answers

Answer:

163.35

__________________________________________________________

We are given:

Mass of the object (m) = 36.3 kg

Velocity of the object (v) = 3 m/s

Kinetic Energy of the object:

We know that:

Kinetic Energy = 1/2(mv²)

KE = 1/2(36.3)(3)²            [replacing the variables with the given values]

KE = 18.15 * 9

KE = 163.35 Joules

Hence, the cart has a Kinetic Energy of 163.35 Joules

An extraterrestrial creature is standing in front of plane mirror. The height of this creature is H and we know that this creature has eyes positioned h below the top of its head. This creature sees its reflection which fit exactly the mirror, it means, this creature can just see the top of head and the bottom of its feet (or whatever it uses for motion). We can conclude that the top of a mirror is exactly:________

a. H/2 above the ground
b. H above the ground
c. (H-h/2) above the ground
d. (H-h) above the ground
e. We can not guess anything without information about the nature of this creature.

Answers

Answer:

c. (H-h/2) above the ground

Explanation:

The mirror must be at least half as tall as the alien, and its base must be located at half of the distance between the alien's eyes and the ground (assuming that the alien doesn't float or levitate).

This question is about the Law of Reflection which states that the angle of reflection = angle of incidence.

I attached an image that can help you understand the concept, although the alien is not included.

Which statement best describes a characteristic of gases?

Gases can be compressed, or squeezed together.
The particles of gases are packed close together.
The particles of gases spread our vertically instead of horizontally.
Gases have a definite shape and volume.

Answers

Answer:

A

Explanation:

I'm pretty sure it's A. That's the one that makes the most sense and checks out

Answer:

Just here to confirm that it is A

Explanation:

Sona wants to install room heater in her living room. She had only two options, either to install heater at the top of the window or near the ground level. . Which method of installing room heater would be the effective way . Why or Why not ?

Answers

Installing the heater near ground level is the better option because heat rises so putting it near the bottom will allow the warmth to rise and warm up more of the air in the room. Especially if it’s below the window because it will heat the cool air seeping through the window.

if you increase the frequency of a wave by 5x whats it’s period?

Answers

We know that Period of a wave is the inverse of its Frequency

So,  Period = 1 / Frequency

From the above, we can say that Period is inversely proportional to Frequency and hence, any change in Frequency will be the inverse change in the period

Therefore, we can say that if the frequency is increased by 5 times, the period will increase by 1/5 times

The speed of a space shuttle is 10 / express this in /�

Answers

Answer:

268.22m/s

Explanation:

Given;

    10mile/min to m/s

We need to convert between the two units;

    1 mile  = 1609.34m

     60s  = 1min

Now;

    10 x [tex]\frac{mile}{min}[/tex] x [tex]\frac{1min}{60s}[/tex] x [tex]\frac{1609.34m}{1mile}[/tex]

  = 268.22m/s

I NEED THIS ASAP!!
Which formula defines the unit for electrical power?

Answers

C option
Power = voltage x current

Answer:

1 W = 1 V x 1 A

Explanation:

Other dude is wrong this is right

what is gathering and analyzing information about an object without physical contact with the object

Answers

Answer:

Remote Sensing

Explanation:

A student throws a 110 g snowball at 6.5 m/s at the side of the schoolhouse, where it hits and sticks.
What is the magnitude of the average force on the wall if the duration of the collision is 0.19 s ?

Answers

Answer:

3.6N

Explanation:

Impulse is the change of momentum of an object when the object is acted upon by a force for an interval of time.

the expression is Ft=mv

where F= force

 m= mass

t= time

v= velocity

Step one:

given data

mass m= 110g= 0.11kg

velocity v= 6.5m/s

time t= 0.19seconds

Step two:

we also know that the force on impulse is given as

Ft=mv

F=0.11*6.5/0.19

F=0.715/0.19

F=3.76N

The magnitude of the average force on the wall if the duration of the collision is 0.19 is 14N

A medicine ball has a mass of 5kg and is thrown with a speed of 3 m/sec what is it's kinetic energy

Answers

KE = 1/2mv^2

KE = 1/2 (5kg)(3m/s)

KE = 22.5 J

If a ball rolls across a table to the left with constant speed, which of the following is true about the force(s) on the ball? (3a1)

Question 10 options:

There cannot be any forces applied to the ball.


There must be exactly one force applied to the ball.


The net force applied to the ball is zero.


The net force applied to the ball is directed to the right.

Answers

Answer:

C. The net force applied to the ball is zero.

Explanation:

From Newton's second law of motion;

F = ma

Where F is the force on an object, m is its mass and a is its acceleration.

Therefore, the force on an object is a product of its mass and acceleration as it travel from one point to another.

Since acceleration relates to the rate of change in velocity to time. Then when the object moves at uniform velocity (especially along a straight path), its acceleration is zero.

So that;

F = m x 0

  = 0

No force is applied on the object.

Therefore for the ball in the given question, the net force applied to the ball is zero because it rolls with constant speed along a straight path.

The propeller of an aircraft accelerates from rest with an angular acceleration α = 7t + 8, where α is in rad/s2 and t is in seconds. What is the angle in radians through which the propeller rotates from t = 1.00 s to t = 6.10 s?

Answers

Answer:

The value  is  [tex]\theta =407.3 \ radian[/tex]

Explanation:

From the question we are told that

    The angular acceleration is  [tex]\alpha = (7t + 8) \ rad/ s^2[/tex]

    The first time is  [tex]t_1 = 1.00 \ s[/tex]

    The second time [tex]t_2 = 6.10 \ s[/tex]

Generally the angular velocity is mathematically represented as

     [tex]w = \int\limits {\alpha } \, dt[/tex]

=>  [tex]w = \int\limits {7t + 8 } \, dt[/tex]

=>  [tex]w =\frac{ 7t^2}{2} + 8 t[/tex]

Generally the angular displacement  is mathematically represented as

[tex]\theta = \int\limits^{t_2}_{t_1} { w} \, dt[/tex]

=>  [tex]\theta = \int\limits^{t_2}_{t_1} { \frac{7t^2}{2} + 8t } \, dt[/tex]

=>  [tex]\theta = { \frac{7t^3}{6} + \frac{8t^2}{2} } | \left \ t_2} \atop {t_1}} \right.[/tex]

=> [tex]\theta = { \frac{7t^3}{6} + 4t^2} } | \left \ 6.10} \atop {1}} \right.[/tex]

=> [tex]\theta =[ { \frac{7}{6}[6.10 ]^3 + 4[6.10]^2} } ] -[ { \frac{7}{6}[1 ]^3 + 4[1]^2} } ][/tex]

=> [tex]\theta =407.3 \ radian[/tex]

We will now determine the indexes of refraction for two Mystery materials, A and B. These materials can be selected from the list of materials on the right. Be sure to set your laser pointer to a frequency of 589 nm. Questions:A. Devise an experiment for determining the indices of refraction for these. Explain your methodology. B. What are the indices of refraction for the two mystery materials, A and B?

Answers

Answer:

A) refraction experiment   n = n₁ sin θ₁ / sin θ₂

B)  n_A = 1.19 ,    n_B = 1.53

Explanation:

A) This exercise is a method to measure the refractive index of materials, a very useful and simple procedure is to create a plate of known thickness from each material, place the material on a paper with angle measurements (protractor), incline the laser beam and measure the angles of incidence and refraction (within the material), repeat for about three different angles of incidence and use the equation of refraction to determine the index

            n₁ sin θ₁ = n₂ sinθ₂

            n₂ = n₁ sin θ₁₁ /sin θ₂

If the medium surrounding the plate is air, its refractive index is n₁ = 1, the final expression is

            n = n₁ sin θ₁ / sin θ₂

B) For this part, no data are given in the exercise, but we can take 50º as the angle of incidence and measure the angle of refraction. Suppose it is 40º for material A and 30º for material B, the refractive index would be

         

material A

             n_A = sin 50 / sin 40

             n_A = 1.19

material B

              n_B = sin 50 / sin30

              n_B = 1.53

On a surface of a planet of radius R and mass M the acceleration due to gravity is 7m/s?. Consider another planet of radius 2R and mass 0.4M. What would the acceleration due to gravity be on this new planet? Show your calculations.

Answers

Answer:

0.7 m/[tex]s^{2}[/tex]

Explanation:

From Newton's law of universal gravitation,

F = [tex]\frac{GMm}{r^{2} }[/tex]

and from Newton's second law of motion,

F = mg

So that;

mg = [tex]\frac{GMm}{r^{2} }[/tex]

⇒ g = [tex]\frac{GM}{r^{2} }[/tex]

For the first planet,

7 = [tex]\frac{GM}{R^{2} }[/tex]

⇒ G = [tex]\frac{7R^{2} }{M}[/tex] .............. 1

For the second planet,

g = [tex]\frac{0.4GM}{(2R)^{2} }[/tex]

   = [tex]\frac{0.4GM}{4R^{2} }[/tex]

⇒ G = [tex]\frac{4gR^{2} }{0.4M}[/tex] ............. 2

Equating 1 and 2, we have;

[tex]\frac{7R^{2} }{M}[/tex] = [tex]\frac{4gR^{2} }{0.4M}[/tex]

g = [tex]\frac{7R^{2} *0.4M}{4R^{2}M }[/tex]

  = [tex]\frac{7*0.4}{4}[/tex]

  = [tex]\frac{2.8}{4}[/tex]

g = 0.7

Therefore, the acceleration due to gravity on the new planet is 0.7 m/[tex]s^{2}[/tex].

When two RF signals on the same frequency arrive at a receiver at the exact same time and their peaks and valleys are in alignment, what is true about these signals?

Answers

Answer:

The two RF signals are in phase.

Explanation:

A wave is a disturbance that travels through a medium which transfers energy from one point to another in the medium without causing any permanent displacement of the particles of the medium.

Characteristics of waves include frequency, wavelength, velocity, etc.

Two types of waves are longitudinal and transverse wave. Radio Frequency (RF) signals travel in the form of transverse waves which have regions of maximum and minimum displacements called crests and troughs.

Travelling waves with  the same frequency may be said to be in phase or out of phase depending on whether their crests/peaks or troughs/valleys are reached at the same instant of time.

When two RF signals on the same frequency arrive at a receiver at the exact same time and their peaks and valleys are in alignment or in step, they are said to be in phase.

The phase of a wave involves the relationship between the position of the amplitude peaks and valleys of two waveforms.

Any five physics problems

Answers

Explanation:

There are still some questions beyond the Standard Model of physics, such as the strong CP problem, neutrino mass, matter–antimatter asymmetry, and the nature of dark matter and dark energy.

What percentage of an iron anchor’s weight will be supported by buoyant force when submerged in salt water?

Answers

Answer:

0.87

Explanation:

To solve this, we use the principle of Archimedes. Archimedes Principal of flotation states that "the buoyant force of an object is equal to the total weight of the fluid it displaces."

In the attachment, I stated the mathemacal formula, of which

F(B) = The buoyant force

w(fl) = The weight of the salt water displaced

p(iron) = density of iron

p(salt) = density of the salt water = 1025 kg/m³

F' = weight of the iron in air

F = weight of the iron in salt water

p(man) = density of man = 7680 kg/m³

The rest are the easy calculations done by substituting the values

A boxer is punching the heavy bag. The impact of the glove with the bag is 0.10s. The mass of the glove and his hand is 3kg. The velocity of the glove just before impact is 25m/s. What is the average impact force exerted on the glove?

Answers

Answer:

750N

Explanation:

Impulse is the change of momentum of an object when the object is acted upon by a force for an interval of time.

the expression is Ft=mv

where F= force

           m= mass

             t= time

            v= velocity

Step one:

given data

mass m=3kg

velocity v= 25m/s

time t= 0.10seconds

Step two:

we also know that the force on impulse is given as

Ft=mv

F=3*25/0.10

F=75/0.10

F=750N

The magnitude of the average force on the heavy bag if the duration of the collision is 0.1 is 750N

In 1993, Wayne Brian threw a spear at a record distance of 201.24 m. (This is not an official sports record because a special device was used to “elongate” Brian’s hand.) Suppose Brian threw the spear at a 35.0° angle with respect to the horizontal. What was the initial speed of the spear? 2. Find the maximum height and time of flight of the spear in problem #1.

I really don't know how to do any of this please help me :(

Answers

Answer:

V₀ = 45.81 m/s

H = 70.45 m

T = 5.36 s

Explanation:

The motion of the spare is projectile motion. Therefore, we will first use the formula of range of projectile:

R = V₀² Sin 2θ/g

where,

R = Range of Projectile = 201.24 m

V₀ = Initial Speed = ?

θ = Launch Angle = 35°

g = 9.8 m/s²

Therefore,

201.24 m = V₀²[Sin 2(35°)]/9.8 m/s²

V₀ = √[(201.24 m)/(0.095 m/s²)

V₀ = 45.81 m/s

Now, for maximum height:

H = V₀² Sin² θ/g

H = (45.81 m/s)² Sin² 35°/9.8 m/s²

H = 70.45 m

For the total time of flight:

T = 2 V₀ Sin θ/g

T = 2(45.81 m/s) Sin 35°/9.8 m/s²

T = 5.36 s

Your teacher placed a 3.5 kg block at the position marked with a “ + ” (horizontally, 0.5 m from the origin) on a large incline outlined on the graph below and let it slide, starting from rest. ***There are two images included!***

Answers

Answer:

 x = 10.75 m

Explanation:

For this problem we will solve it in two parts, the first using energy and the second with kinematics

Let's use the energy work relationship to find the velocity of the block as it exits the ramp

       W = [tex]Em_{f}[/tex] - Em₀

Starting point. Higher

       Em₀ = U = m g h

the height from the edge of the ramp of the graph has a value

        h = 9-3 = 6 m

Final point. At the bottom of the ramp

       Em_{f} = K = ½ m v²

Friction force work

      W = - fr  d

The friction force has the formula

      fr = μ N

 

On the ramp, we can use Newton's second law

         N - W cos θ = 0

         N = W cos θ

where the angle is obtained from the graph

         tan θ = (9-3) / (0.5-4) = -6 / 3.5

         θ = tan⁻¹ (-1,714)

         θ = -59.7º

the distance d is

         d = √ (Δx² + Δy²)

         d = √ [(0.5-4)² + (9-3)²]

         d = 6.95 m

for which the work is

       W = - μ mg cos 59.7 d

we substitute

        W = Em_{f} -Em₀

        - μ mg cos 59.7 d = ½ m v² - m g h

In the graph o text the value of the friction coefficient is not observed, suppose that it is μvery = 0.2

        - μ g cos 59.7 d = ½ v² - g h

         v² = 2g (h - very d coss 59.7)

let's calculate

         v² = 2 9.8 (6 - 0.2 6.95 cos 59.7)

         v = √ 103.8546

         v = 10.19 m / s

in the same direction as the ramp

in the second part we use projectile launch kinematics

       

let's look for the components of velocity

         v₀ₓ = vo cos -59.7

         [tex]v_{oy}[/tex] = vo sin (-59,7)

         v₀ₓ = 10.19 cos (-59.7) = 5.14 m / s

         v_{oy} = 10.19 if (-59.7) = -8.798 m / s

Let's find the time to get to the floor (y = o)

          y = y₀ + v_{oy} t - ½ g t²

to de groph y₀=3 m

          0 = 3 - 8.798 t - ½ 9.8 t²

          t² - 1.796 t - 0.612 = 0

we solve the quadratic equation

          t = [1.796 ±√(1.796² + 4 0.612)] / 2

          t = [1,795 ± 2,382] / 2

          t₁ = 2.09 s

          t₂ = -0.29 s

since time must be a positive quantity the correct value is t = 2.09 s

we calculate the horizontal displacement

          x = v₀ₓ t

          x = 5.14 2.09

          x = 10.75 m

The motion of the box, after it exits the incline is the motion and trajectory

of a projectile.

Horizontal distance from the right-hand edge of the incline to the point of

contact with the floor is approximately 1.24613 m.

Reasons:

Mass of the block,  m = 3.5 kg

Coefficient of kinetic friction, μ = 1.2

Location of the = 0.5 m from the origin

Required:

Horizontal distance between the block's point of contact with the floor and

the bottom right-hand edge of the incline.

Solution:

Let θ represent the angle the incline make with the horizontal.

The normal reaction of the incline on the block, [tex]F_N[/tex] = m·g·cos(θ)

Work done on friction = [tex]F_N[/tex]×μ×Length of the incline, L

Rise of the incline = 10 - 3 = 7

Run of the incline = 4

L = √(6.125² + 3.5²) = [tex]\dfrac{7 \times \sqrt{65} }{8}[/tex]

Let ΔP.E.₁  represent the potential energy transferred to kinetic energy

and work along the incline, we have;

Energy of the block at the bottom of the incline, M.E.₂, is found as follows;

K.E.₂ = mgh - m·g·μ·cos(θ)·L

[tex]K.E. =\frac{1}{2} \times 3.5 \times v^2 = 3.5 \times 9.81 \times 6.125 - 3.5 \times 9.81 \times 1.2 \times \dfrac{4}{\sqrt{65} } \times \dfrac{7 \times \sqrt{65} }{8}[/tex]

v ≈ 6.1456 m/s

The vertical component of the velocity is therefore;

[tex]v_y = v \cdot sin(\theta)[/tex]

[tex]v_y = 6.1456 \times \dfrac{7}{\sqrt{65} } \approx 5.33588[/tex]

From the equation, h = u·t + 0.5·g·t² derived from Newton's Laws of motion, we have;

ΔP.E.₁ = 3.5×9.81×7

3 = 5.33588·t + 0.5×9.81·t²

Factorizing, the above quadratic equation, we get;

The time it takes the block to reach the floor, t ≈ 0.40869 seconds

Horizontal component of the velocity is [tex]v_x \approx 6.1456 \times \dfrac{4}{\sqrt{65} } \approx 3.04908[/tex]

The horizontal distance, x = vₓ × t

∴ x = 3.04908 × 0.40869 ≈ 1.08194

Horizontal distance from the right-hand edge of the incline to the point of

contact with the floor, x ≈ 1.24613 m.

Learn more here:

https://brainly.com/question/24888457

A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2?

A 0.31 s
B 0.56 s
C 4.3s
D 70s

Answers

Answer:

C. 4.3 seconds

Explanation:

B 0.56 s is the time period of a twirlers baton.

What is Centripetal Acceleration?

Centripetal acceleration is defined as the property of the motion of an object which traversing a circular path.

Any object that is moving in a circle and has an acceleration vector pointed towards the center of that circle is known as Centripetal acceleration.

The centripetal acceleration is given by:

a = 4π²R/T²

Given values are:

a = 47.8 m/s²

D = 0.76 m  so , R = 0.76/2 = 0.38m

Using this formula,

47.8*T² = 4π² x0.38

T² = [tex]\frac{4*3.14^2*0.38}{47.8}[/tex]

T = 0.56 s

Therefore,

A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2 which have  time period of 0.56 s.

Learn more about Centripetal acceleration here:

https://brainly.com/question/14465119

#SPJ5

.

Select the correct answer.
The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car's motion?

A. It's not moving.

B.It's moving at a constant speed.

C.It's moving at a constant velocity

D.It's speeding up.

Answers

Answer:

It isn't moving

Explanation:

Having established that a sound wave corresponds to pressure fluctuations in the medium, what can you conclude about the direction in which such pressure fluctuations travel?A) The direction of motion of pressure fluctuations is independent of the direction of motion of the sound wave.B) Pressure fluctuations travel perpendicularly to the direction of propagation of the sound wave.C) Pressure fluctuations travel along the direction of propagation of the sound wave.D) Propagation of energy that passes through empty spaces between the particles that comprise the mediumDoes air play a role in the propagation of the human voice from one end of a lecture hall to the other?a) yesb) no

Answers

Answer:

None of them: the direction of the pressure fluctuations is parallel to the direction of motion of the wave

Explanation:

What is the volume of a box if he has Length=7 cm Width=5cm , Height=10cm ?

Answers

Answer:

Volume of Cuboid = Height*Width*Length

Explanation:

Volume of Cuboid = 10*5*7

= 350 cu² cm

Answer:

Diagram:-

[tex]\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(5.6,5.4){\bf A}\put(11.1,5.4){\bf B}\put(11.2,9){\bf C}\put(5.3,8.6){\bf D}\put(3.3,10.2){\bf E}\put(3.3,7){\bf F}\put(9.25,10.35){\bf H}\put(9.35,7.35){\bf G}\put(3.5,6.1){\sf 5cm}\put(7.7,6.3){\sf 7cm}\put(11.3,7.45){\sf 10cm}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}[/tex]

Required Answer:-

It is a cuboid

where

length =l=7cmwidth=b=5cmheight =h=10cm

As we know that in a cuboid

[tex]{\boxed{\sf Volume=lbh}}[/tex]

Substitute the values

[tex]{:}\longrightarrow[/tex][tex]\sf Volume =7×5×10 [/tex]

[tex]{:}\longrightarrow[/tex][tex]\sf Volume=35×10 [/tex]

[tex]{:}\longrightarrow[/tex][tex]\sf Volume=350cm^3 [/tex]

The components of lifetime fitness include all of the following components except

Answers

Answer:it’s A

Explanation:

because i took the quiz

Answer:

D is the correct answer, not A

Explanation:


A material that provides resistance to the flow of electric current is called a(n):

circuit

conductor

insulator

resistor

Answers

Answer:

it's an insulator

Explanation:

Insulators provides resistance

Answer:

C. insulator

Explanation:

explain the relationship among visible light, the electromagnetic spectrum, and sight.

Answers

Explanation:

The electromagnetic spectrum is the name given to the full range of frequencies and/or wavelengths that electromagnetic phenomena may have.

Human eyes respond to a small range of wavelengths in that spectrum. That response is called sight. Because humans can see that electromagnetic energy, it is called visible light.

Other Questions
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