which phenomenon supports the particular model of light?​

The farmer uses this technique to protect the produce from freezing because as the water in the tubs begins to freeze, it releases heat. This heat can help to keep the surrounding air in the shed above freezing, protecting the produce from freezing as well. Additionally, as the water in the tubs freezes, it releases latent heat of fusion, which is the energy required to change a substance from a solid to a liquid. This energy is also released in the form of heat, further helping to keep the produce from freezing.

Answer:

A. the motion of an object along a curved path

Answer:

The wavelength of a wave with the frequency of 330hz and a speed of 343m/s would be 1.04m

Explanation:

You can get the wavelength of a wave by dividing the speed of the wave by its frequency, which in this case would be:

343/300, which as a decimal number, it'd be 1.04.

I hope I helped you, and a "Brainliest" is always appreciated! ☺

The factors that affect the frequency at which a string or stringed instrument vibrates are:

• The length of the string

• The tension in the string

The frequency of vibration of a string is given by the formula:

Where:

T represents the tension of the string

l represents the length of the string

μ represents the mass per unit length

From the formula written above:

The frequency of vibration of the string or stringed instruments depends on:

• The length of the string

• The tension in the string

Answer:

Force/Area in fact Pascal is defined as Newton/m^2

The approximate circumference of circle C is 62.8 mm. Where the value of π is 3.14.

Given that PQ is the diameter of the circle and its length is 20 mm

The circumference of a circle can be calculated using the formula:

Circumference (C) = π × Diameter

Circumference (C) = 3.14 × 20

Circumference (C) = 62.8 mm

So, the approximate circumference of circle C is 62.8 mm.

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The complete question is:

What is the approximate circumference of circle c where PQ is the diameter of the circle (PQ = 20mm)? (use π = 3.14).

Answer:

The frequency of light can be calculated using the formula:

`c = λv`

Where `c` is the speed of light in a vacuum, `λ` is the wavelength of light, and `v` is the frequency of light.

The speed of light in a vacuum is `3.00 × 10^8 m/s`.

To convert the wavelength from nanometers to meters, we need to divide by `1 × 10^9`.

Thus, the frequency of light with a wavelength of 655 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(655 × 10^-9 m)`

`v = 4.58 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 655 nm is `4.58 × 10^14 Hz`.

Similarly, the frequency of light with a wavelength of 515 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(515 × 10^-9 m)`

`v = 5.83 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 515 nm is `5.83 × 10^14 Hz`.

Finally, the frequency of light with a wavelength of 475 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(475 × 10^-9 m)`

`v = 6.32 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 475 nm is `6.32 × 10^14 Hz`.

So, the frequency of light with a wavelength of 655 nm is `4.58 × 10^14 Hz`, the frequency of light with a wavelength of 515 nm is `5.83 × 10^14 Hz` and the frequency of light with a wavelength of 475 nm is `6.32 × 10^14 Hz`.

The kinetic energy of the block  when it reaches the bottom is 39.2 J.

Apply the principle of conservation of energy.

K.E(bottom) = P.E(top)

P.E(top) = mgh

where;

P.E(top) = 5 x 9.8 x 0.8

P.E(top) = K.E(bottom) = 39.2 J

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Answer: you have to divide and multiple fam

Explanation:

5.48 V is the potential drop across the cable for a copper transmission cable of 50.0 km long and 10.0 cm in diameter carries a current of 105 A

Ohm's Law states that the potential drop is determined by the equation: V = IR, where I is the current and R is the wire resistance.
R=PL/A
Under the assumption that all physical parameters and temperatures remain constant, Ohm's law asserts that the voltage across a conductor is directly proportional to the current flowing through it.

Only when the given temperature and the other physical variables remain constant does Ohm's law apply. Increasing the current causes the temperature to rise in some components. The filament of a light bulb serves as an illustration of this, where the temperature increases as the current increases. Ohm's law cannot be applied in this situation. The filament of the lightbulb defies Ohm's Law.

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A)  The resistance is 2000 ohm. B) 6 times the current through the muscle. C) The voltage is 1 volt. D) The resistance is 1000 ohms.

A)

r = resistance of muscle

R = resistance of fat

given that: R = 3r

Rtotal = total resistance = R × r /(R + r) = 3r2 /4r= 0.75 r

V = potential difference = 1 Volts

I = current = 1 mA = 0.001 A

Using Ohm's law:

V = I × R(total)

0.5 = (0.001) × (0.75r)

r = 666.67

R = 3r = 3 x 666.67 = 2000

Hence, The resistance is 2000 ohm.

b)

1/6 times the current through the muscle.

since muscle and fat are in parallel, they have the same voltage across each, hence

i(muscle) × R(muscle) = i(fat) × R(fat)

i(muscle) × R(muscle) = i(fat) × 6 × R(muscle)

i(fat) = (1/6) i(muscle)

Therefore, 6 times the current through the muscle.

c)

since fat and muscle are in parallel

Hence, V(fat) = 1 Volts

Hence, The voltage is 1 volt.

d)

R(arm) = arm resistance = 750

R(fat) = fat resistance

R(muscle) = muscle resistance

given that

R(fat) = 3 R(muscle)

since fat and muscle are in parallel their combination is given as

R(arm) = R(fat) ×  R(muscle) ÷ (R(fat) + R(muscle) )

750 = (3 R(muscle) )R(muscle) ÷ ((3 R(muscle))+ R(muscle) )

750 = 3 R(muscle) /4

R(muscle) = 1000

Therefore, The resistance is 1000 ohms.

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Answer:

The speed of light traveling through a vacuum is exactly 299,792,458 meters (983,571,056 feet) per second. That's about 186,282 miles per second — a universal constant known in equations as "c," or light speed.

\(s\frac{d}{t}\)

Explanation:

hope this helps you my friend

Answer:

no se pregúntale a otro gracias cuidate :)

The bullet is in the barrel for 4.5610^-5 s, experiences a force of 16713 N, an impulse of 0.763 Ns, and has a momentum of mv = 34.5 kg*m/s as it leaves the barrel.

What are the time, force, impulse, and momentum of a bullet fired from a gun with given mass, barrel length, and muzzle velocity

(a) The time the bullet spends in the barrel can be found using the equation for distance traveled by an object with constant acceleration, where the initial velocity is zero and the distance is the barrel length:

d = 1/2 at^2

Solving for time t, we get:

t = √(2d/a)

where a is the acceleration of the bullet as it is propelled out of the barrel. To find the acceleration, we can use the equation for force:

F = ma

where F is the force exerted on the bullet, m is the mass of the bullet, and a is its acceleration. Since the bullet is in the barrel, the force is the force of the expanding gases from the gunpowder. We can find this force using the ideal gas law:

PV = nRT

where P is the pressure of the gases, V is the volume of the barrel, n is the number of moles of gas produced by the gunpowder, R is the ideal gas constant, and T is the temperature of the gases. Solving for P, we get:

P = nRT/V

We can assume that the volume of the barrel remains constant during the firing of the bullet, so we can write:

P = k nT

where k is a constant that depends on the volume of the barrel and the gas constant R. Since the bullet is fired in a fraction of a second, we can assume that the temperature of the gases remains constant, so we have:

P = k n

The force exerted on the bullet is then:

F = PA = k nA

where A is the cross-sectional area of the barrel.

a = F/m = k nA/m = k P A/m = (kRT/V)(πr^2)/m

where r is the radius of the barrel. we get:

a = (k)(8.31 J/molK)(293 K)/(0.00066 m^3)(π(0.0033 m)^2)/(0.075 kg) = 2.4110^7 m/s^2

Plugging this into the equation for time, we get:

t = √(2d/a) = √(20.66 m/2.4110^7 m/s^2) = 4.56*10^-5 s

So the bullet is in the barrel for 4.56*10^-5 s.

(b) The force on the bullet while it is in the barrel is the force exerted by the expanding gases from the gunpowder, which we found to be:

F = k nA

Substituting in the given values, we get:

F = (k)(8.31 J/mol*K)(293 K)/(0.00066 m^3)(π(0.0033 m)^2) = 16713 N

So the force on the bullet while it is in the barrel is 16713 N.

(c) The impulse exerted on the bullet while it is in the barrel is the product of the force and the time:

J = FΔt = (16713 N)(4.5610^-5 s) = 0.763 Ns

So the impulse exerted on the bullet while it is in the barrel is 0.763 N*s.

(d) The momentum of the bullet as it leaves the barrel can be found using the equation:

p = mv

where p is the momentum, m is the mass of the bullet, and v is its velocity. We can find the velocity using conservation of energy, since the gun and bullet

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The cat’s speed when it slid off the table will be 6.552 m/s

The branch of physics that defines motion with respect to space and time, ignoring the cause of that motion, is known as kinematics. Kinematics equations are a set of equations that can derive an unknown aspect of a body’s motion if the other aspects are provided.

a = -g = 9.8 m\(/s^{2}\)

using equation of motion

x = u(horizontal )*t + 1/2 * a (horizontal) * \(t^{2}\)

since , a (horizontal) = 0

x = u(horizontal )*t

u   = x / t                   equation 1

similarly

y = u(vertical)*t + 1/2 * a (vertical) * \(t^{2}\)

u(vertical) = 0

t = \(\sqrt{2y / a}\)                               equation 2

substituting the value of equation 2  in equation 1  

u = x /  \(\sqrt{2y / a}\)  

= \(\sqrt{\frac{-9.81}{2*-0.66} } * 2.4\)

= 6.552 m/s

The cat’s speed when it slid off the table will be 6.552 m/s

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(a) The time taken for the ball to reach the height of the fence is 4.1 s.

(b) The position of the ball from the batter at the time is 111.33 m.

(c) Henry will have a home run.

The time taken for the ball to reach the height of the fence is calculated as follows;

\(h = v_0_y t - \frac{1}{2} gt^2\\\\(3 - 1.2) = (34 \times sin37)t - \frac{1}{2} (9.8)t^2\\\\1.8 = 20.46t- 4.9t^2\\\\4.9t^2 - 20.46t+ 1.8 = 0\\\\a = 4.9, \ b = -20.46, \ c = 1.8\\\\t = \frac{-b\pm \sqrt{b^2 - 4ac} }{2a} \\\\t = 4.1 \ s\)

The position of the ball at the time is calculated as follows;

\(X = v_x t\\\\X = (34 \times cos37) \times 4.1\\\\X = 111.33 \ m\)

Since, 111.33 m is greater than 106 m, Henry will have a home run.

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If the force provided by the engine of a toy car decreases, the car will slow down and eventually come to a stop.

1. The force provided by the engine of a toy car is responsible for propelling it forward at a constant speed. This force overcomes any friction or resistance acting on the car.

2. When the force provided by the engine decreases, there is a reduction in the overall force acting on the car. As a result, the force can no longer counteract the resistance and friction effectively.

3. The resistance and friction acting on the toy car, such as air resistance and the friction between the wheels and the surface, start to have a greater impact on the car's motion.

4. With a reduced force from the engine, the car begins to slow down gradually. The deceleration occurs because the opposing forces now have a greater influence on the car's motion.

5. As the force continues to decrease, the opposing forces eventually surpass the remaining force from the engine. Consequently, the toy car slows down even more and eventually comes to a complete stop.

6. If the force provided by the engine becomes extremely low or nonexistent, the opposing forces will completely overpower the forward force, causing the toy car to stop moving altogether.

In summary, when the force provided by the engine of a toy car decreases, the car's speed decreases, and it eventually comes to a stop due to the increased influence of opposing forces.

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Answer:

I don't have friendly brother

Explanation:

So no matter if I write he is gonna throw it out

Answer:

C. Mass

Explanation:

If something has the same weight than it has the same mass

Answer: C

Explanation: the mass is the same as weight so if they have weight they also must have mass.

The vectors can be expressed using unit vectors as follows:

Vector A = 96N i + 96N j + 0N k

Vector B = 76.6N i + 64.3N j + 0N k

Vector C = 64N i + 48N j + 0N k.

In the given diagram, let's consider the vectors as follows:

Vector A with magnitude 120N, Vector B with magnitude 100N, and Vector C with magnitude 80N.

To express these vectors using unit vectors i, j, and k, we need to determine their respective components in the x, y, and z directions.

For Vector A (120N), we have the following information:

Magnitude = 120N

Direction: X-axis (cosine component) with an angle of 37° and Y-axis (sine component) with an angle of 53°.

Using the trigonometric values provided:

cos37° = 0.8 and sin53° = 0.8

Therefore, the components of Vector A are:

Ax = 120N * 0.8 = 96N (in the X-direction)

Ay = 120N * 0.8 = 96N (in the Y-direction)

Az = 0N (no component in the Z-direction)

Thus, Vector A can be written as 96N i + 96N j + 0N k.

Similarly, using the trigonometric values for Vector B and Vector C, we can calculate their components:

For Vector B (100N):

Bx = 100N * cos40° = 100N * 0.766 = 76.6N

By = 100N * sin40° = 100N * 0.643 = 64.3N

Bz = 0N

Vector B can be expressed as 76.6N i + 64.3N j + 0N k.

For Vector C (80N):

Cx = 80N * cos37° = 80N * 0.8 = 64N

Cy = 80N * sin37° = 80N * 0.6 = 48N

Cz = 0N

Vector C can be written as 64N i + 48N j + 0N k.

In summary, the vectors can be expressed using unit vectors as follows:

Vector A = 96N i + 96N j + 0N k

Vector B = 76.6N i + 64.3N j + 0N k

Vector C = 64N i + 48N j + 0N k.

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Answer:

Explanation:

Oh sorry I thought I knew this one I guess not but you can prob look it up on safari

The total power received by the antenna is 2.14 × 10^-14 W, and  a source at a distance of 17000 ly would need to emit a power of 1.35 × 10^-26 W to provide a signal that could be detected by the radio telescope.

A radio wave is a type of electromagnetic wave with a frequency between about 3 kHz and 300 GHz. Radio waves are used for a variety of purposes, including communication, broadcasting, and radar. They are produced by oscillating electrical charges, and can travel through air and other materials, including space.

Radio waves are characterized by their frequency and wavelength. The frequency of a radio wave is the number of cycles per second, measured in hertz (Hz). The wavelength is the distance between two adjacent peaks of the wave, and is inversely proportional to the frequency. Radio waves with lower frequencies have longer wavelengths, while those with higher frequencies have shorter wavelengths.

Radio waves can be generated by a variety of sources, including antennas, electrical circuits, and certain types of electronic devices. They are used for communication between devices, such as radios and cell phones, as well as for broadcasting television and radio signals. Radio waves are also used for remote sensing, including radar and satellite communication.

Here in the Question,

(a) The power per unit area received by the antenna is given by:

P/A = 0.95 pW / 6.1 × 10^17 m^2

The total power received by the antenna is equal to the power per unit area multiplied by the area of the antenna:

P = (P/A) × π(0.5d)^2

where d is the diameter of the antenna. Substituting d = 380 m, we get:

P = (0.95 pW / 6.1 × 10^17 m^2) × π(0.5 × 380)^2

P = 2.14 × 10^-14 W

Therefore, the total power received by the antenna is 2.14 × 10^-14 W.

(b) To find the power of a source at 17000 ly distance that could provide such a signal, we need to consider the power spreading over the surface of a sphere with a radius of 17000 ly. The power density of the signal decreases with the square of the distance from the source, so the power received by the antenna from a source at a distance of 17000 ly is:

P = P0 × (d0 / d)^2

where P0 is the power emitted by the source, d0 is the distance from the source to the antenna (i.e., 17000 ly), and d is the diameter of the antenna.

Substituting the values given, we get:

P = (0.95 pW / 6.1 × 10^17 m^2) × π(0.5 × 380)^2 × (17000 × 9.461 × 10^15 m / 380 m)^2

P = 1.35 × 10^-26 W

Therefore, a source at a distance of 17000 ly would need to emit a power of 1.35 × 10^-26 W to provide a signal that could be detected by the radio telescope.

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Answer:

ke=648J

Explanation:

ke=1/2mv²

m=9kg

v=12m/s

1/2(9)(12)²

=648J

Given:

The radius of the circular loop is,

The magnetic field is,

The normal to the plane of the loop is parallel to the field.

To find:

The magnetic flux

Explanation:

The magnetic flux is given as,

Here, A is the area of the loop and

is the angle between the area vector of the loop and the magnetic field.

As we know, the area vector is always perpendicular to the plane of the loop, so the angle in this case is,

The area of the loop is,

The magnetic flux is,

Hence, the magnetic flux is 0.525 Wb.

Answer:

Because gravity makes you able to stand and move around

Gravity can't fail

it can fail only in astronuts camp and in space

Explanation:

✌❤

Yh....

:)

The force exerted by the man is 1500 N

The work done is given as  - 4200 J

The work done on the piano by gravity is 4200 J

The force exerted by the man

= mgsinθ

= 340 * 9.8 * sin 27

= 1512.7 N

= 1500 N in two significant figures

b. The work done is given as - Fs

= - 1512.7 N * 2.8 m

= - 4235.56

= - 4200 J in two significant figures

c. The work done on the piano by gravity

Wg = -W

= 4200 J

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Answer:

The moon does not have an atmosphere because it does not have enough gravity to hold onto the gases that make up an atmosphere. The moon also does not have plate tectonic activity because it does not have enough internal heat to drive the movement of the plates.

Explanation:

The mass of soccer ball is 5 kg. A ball got a force (50 N) which caused change in momentum for 25 second.

The value of the change in momentum can be described below :

The change in momentum is called as Impuls.

The formula of Impuls --> I = F × t

I = Impuls, F = Force (N), t = time (s)

I = F × t

I = 50 × 25 = 1250 Ns

So, the change in momentum of 5 kg ball is 1250 Ns.

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Answer:

a = 1 m/s^2

Explanation:

if we assume that right is the positive x direction and y is the negative x-direction then we get:

Fr = 8N

Fl = -3N

We then find the net force acting upon the object

Ft = Fr + Fl

Ft = 8N - 3N

Ft = 5N

After we find net force acting upon the object we can then use Newton's 2nd Law to find its acceleration.

Ft = m*a

Isolate accerleration

a = Ft / m

a = 5N / 5kg

a = 1 m/s^2

Answer:

8.45 m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 90 Kg

Initial velocity (u) = 13 m/s

Final velocity (v) = 0 m/s

Height (h) =?

NOTE: Acceleration due to gravity (g) = 10 m/s²

The height of the hill can be obtained as follow:

v² = u² – 2gh (since the cart is going against gravity)

0² = 13² – (2 × 10 × h)

0 = 169 – 20h

Rearrange

20h = 169

Divide both side by 20

h = 169/20

h = 8.45 m

Therefore, the height of the hill is 8.45 m