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3 years ago

Two energy transfers take place when a book hits the ground, Which type of energy transfers are those

Physics

1 answer:

bazaltina [42]3 years ago

6 0

Answer:

I think that when a book hits the ground its potential energy converts into kinetic energy and then kinetic energy is transformed into sound and heat energy.

Explanation:

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What is the difference between a solar cell and a solar panel

dexar [7]

Solar cells and solar panels are both integral, and closely related, parts of a solar energy system. When reading about solar energy systems, it may seem as if these titles are almost interchangeable. Writers refer to them both when discussing energy production and output, and often do so without explanation of how these parts work. However, each plays a distinct role. Solar cells contain all the parts necessary to convert sunlight to electricity. Solar panels combine and direct all of that energy output.

5 0

3 years ago

Due to the wave nature of light, light shined on a single slit will produce a diffraction pattern? Green light (520 nm) is shine

TiliK225 [7]

Answer:

Yes, it will produce a diffraction pattern.

a. 3.9 mm b. 1.95 mm

Explanation:

The light shined from a single slit will produce a diffraction pattern because, the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.

Given

wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m

width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m

Distance of slit from central maximum , D = 1.65 m

Distance of first minimum from central maximum, y = ?

a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...

The relationship between y and D is given by $tanθ = y/D$

Since θ is small, sinθ ≈ θ ≈ tanθ

so, dθ = mλ ⇒ θ = mλ/d = y/D

Therefore, y = mλD/d

Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ = λD/d and y₋₁ = -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d

Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ = 3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm

b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ = λD/d and y₂ = 2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have

λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm

7 0

3 years ago

When you answer someones question how do you make it so your answer is verified??

malfutka [58]

Well YOU cant make your questions verified. if theres more than one answer to the question the person who asked the question will choose the best answer. usually based upon like how well it explains and if its the right answer

6 0

3 years ago

Read 2 more answers

A person can jump a maximum horizontal distance (by using a 45◦ projectile angle) of 5 m on Earth. The acceleration of gravity i

snow_lady [41]

Answer:30 m

Explanation:

Given

Maximum Horizontal distance is 5 m on earth

launching angle $=45^{\circ}$

Acceleration due to gravity on earth is $9.8 m/s^2$

Acceleration due to gravity on moon is $\frac{9.8}{6}=1.63 m/s^2$

Range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

$R_{earth}=\frac{u^2\sin 2\theta }{g}=5$ ----1

$R_{moon}=\frac{u^2\sin 2\theta }{\frac{g}{6}}$ -----2

Divide 1 & 2

$\frac{5}{R_{moon}}=\frac{1}{6}$

$R_{moon}=30 m$

4 0

3 years ago

Physical data is often used in the court system. In fact, police officers use radar to determine your speed when you are driving

Mashcka [7]

The driver is telling the truth, the radar gun must have been set incorrectly to record relative velocity.

The given parameters:

Speed of the driver observed by the stationary police officer, Vo = 44.7 m/s
Speed of the driver, V = 26.8 m/s.
Speed limit = 60 mph

The speed limit of the driver in meter per second is calculated as follows;

$= 60 \ \times \frac{miles}{hour} \times \frac{1609.34 \ m}{1 \ mile} \times \frac{1 \ hr}{3600 \ s}\\\\= 26.82 \ m/s$

From the speed limit, it is obvious that the driver's speed is within the limit. Thus, we can conclude that the driver is telling the truth, the radar gun must have been set incorrectly to record relative velocity.

Learn more about relative velocity here: brainly.com/question/17228388

6 0

2 years ago