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sp2606 [1]
3 years ago
12

You are riding on your bike and stop pedaling , coasting along the road. eventually, your bike slows down and stops. what happen

ed to the energy of your motion when your bike stopped?
Physics
1 answer:
andrew11 [14]3 years ago
8 0
<span>When still rolling your bike had kinetic energy, or energy in the form of motion. The forces resisting this motion--a mixture of many factors including air resistance, tire friction, friction in the hubs of the wheels, etc...--eventually decrease the kinetic energy of the bike to zero as it reaches a complete stop.</span>
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Many of the whales in the ocean rely upon tiny marine organisms, called plankton, for food. If all of the plankton suddenly died
exis [7]

Answer:

I believe the answer to be B.

Explanation:

Without food, the whales would die.

8 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
A long straight wire carries a conventional current of 0.7 A. What is the approximate magnitude of the magnetic field at a locat
vfiekz [6]

Answer:

2.64 x 10⁻⁶T

Explanation:

The magnitude of the magnetic field produced by a long straight wire carrying current is given by Biot-Savart law as follows: "The magnetic field strength is directly proportional to the current on the wire and inversely proportional to the distance from the wire". This can be written mathematically as;

B = (μ₀ I) / (2π r)                ----------------(i)

B is magnetic field

I is current through the wire

r is the distance from the wire

μ₀  is the magnetic constant = 4π x 10⁻⁷Hm⁻¹

From the question;

I = 0.7A

r = 0.053m

Substitute these values into equation (i) as follows;

B =  (4π x 10⁻⁷ x 0.7) / (2π x 0.053)

B = 2.64 x 10⁻⁶T

Therefore the approximate magnitude of the magnetic field at that location is 2.64 x 10⁻⁶T

5 0
3 years ago
He conducted experiments in combining elements
Wewaii [24]
Where is the rest .........
3 0
3 years ago
Starting from rest, a crate of mass m is pushed up a frictionless slope of angle theta by a horizontal force of magnitude F. Use
ludmilkaskok [199]

Answer:

v =  sqrt[2*(F*h*cot(theta)-mgh)/m]

Explanation:

Work  = KE + Ug

F*r=1/2mv^2+mgh

1/2mv^2=F*r-mgh

v=sqrt[2(F*r-mgh)/m]

r=h/tan(theta)=h*cot(theta)

6 0
3 years ago
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