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3 years ago

15

A cyclist is riding his bike up a mountain trail. When he starts up the trail, he is going 8 m/s. As the trail gets steeper, he

slows to 3 m/s in 1 minute. What is the cyclist’s acceleration? –0.08 m/s2 0.08 m/s2 –12 m/s2 12 m/s2

1 answer:

aalyn [17]3 years ago

7 0

Answer:

-0.08m/s²

Explanation:

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A spring has a spring constant of 53N/m. How much elastic potential energy is stored in the spring in the spring when it is comp

IRISSAK [1]

Answer:11686.5 joules

Explanation:

elastic constant(k)=53N/m

extension(e)=21m

Elastic potential energy=(k x e^2)/2

Elastic potential energy=(53 x (21)^2)/2

Elastic potential energy=(53x21x21)/2

Elastic potential energy=23373/2

Elastic potential energy=11686.5

Elastic potential energy is 11686.5 joules

5 0

3 years ago

In a particular beam of radiation, which is traveling in a vacuum, the amounts of energy per second at an ultraviolet wavelength

Darina [25.2K]

Answer: d)

Explanation: In order to justify the answer we have to consider that the energy of photons directely depent on the frequency so the energy is inverselly dependent of the wavelegth.

If both beams have the same power, this means Energy/time so the number of photons per second must be different. As consequence a) is wrong as b) since it is not posible since UV photon have more energy that IR photons. c) It is no necessary know the frequency since the wavelength is related in the form:

c=λν c is the speed of light, λ the wavelegth and ν the frequency.

d) Certainly will be more more IR photons than UV photons to get the same beam power.

8 0

3 years ago

A 20 kg truck drives in a circle of radius 4 m at 10m/s. What is the centripetal acceleration of the truck?

asambeis [7]

Answer:

B. 25 m/s/s

Explanation:

Centripetal acceleration is the square of the tangential velocity divided by the radius of curvature.

a = v² / r

Given v = 10 m/s and r = 4 m:

a = (10 m/s)² / 4 m

a = 25 m/s²

4 0

2 years ago

You wind a small paper tube uniformly with 153 turns of thin wire to form a solenoid. The tube\'s diameter is 5.11 mm and its le

lina2011 [118]

The concept required to solve this problem is linked to inductance. This can be defined as the product between the permeability in free space by the number of turns squared by the area over the length. Recall that Inductance is defined as the opposition of a conductive element to changes in the current flowing through it. Mathematically it can be described as

$L = \frac{\mu N^2 A}{l}$

Here,

$\mu$ = Permeability at free space

N = Number of loops

A = Cross-sectional Area

l = Length

Replacing with our values we have,

$L = \frac{(4\pi *10^{-7})(153)^2(\pi (\frac{5.11*10^{-3}}{2})^2)}{2.47*10^{-2}}$

$L = 0.00002442H$

$L = 24.42\mu H$

Therefore the Inductance is $24.42\mu H$

3 0

3 years ago

Suppose you observe two stars and you know they have the same luminosity. If one star is twice as far away as the other, the mor

rosijanka [135]

Answer:

The farther star will appear 4 times fainter than the star that is near to the observer.

Explanation:

Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time

Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)

This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by

$e_{1}=\frac{E}{4\pi r^{2}}$

For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by

$e_{2}=\frac{E}{4\pi (2r)^{2}}=\frac{E}{8\pi r^{2}}=\frac{e_{1}}{4}$

Hence we sense it as 4 times fainter than the nearer star.

5 0

2 years ago