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Flauer [41]
3 years ago
9

A cyclist rides in a circle with speed 8.1 m/s. What is his centripetal acceleration if the circle has a radius of 27 m? A. 3.3

m/s^2 B. 2.4 m/s^2 C. 0.3 m/s^2 D. 1.8 m/s^2
Physics
1 answer:
GarryVolchara [31]3 years ago
5 0
The answer would be B. 2.4 

You're welcome.
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A tank of gasoline (n = 1.40) is open to the air (n = 1.00). A thin film of liquid floats on the gasoline and has a refractive i
klio [65]

Answer:

1.08

Explanation:

This is the case of interference in thin films in which interference bands are formed due to constructive interference of two reflected light waves , one from upper layer and the other from lower layer . If t be the thickness and μ be the refractive index then

path difference created will be 2μ t.

For light coming from rarer to denser medium , a phase change of π occurs additionally after reflection from denser medium, here, two times, once from upper layer and then from the lower layer ,  so for constructive interference

path diff = nλ , for minimum t , n =1

path diff = λ

2μ t. =  λ

μ = λ / 2t

= 626 / 2 x 290

= 1.08

5 0
3 years ago
un resorte de 10cm de longitud recibe una magnitud de fuerza que lo estira hasta medir 15cm ¿cual es la magnitud de la tension u
Ber [7]

Answer: 0.5

Explanation:

The modulus of elasticity (called <em>"alargamiento unitario"</em> in spanish) \epsilon of a spring is given by the following formula:

\epsilon=\frac{\Delta L}{L}

Where:

L=10 cm  is the original length of the spring

\Delta L=L_{f}-L  is the elongation of the spring, being L_{f}=15 cm the length of the spring after a force is applied to it.

\epsilon=\frac{L_{f}-L}{L}=\frac{15 cm - 10 cm}{10 cm}

Then:

\epsilon=0.5

8 0
3 years ago
A linear network has a current input 7.5 cos(10t + 30°) A and a voltage output 120 cos(10t + 75°) V. Determine the associated im
Leona [35]

Answer:

16∠45° Ω

Explanation:

Applying,

Z = V/I................... Equation 1

Where Z = Impedance, V = Voltage output, I = current input.

Given: V = 120cos(10t+75°), = 120∠75°,  I = 7.5cos(10t+30) = 7.5∠30°

Substitute these values into equation 1

Z = 120cos(10t+75°)/7.5cos(10t+30)

Z = 120∠75°/ 7.5∠30°

Z = 16∠(75°-30)

Z = 16∠45° Ω

Hence the impedance of the linear network is 16∠45° Ω

8 0
3 years ago
Ediment deposited where a river flows into an ocean or lake
Triss [41]
The answer is Delta.
3 0
3 years ago
Two parallel plate capacitors 1 and 2 are identical except that capacitor 1 has charge +q on one plate and charge −q on the othe
Grace [21]

Answer:

a) the capacitance is the same for both capacitors.

b) The potential difference between the plates for the capacitor with charge +2q, is double of the one for the capacitor  with charge +q.

c) The electric field magnitude between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q

d) The energy stored between the plates for the capacitor with charge +2q, is 4 times the value for the one with charge +q  

Explanation:

a) The capacitance of a capacitor, by definition, is as follows:

C = \frac{q}{V}

Appying Gauss' Law to one of plates, it can be showed, that the capacitance (for a parallel plates capacitor) can be  expressed as follows:

C = ε*A / d

As it can be seen, it does not depend on the charge. so we conclude that the capacitance must be the same for both capacitors, due to they are identical except for the value of the charge on the plates.

b) By definition, as we said above, the capacitance is equal to the proportion between the charge of one of the plates, and the potential difference between them.

If this proportion must remain the same, and one of the capacitors has the double of  the charge than the other, the potential difference must be the double also.

c) Applying Gauss' law, to the surface of one of  the plates, and assuming a constant surface charge density σ, it can be  showed that the  electric field can be calculated as follows:

E*A = Q/ε₀ as σ=Q/A

⇒ E = σ/ε₀

As σ is directly proportional to the charge (being the area A the same), we conclude that the electric field for the capacitor with charge +2q must be the double than the one for the capacitior with charge +q.

c) The electric potential energy, stored between plates of a capacitor, can be written as follows:

Ue = \frac{1}{2} *\frac{q^{2}}{C}

If the capacitance remains the same, we can conclude that the electric potential energy for the capacitor with charge +2q, as the charge is raised to the 2nd power, must be 4 times the one for the capacitor with charge +q.

4 0
3 years ago
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