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pav-90 [236]
2 years ago
9

_______ drives the energy and motion of waves? ______

Physics
2 answers:
Papessa [141]2 years ago
7 0

Answer:

Wind-driven waves, or surface waves,

Explanation:

____ [38]2 years ago
5 0

Answer:

Kinetic Energy(?)

<u><em>I hope this helped at all. sorry if it didn't</em></u>

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Which of the following situations would violate the second law of
Sonja [21]

Answer:

c

Explanation:i don’t know

6 0
3 years ago
A solar cell generates a potential difference of 0.25 V when a 550 Ω resistor is connected across it, and a potential difference
Andre45 [30]

a) 400 \Omega

b) 0.43 V

c) 0.44 %

Explanation:

a)

For a battery with internal resistance, the relationship between emf of the battery and the terminal voltage (the voltage provided) is

V=E-Ir (1)

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

In this problem, we have two situations:

1) when R_1=550 \Omega, V_1=0.25 V

Using Ohm's Law, the current is:

I_1=\frac{V_1}{R_1}=\frac{0.25}{550}=4.5\cdot 10^{-4} A

2) when R_2=1000 \Omega, V_2=0.31 V

Using Ohm's Law, the current is:

I_2=\frac{V_2}{R_2}=\frac{0.31}{1000}=3.1\cdot 10^{-4} A

Now we can rewrite eq.(1) in two forms:

V_1 = E-I_1 r

V_2=E-I_2 r

And we can solve this system of equations to find r, the internal resistance. We do it by substracting eq.(2) from eq(1), we find:

V_1-V_2=r(I_2-I_1)\\r=\frac{V_1-V_2}{I_2-I_1}=\frac{0.25-0.31}{3.1\cdot 10^{-4}-4.5\cdot 10^{-4}}=400 \Omega

b)

To find the electromotive force (emf) of the solar cell, we simply use the equation used in part a)

V=E-Ir

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

Using the first set of data,

V=0.25 V is the voltage

I=4.5\cdot 10^{-4}A is the current

r=400\Omega is the internal resistance

Solving for E,

E=V+Ir=0.25+(4.5\cdot 10^{-4})(400)=0.43 V

c)

In this part, we are told that the area of the cell is

A=4.0 cm^2

While the intensity of incoming radiation (the energy received per unit area) is

Int.=5.5 mW/cm^2

This means that the power of the incoming radiation is:

P=Int.\cdot A=(5.5)(4.0)=22 mW = 0.022 W

This is the power in input to the resistor.

The power in output to the resistor can be found by using

P'=I^2R

where:

R=1000 \Omega is the resistance of the resistor

I=3.1\cdot 10^{-4} A is the current on the resistor (found in part A)

Susbtituting,

P'=(3.1\cdot 10^{-4})^2(1000)=9.61\cdot 10^{-5} W

Therefore, the efficiency of the cell in converting light energy to thermal energy is:

\epsilon = \frac{P'}{P}\cdot 100 = \frac{9.6\cdot 10^{-5}}{0.022}=0.0044\cdot 100 = 0.44\%

7 0
3 years ago
Total distance between Karachi and Hyderabad is 120 km , if a car speed is 40 km/h, In how many hours it can travel back to Hyde
Nuetrik [128]

it takes 6 hours to travel back to Hyderabad

6 0
3 years ago
Read 2 more answers
Suppose the value of one division of vernier scale is 0.5mm and the value of one main scale
gregori [183]

Answer:

-0.01 mm

Explanation:

We are given that

The value of one division of vernier scale =0.5 mm

The value of one main scale division=0.49 mm

We have to find the value of least count of the instrument in mm.

We know that

Leas count of vernier caliper=1 main scale division-1 vernier scale division

Least count of vernier caliper=0.49-0.50=-0.01 mm

Hence, the least count of the instrument=-0.01 mm

Answer: -0.01 mm

8 0
3 years ago
An electron has an initial velocity of (19.0 j + 18.0 k) km/s and a constant acceleration of (3.00 ✕ 1012 m/s2)i in a region in
asambeis [7]

Answer:

E=(-17.08 i +7.2 j -7.6 k )N/C

Explanation:

v= (19.0j+18.k)km/s

a=3.0x10^{13}m/s^2

\beta= 400x10{-6}T

Electron information needed to solve the question:

m_e=9.11x10^{-31}kg

q=-1.6x10{-19}C

F=F_E+F_B=q*(E+Vx \beta)

F=m*a

m*a=q*(E+Vx\beta)

E=\frac{m*a}{q}-(Vx\beta )

E=\frac{9.11x10{-31}kg*3.0x10^{12}m/s^2}{-1.6x10{-19}C}-[(19.0x10^3mj+18.0x10^3m)xi(400x10^{-6}T)]

E=-i17.08N/C-[7.6(-k)+7.2(j)]N/C

E=(-17.08 i +7.2 j -7.6 k )N/C

6 0
3 years ago
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