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3 years ago

14

I give a ball a push on an acclivity. The "start velocity" is on 7m/s. The time it took the ball to get back to me was 10 second

s. The acceleration is a constant. How much is the acceleration?

2 answers:

kozerog [31]3 years ago

8 0

Answer:

-1.4 m/s²

Explanation:

The initial velocity is 7 m/s. When the ball rolls back, its velocity is -7 m/s.

Acceleration is change in velocity over change in time.

a = Δv / Δt

a = (-7 m/s − 7 m/s) / 10 s

a = -1.4 m/s²

Degger [83]3 years ago

3 0

Answer:

7÷10

Explanation:

initial velocity=7m/s

final velocity=0m/s

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The primary purpose of a mirror is to _________ light rays.

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a. reflect (I found this using prior knowledge and process of illumination it can't be absorb cuz you wouldn't see anything it can't be refract because it doesn't reverse the image and it isn't transmit )

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3 years ago

A sample of an ideal gas initially occupies a volume of 6 L. The pressure of the sample is then doubled while it is cooled to on

Troyanec [42]

Answer:

V₂= 1 L

Explanation:

Given that

Volume occupies V₁= 6 L

Initial pressure = P₁

Initial temperature = T₁

The final pressure =P₂ = 2 P₁

Final volume =V₂

Final temperature = T₁/3

As we know that equation for ideal gas

P V = m R T

P=pressure, V=volume, T=temperature

m=mass ,R=gas constant

Now from mass conservation

$m=\dfrac{P_1V_1}{RT_1}=\dfrac{P_2V_2}{RT_2}$

$\dfrac{P_1V_1}{RT_1}=\dfrac{P_2V_2}{RT_2}$

$\dfrac{P_1\times 6}{RT_1}=3\times \dfrac{2P_1V_2}{RT_1}$

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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:

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

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2 years ago

What's the airplane velocity when it flies 100 miles in 20 seconds<br><br>

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Answer : 5m/s

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2 years ago

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BaLLatris [955]

Answer:

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Explanation:

Kepler's laws establish that:

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A planet describes equal areas in equal times (Kepler's second law).

The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).

$T^{2} = a^{3}$

Where T is the period of revolution and a is the semi-major axis.

Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.

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3 years ago