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2 years ago

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1) A person having a mass of 59.1 kg stands before a flight of 30 stairs each of which is 25.0 cm high. He runs up 20 stairs, tu

rns around, walks down 10 stairs, changes his mind and goes up the remaining 20. How much work did he do overcoming gravity? Don’t forget to change the cm to meters.

1 answer:

poizon [28]2 years ago

3 0

Answer:

P = 147,75 W

Explanation:

A man whose mass is 59.1kg climbs up 30 steps of a stair each step is 25 cm high

Height at 30 steps , h=30×2.5= 7.5 m

Change in potential energy , =mgh=59.1×10×7.5 = 4432.5 J

So, Work done by the man , W= 4432.5J

Power used , P= $\frac{W}{T}$

P = 4432.5 /30

P = 147,75 W

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A marble rolls with a speed of 15 m/s and has a momentum of 0.15 kg*m/s. What is its mass?<br>

prisoha [69]

Answer:

m = 0.01 kg

Explanation:

Given that,

Momentum of the marble, p = 0.15 kg-m/s

Speed of the marble, v = 15 m/s

We need to find its mass. We know that,

Momentum, p = mv

Where

m is the mass

$m=\dfrac{p}{v}\\\\m=\dfrac{0.15}{15}\\\\m=0.01\ kg$

So, the mass of the marble is equal to 0.01 kg.

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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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I think it's 3. within an outer arm

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Which of these is not true about the proton?

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Answer:

Option C is the untrue statement.

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4 years ago

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A sinusoidal voltage is given by the expression ????(????)=20cos(5π×103 ????+60°) V. Determine its (a) frequency in hertz, (b) p

MA_775_DIABLO [31]

<em>There are some placeholders in the expression, but they can be safely assumed</em>

Answer:

(a) $f=1617.9\ Hz$

(b) $T=0.618\ ms$

(c) $A=20 \ Volts$

(d) $\varphi=60^o$

Explanation:

<u>Sinusoidal Waves </u>

An oscillating wave can be expressed as a sinusoidal function as follows

$V(t)&=A\cdot \sin(2\pi ft+\varphi )$

Where

$A=Amplitude$

$f=frequency$

$\varphi=Phase\ angle$

The voltage of the question is the sinusoid expression

$V(t)=20cos(5\pi\times 103t+60^o)$

(a) By comparing with the general formula we have

$f=5\pi\times 103=1617.9\ Hz$

$\boxed{f=1617.9\ Hz}$

(b) The period is the reciprocal of the frequency:

$\displaystyle T=\frac{1}{f}$

$\displaystyle T=\frac{1}{1617.9\ Hz}=0.000618\ sec$

Converting to milliseconds

$\boxed{T=0.618\ ms}$

(c) The amplitude is

$\boxed{A=20 \ Volts}$

(d) Phase angle:

$\boxed{\varphi=60^o}$

4 0

3 years ago