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4 years ago

14

A motor-driven winch pulls a 50.0 kg student 5.00 m up the rope at a constant speed of 1.25 m/s. how much power does the motor u

se in raising the student? how much work does the motor do on the student?

1 answer:

nadya68 [22]4 years ago

7 0

Power is the rate work done given by dividing work done by unit time. It is measured in watts equivalent to J/s.
In this case the force by the student is mg = 490 N (taking g as 9.8m/s²)
Work done is given by force × distance,
Therefore, Power =(force × distance)/ time, but velocity/speed =distance/time
Thus, Power = force × speed/velocity
= 490 N × 1.25
= 612.5 J/S (Watts)
Hence, power will be 612.5 Watts.

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3 years ago

The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o

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Answers:

(a) $0.0073kg$

(b) Volume gold: $3.79(10)^{-7}m^{3}$ , Volume cupper: $7.6(10)^{-8}m^{3}$

(c) $17633.554kg/m^{3}$

Explanation:

<h2>(a) Mass of gold </h2><h2 />

We are told the total mass $M$ of the coin, which is an alloy of gold and copper is:

$M=m_{gold}+m_{copper}=7.988g=0.007988kg$ (1)

Where $m_{gold}$ is the mass of gold and $m_{copper}$ is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats $K$ and mass is:

$K=24\frac{m_{gold}}{M}$ (2)

Finding ${m_{gold}$ :

$m_{gold}=\frac{22}{24}M$ (3)

$m_{gold}=\frac{22}{24}(0.007988kg)$ (4)

$m_{gold}=0.0073kg$ (5) This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density $\rho$ of an object is given by:

$\rho=\frac{mass}{volume}$

If we want to find the volume, this expression changes to: $volume=\frac{mass}{\rho}$

For gold, its volume $V_{gold}$ will be a relation between its mass $m_{gold}$ (found in (5)) and its density $\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}$ :

$V_{gold}=\frac{m_{gold}}{\rho_{gold}}$ (6)

$V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}$ (7)

$V_{gold}=3.79(10)^{-7}m^{3}$ (8) Volume of gold in the coin

For copper, its volume $V_{copper}$ will be a relation between its mass $m_{copper}$ and its density $\rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}$ :

$V_{copper}=\frac{m_{copper}}{\rho_{copper}}$ (9)

The mass of copper can be found by isolating $m_{copper}$ from (1):

$M=m_{gold}+m_{copper}$

$m_{copper}=M-m_{gold}$ (10)

Knowing the mass of gold found in (5):

$m_{copper}=0.007988kg-0.0073kg=0.000688kg$ (11)

Now we can find the volume of copper:

$V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}$ (12)

$V_{copper}=7.6(10)^{-8}m^{3}$ (13) Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density $\rho_{coin$ will be a relation between its total mass $M$ and its total volume $V$ :

$\rho_{coin}=\frac{M}{V}$ (14)

Knowing the total volume of the coin is:

$V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3}$ (15)

$\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}}$ (16)

Finally:

$\rho_{coin}=17633.554kg/m^{3}}$ (17) This is the total density of the British sovereign coin

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How can you measure mold on food?

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At one instant a bicyclist is 21.0 m due east of a park's flagpole, going due south with a speed of 13.0 m/s. Then 21.0 s later,

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Answer:

Part a)

$d = 21\sqrt2 = 29.7 m$

Part b)

Direction is 45 degree North of West

Part c)

$v_{avg} = 1.41 m/s$

Part d)

direction of velocity will be 45 Degree North of West

Part e)

$a = 0.875 m/s^2$

Part f)

$\theta = 45 degree$ North of East

Explanation:

Initial position of the cyclist is given as

$r_1 = 21.0 m$ due East

final position of the cyclist after t = 21.0 s

$r_2 = 21.0 m$ due North

Part a)

for displacement we can find the change in the position of the cyclist

so we have

$d = r_2 - r_1$

$d = 21\hat j - 21\hat i$

so magnitude of the displacement is given as

$d = 21\sqrt2 = 29.7 m$

Part b)

direction of the displacement is given as

$\theta = tan^{-1}\frac{y}{x}$

$\theta = tan^{-1}\frac{21}{-21}$

so it is 45 degree North of West

Part c)

For average velocity we know that it is defined as the ratio of displacement and time

so here the magnitude of average velocity is defined as

$v_{avg} = \frac{\Delta x}{t}$

$v_{avg} = \frac{29.7}{21}$

$v_{avg} = 1.41 m/s$

Part d)

As we know that average velocity direction is always same as that of average displacement direction

so here direction of displacement will be 45 Degree North of West

Part e)

Here we also know that initial velocity of the cyclist is 13 m/s due South while after t = 21 s its velocity is 13 m/s due East

So we have

change in velocity of the cyclist is given as

$\Delta v = v_f - v_i$

$\Delta v = 13\hat i - (-13\hat j)$

now average acceleration is given as

$a = \frac{\Delta v}{\Delta t}$

$a = \frac{13\hat i + 13\hat j}{21}$

so the magnitude of acceleration is given as

$a = \frac{13\sqrt2}{21} = 0.875 m/s^2$

Part f)

direction of acceleration is given as

$\theta = tan^{-1}\frac{y}{x}$

$\theta = tan^{-1}\frac{13}{13}$

$\theta = 45 degree$ North of East

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Kipish [7]

Answer:

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