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3 years ago

A 0.10 kg stone is thrown from the edge of an ocean cliff with an initial speed

Physics

1 answer:

AfilCa [17]3 years ago

6 0

Answer:

79.20 joule

Explanation:

A 0.10kg stone is thrown from the edge of an ocean cliff with an initial speed of 21m/s. When it strikes the water below, it is traveling at 45m/s. What is the change in kinetic energy of the stone?

ΔEk = m/2*(V2^2-V1^2) = 0.05*(45^2-21^2) = 79.20 joule

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Can I have some help please :)

Anastasy [175]

Answer:

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Explanation:

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2 years ago

Suppose a square wave signal has a 65 percent duty cycle and an on-state voltage of 40 volts DC. What is the average DC voltage

Ierofanga [76]

Answer:

The voltage is $\= DC _v = 2.6 \ V$

Explanation:

From the question we are told that

The duty cycle is p = 65% = 0.65

The on - state voltage is $V = 40 \ volt$

Generally the average DC voltage is mathematically represented as

$\= DC _v = p * V$

=> $\= DC _v = 40 * 0.65$

=> $\= DC _v = 2.6 \ V$

6 0

2 years ago

A cell phone weighing 80 grams is flying through the air at 15 m/s. What is its kinetic energy

Contact [7]

Answer:

The kinetic energy of the cell phone is 9J

Explanation:

The kinetic energy is the energy possessed by a body by virtue of motion.

The kinetic energy is expressed as

KE= 1/2m(v)²

Given data

Mass of cell phone m= 80g--to kg=80/1000= 0.08kg

Velocity of cell phone v= 15m/s

Substituting our given data we have

KE= 1/2*0.08(15)²

KE= (0.08*225)/2

KE=18/2

KE= 9J

8 0

2 years ago

A 1kg mass is thrown to a height of 2cm. what is the potential energy

prohojiy [21]

Mass=m=1kg
Height=h=2cm=0.02m
Acceleration due to gravity=g=10m/s^2

$\\ \tt\hookrightarrow P.E=mgh$

$\\ \tt\hookrightarrow PE=1(10)(0.02)$

$\\ \tt\hookrightarrow PE=0.2J$

4 0

2 years ago

Read 2 more answers

A power plant produces 1000 MW to supply a city 40 km away. Current flows from the power plant on a single wire of resistance 0.

nikitadnepr [17]

Answer:

Current = 8696 A

Fraction of power lost = $\dfrac{80}{529}$ = 0.151

Explanation:

Electric power is given by

$P=IV$

where I is the current and V is the voltage.

$I=\dfrac{P}{V}$

Using values from the question,

$I=\dfrac{1000\times10^6 \text{ W}}{115\times10^3\text{ V}} = 8696 \text{ A}$

The power loss is given by

$P_\text{loss} = I^2R$

where R is the resistance of the wire. From the question, the wire has a resistance of $0.050\Omega$ per km. Since resistance is proportional to length, the resistance of the wire is

$R = 0.050\times40 = 2\Omega$

Hence,

$P_\text{loss} = \left(\dfrac{200000}{23}\right)^2\times2$

The fraction lost = $\dfrac{P_\text{loss}}{P}=\left(\dfrac{200000}{23}\right)^2\times2\div (1000\times10^6)=\dfrac{80}{529}=0.151$

3 0

3 years ago