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mariarad [96]
3 years ago
6

Suppose a 4,000-kg elephant is hoisted 20 m above Earth’s surface. Use a calculator and follow the steps below to find the eleph

ant’s gravitational potential energy.
Step 1:
Calculate the elephant’s weight
(w = mg)

Step 2:
What is the GPE of the elephant?
(GPE = w * h)
Physics
1 answer:
stiv31 [10]3 years ago
4 0
GPE = 78,380 J
w = 39,240 N

First list what you know. You know the elephants mass and it’s height. You also know gravity on Earth. I will use g = 9.81.
m = 4,000 kg
h = 20 m
g = 9.81 m/s^2

You need to find the elephants weight. Weight = mass x gravity
w = mg
w = (4000 kg)(9.81 m/s^2)
w = 39,240 N (N = newtons)

Now, knowing the elephants weight, you can calculate its GPE.
Gravitational Potential Energy = weight x height

GPE = wh
GPE = (39,240N)(20m)
GPE = 78,380 J (J = joules)
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A two-turn circular wire loop of radius 0.63 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0.219 T. I
Basile [38]

Answer:

The magnitude of the average induced emf in the wire during this time is 9.533 V.

Explanation:

Given that,

Radius r= 0.63 m

Magnetic field B= 0.219 T

Time t= 0.0572 s

We need to calculate the average induce emf in the wire during this time

Using formula of induce emf

E=-\dfrac{d\phi}{dt}

E=-B\dfrac{dA}{dt}

E=-B\dfrac{A_{2}-A_{1}}{dt}

E=B\dfrac{A_{1}-A_{2}}{dt}.....(I)

In reshaping of wire, circumstance must remain same.

We calculate the length when wire is in two loops

l=2\times 2\pi\times r_{1}

l=2\times 2\pi\times 0.63

l=7.916\ m

The length when wire is in one loop

l=2\pi\times r_{2}

7.916=2\times \pi\times r_{2}

r_{2}=\dfrac{7.916}{2\times \pi}

r_{2}=1.259\ m

We need to calculate the initial area

A_{1}=N\times\pi\times r_{1}^2

Put the value into the formula

A_{1}=2\times3.14\times(0.63)^2

A_{1}=2.49\ m^2

The final area is

A_{2}=N\times\pi\times r_{2}^2

A_{2}=1\times\pi\times(1.259)^2

A_{2}=4.98\ m^2

Put the value of initial area and final area in the equation (I)

E=0.219\dfrac{2.49-4.98}{0.0572}

E=-9.533\ V

Negative sign shows the direction of induced emf.

Hence, The magnitude of the average induced emf in the wire during this time is 9.533 V.

6 0
2 years ago
if the same amount of work needed to keep a 100W light bulb lit for one hour could be used to lift a 2000 kg SUV, what height co
yulyashka [42]

(100 watts) = 100 joules/second

(100 joules/second) x (3,600sec/hour) = 360,000 joules/hour

360,000 joules = 360,000 newton-meters

Weight of 2,000 kg on Earth = (2,000 kg) x (9.8 m/s²) = 19,600 newtons

(360,000 newton-meters) / (19,600 newtons) = <em>18.4 meters off the ground</em>


5 0
3 years ago
The grocery store is 20 miles away
adoni [48]

Answer:

40mph

Explanation:

20miles  .5 miles away if you go 20 mph you get there in a hour soo if you double to 40 mpg you get there in 0.5

5 0
2 years ago
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An electric hoist does 56,447 J of work in raising 115 kg load. How high (in meters) was the load lifted?
Ugo [173]

\\ \rm\rightarrowtail W=mgh

\\ \rm\rightarrowtail 56447=1150h

\\ \rm\rightarrowtail h=56447/11150

\\ \rm\rightarrowtail h=49.1m

8 0
2 years ago
In a butcher shop, a horizontal steel bar of mass 4.94 kg and length 1.46 m is supported by two vertical wires attached to its e
mojhsa [17]

Answer:

Tension in right wire = 25.9N

Explanation:

I have attached a free body diagram to depict this question.

From the diagram, i have labelled the tensions in the strings T1 and T2.

While i labelled the weight of the bar as Wb and weight of sausage as Ws.

Now, when solving a problem like this we want to first remember that the beam is static; meaning it is not moving. From simple physics, this means that the sum of the forces in the y direction equals zero (i.e. the total downward forces equal the total upward forces)

Thus, from the diagram, the upward forces are T1 and T2 while the downward forces are Ws and Wb.

Thus;

T1 + T2 = Wb + Ws

We know that mass of bar = 4.94kg. Thus, Weight of bar(Wb) = mg = 4.94 x 9.81 = 48.46N

Also, weight of sausage (Ws) = mg = 2.49 x 9.81 = 24.43N

Thus,

T1 + T2 = 48.46N + 24.43

T1 + T2 = 72.89N - - - - - (eq 1)

Now, let's take moments about the left end of the bar.

The maximum weight of the bar will act at the centre, so distance from the Wb to left end = 1.46/2 = 0.73m

So, moments about left end;

T2 x 1.46 = (Wb x 0.73) + (Ws x 0.1)

1.46T2 = (48.46 x 0.73) + (24.43 x 0.1)

1.46T2 = 35.373 + 2.443

1.46T2 = 37.816

T2 = 37.816/1.46 = 25.9N

3 0
2 years ago
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