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3 years ago

How much work does a motor do as it lifts a 675-kilogram elevator from a height of 15 meters to a height of 35 meters?

1 answer:

3 0

Work is defined as Force x Distance

W = Fd

so (675kg)(9.8m/s^2) = 6615 N

d = 35-15 = 20m

so

W = Fd
(6615N)(20m)

132,300 J = W

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What are two parts of an atom

seraphim [82]

Answer:

Nucleus And electron cloud

Explanation:

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3 years ago

Why do electrons move from the negative end of the tube to the positive end

defon

Due to attraction ... of opposite charges

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3 years ago

A race car has a maximum speed of 0.104 km/s .What is this speed in miles per hour ?

sweet [91]

Answer:

232.641374 mph

Explanation:

A race car has a maximum speed of 0.104km/s

Let X represent the speed in miles per hour

Therefore the speed in miles per hour can be calculated as follows

1 km/s = 2,236.936292 mph

0.104km/s = X

X = 0.104 × 2,236.936292

X = 232.641374

Hence the speed in miles per hour is 232.641374 mph

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3 years ago

A spaceship hovering over the surface of Venus drops an object from a height of 17 m. How much longer does it take to reach the

Paraphin [41]

1.96s and 1.86s. The time it takes to a spaceship hovering the surface of Venus to drop an object from a height of 17m is 1.96s, and the time it takes to the same spaceship hovering the surface of the Earth to drop and object from the same height is 1.86s.

In order to solve this problem, we are going to use the motion equation to calculate the time of flight of an object on Venus surface and the Earth. There is an equation of motion that relates the height as follow:

$h=v_{0} t+\frac{gt^{2}}{2}$

The initial velocity of the object before the dropping is 0, so we can reduce the equation to:

$h=\frac{gt^{2}}{2}$

We know the height h of the spaceship hovering, and the gravity of Venus is $g=8.87\frac{m}{s^{2}}$ . Substituting this values in the equation $h=\frac{gt^{2}}{2}$ :

$17m=\frac{8.87\frac{m}{s^{2} } t^{2}}{2}$

To calculate the time it takes to an object to reach the surface of Venus dropped by a spaceship hovering from a height of 17m, we have to clear t from the equation above, resulting:

$t=\sqrt{\frac{2(17m)}{8.87\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{8.87\frac{m}{s^{2} } } }=1.96s$

Similarly, to calculate the time it takes to an object to reach the surface of the Earth dropped by a spaceship hovering from a height of 17m, and the gravity of the Earth $g=9.81\frac{m}{s^{2}}$ .

$t=\sqrt{\frac{2(17m)}{9.81\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{9.81\frac{m}{s^{2} } } }=1.86s$

8 0

3 years ago

Read 2 more answers

an airplane flies at a speed of 100 m/s and starts to accelerate constantly at a rate of 50 m/s2. how fast is the plane flying a

mart [117]

Answer:

331.7m/s

Explanation:

Given parameters:

Initial velocity = 100m/s

Acceleration = 50m/s²

Distance = 1km = 1000m

Unknown:

Final velocity = ?

Solution:

To solve this problem, we have to apply the right motion equation shown below;

v² = u² + 2aS

v is the final velocity

u is the initial velocity

a is the acceleration

S is the distance

Now insert the parameters and solve;

v² = 100² + (2 x 50 x 1000)

v² = 110000

v = √110000 = 331.7m/s

4 0

3 years ago