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Viktor [21]
4 years ago
15

What is the rate at which work is done in lifting a 35-kilogram object vertically at a constant speed of 5.0 meters per second?

Physics
1 answer:
svet-max [94.6K]4 years ago
8 0

Answer:

1.75 kW

Explanation:

First, the rate at which work is done is simply termed Power

m = 35kg

v = 5.0 m/s

since the undergoes a vertical motion g = 10 m/s²

F = W = mg = 35×10 = 350 N

Power = Fv = 350×5 = 1750 Watt

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Over a span of 6.0 seconds, a car changes it's speed from 89 km/h to 37 km/h. What is its average acceleration in meters per sec
scoundrel [369]

Acceleration = (change in speed) / (time for the change)

change in speed = (speed at the end) - (speed at the beginning)

change in speed = (37 km/hr) - (89 km/hr) = -52 km/hr

Acceleration = (-52 km/hr) / (6 sec)

Acceleration = (-26/3) km/(hr·sec)

Units: (1/hr·sec) · (hr/3600 sec) = 1 / 3600 sec²

(-26/3) km/(hr·sec) = (-26/3) km/(3600 sec²)

= -26,000/(3 · 3600) m/s²

<em>Acceleration = -2.41 m/s²</em>

3 0
4 years ago
Ex 1) A Major League baseball pitcher throws a baseball straight up into the air and the ball travels for 4
nlexa [21]

Answer:

52 / 4 = 13 m/s

Explanation:

7 0
3 years ago
Joe is hiking through the woods when he decides to stop and take in the view. He is particularly interested in three objects: a
Novosadov [1.4K]

Answer:

A) correct answer is C,   B)   correct answer is b  and C) The correct answer is b

Explanation:

In the exercises of geometric optics, the equation of the constructor tells us the location of the image.

        1 / f = 1 / p + 1 / q

where f is the focal length of the cornea-crystalline system, p and q are the distances to the object and the image.

In this case, the distance to the image on the retina is constant, about 3 cm. Therefore depending on the distance to the object) p = the focal length must change

        1 / q = 1 / f 1 / p

let's apply this expression to our case

A) indicates that the tree is at a medium distance

so that the image is formed on the retina THE SAME AS

correct answer is C

B) The squirrel is at a smaller distance (p ') than the tree (p), therefore if we substitute in the equation above we find that q must decrease. Consequently the image is in front of the retina

The mountain is very far, suppose in infinity, so the image is BEHIND THE RETINA

therefore the correct answer is b

C) The squirrel is very close so the curvature of the lens INCREASES, resulting in a DECREASE in the focal length

The correct answer is b

6 0
3 years ago
2. The substances that undergo change in a chemical reaction are called
AVprozaik [17]
Reactants

In a chemical reaction, the substances that undergo change are called reactants. The new substances formed as a result of that change are called products.
4 0
3 years ago
A 0.580-kg rock is tied to the end of a string and is swung in a circle with a radius of 0.500 meters. The velocity of the rock
gtnhenbr [62]
F = m  \frac{v^2}{r}

F= 0.58  \frac{4.5^2}{0.5}  = 23.49 N

5 0
3 years ago
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