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uranmaximum [27]
3 years ago
8

A total work of 3200 J is done to lift a container a distance of 2.40 m. What was the weight of the container? What was the mass

of the container, in kilograms?
Physics
1 answer:
Diano4ka-milaya [45]3 years ago
5 0

1) The weight of the object is 1333 N

2) The mass of the object is 136 kg

Explanation:

1)

The work done when lifting an object is equal to the change in gravitational potential energy of the object:

W=Fd

where

F is the weight of the object

d is the distance through which the object has been moved

In this problem, we know that

W = 3200 J is the work done

d = 2.40 m is the distance through which the object has been lifted

Solving for F, we find the weight of the object:

F=\frac{W}{d}=\frac{3200}{2.40}=1333 N

2)

The weight of an object is given by

F=mg

where

F is the weight

m is the mass of the object

g is the acceleration of gravity

In this problem, we know that

F = 1333 N is the weight of the container

g=9.8 m/s^2 is the acceleration of gravity

Solving for m, we find the mass of the container:

m=\frac{F}{g}=\frac{1333}{9.8}=136.0 kg

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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2 years ago
A pitcher throws an overhand fastball from an approximate height of 2.65 m and at an angle of 2.5° below horizontal. The catcher
rodikova [14]

Answer:

The initial velocity of the pitch is approximately 36.5 m/s

Explanation:

The given parameters of the thrown fastball are;

The height at which the pitcher throws the fastball, h₁ = 2.65 m

The angle direction in which the ball is thrown, θ = 2.5° below the horizontal

The height above the ground the catcher catches the ball, h₂ = 1.02 m

The distance between the pitcher's mound and the home plate = 18.5 m

Let 'u' represent the initial velocity of the pitch

From h = u_y·t + 1/2·g·t², we have;

u_y = The vertical velocity = u·sin(θ) = u·sin(2.5°)

h = 2.65 m - 1.02 m = 1.63 m

uₓ·t = u·cos(θ) = u·cos(2.5°) × t = 18.5 m

∴ t = 18.5 m/(u·cos(2.5°))

∴ h = u_y·t + 1/2·g·t² =  (u·sin(2.5°))×(18.5/(u·cos(2.5°))) + 1/2·g·t²

1.63 = 8.5·tan(2.5°) + 1/2 × 9.8 × t²

t² = (1.63 - 8.5·tan(2.5°))/(1/2 × 9.8) = 0.25691469087

t = √(0.25691469087) ≈ 0.50686752763

t ≈ 0.50686752763 seconds

u = 18.5 m/(t·cos(2.5°)) = 18.5 m/(0.50686752763 s × cos(2.5°)) = 36.5334603 m/s ≈ 36.5 m/s

The initial velocity of the pitch = u ≈ 36.5 m/s.

3 0
3 years ago
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