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dsp73
1 year ago
12

A crowbar 27 in. long is pivoted 8 in. from the end. What force must be applied at the long end in order to lift a 600 lb object

at the short end?
Physics
1 answer:
Fynjy0 [20]1 year ago
5 0

pivot = 8 in

Balance lenght = 27 in - 8 in = 19 in

Load x load arm = Force x force arm

Replacing:

600 lb * 8 in = F * 19

Solve for F

F = 600 * 8 / 19

F= 252.6 lb

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60ml

Explanation:

I'm going to assume you mean 210ml not centimeters. To find the volume all we do is subtract both values or with the formula [ f - i = v ]  where f = final amount and i = initial amount.

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A projectile is launched horizontally at a speed of 45.0 m/s from a
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2 years ago
Shawn uses 45 N of force to stop the cart 27 meter from running his foot over. How much work does he do?
Ede4ka [16]

Answer:

W=f×d

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3 0
3 years ago
A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it w
Bumek [7]

Answer:

Explanation:

Given that,

Mass attached m = 0.95kg

Spring constant k = 16N/m

Instantaneous speed v = 36cm/s = 0.36m/s

Amplitude A=?

When x = 0.7A

Using conservation of energy

∆K.E + ∆P.E = 0

K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0

At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.

Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0

So, the equation becomes

— K.E(initial) + P.E(final) = 0

K.E(initial) = P.E(final)

½mv² = ½kA²

mv² = kA²

0.95 × 0.36² = 16×A²

0.12312 = 16•A²

A² = 0.12312/16

A² = 0.007695

A = √0.007695

A = 0.088 m

A = 8.8cm

B. Speed at x = 0.7A

Using the same principle above

K.E(initial) = P.E(final)

½mv² = ½kA²

Where A = 0.7A = 0.7 × 0.088 = 0.0614m

Then,

½× 0.95 × v² = ½ × 16 × 0.0614²

0.475v² = 0.0310644

v² = 0.0310644/0.475

v² = 0.0635

v = √0.0635

v = 0.252 m/s

v = 25.2 cm/s

8 0
3 years ago
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