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3 years ago

How much power would be needed to lift a 1.0x10^3N crate 1.5m in 5 seconds?

Physics

2 answers:

yulyashka [42]3 years ago

6 0

Answer:

300 watts

Explanation:

Power = energy / time

First solve for energy

Energy = force x distance

Force = 1.0x10^3N

Distance = 1.5m

Energy = 1.0x10^3 x 1.5

= 1500j

Time = 5 secs

Therefore,

Power = 1500/5

300watts

lina2011 [118]3 years ago

5 0

Answer:power =300watts

Explanation:

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What is one of the principles of charges in a conductor?

nevsk [136]

Answer:

A conductor allows free charges to move about within it. The electrical forces around a conductor will cause free charges to move around inside the conductor until static equilibrium is reached. Any excess charge will collect along the surface of a conductor. Conductors with sharp corners or points will collect more charge at those points.

Explanation:

3 0

3 years ago

Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 2.88 kg and rotate with

Tanya [424]

Answer:

(a) $K_{small}=4839.3J$

(b) $K_{larger}=17406.4J$

Explanation:

Given data

The angular velocity of two cylinders ω=257 rad/s

The mass of the two cylinders m=2.88 kg

The radius of small cylinder r₁=0.319 m

The radius of larger cylinder r₂=0.605 m

For Part (a)

The rotational kinetic energy of the cylinder is given by:

$K=\frac{1}{2}Iw^2$

Where I is rotational of inertia of solid cylinder about its central axis.

$K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr^2)w^2$

Substitute the given values

$K_{small}=\frac{1}{4}(2.88kg)(0.319)^2(257rad/s)^2 \\K_{small}=4839.3J$

For Part (b)

$K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr_{2}^2)w^2$

Substitute the given values

$K_{larger}=\frac{1}{4}mr_{2}^2w^2\\ K_{larger}=\frac{1}{4}(2.88kg)(0.605m)^2(257rad/s)^2\\ K_{larger}=17406.4J$

8 0

3 years ago

What is the potential energy of a spring that is compressed 0.65 m by a 25 kg block if the spring constant is 95 N/m?

Sonbull [250]

E=F*d/2 = M*g*d/2 = 25kg * 9.8 N/kg *0.65m / 2 = 79.625J

7 0

3 years ago

Read 2 more answers

When the bug is stationary and creating waves, how does the frequency of the wave some distance away from the bug compare with t

Brrunno [24]

Answer:

The frequency is the same

Explanation:

When a wave is created by a source which is vibrating at a certain frequency, the frequency of the wave itself is equal to the frequency of the source.

This occurs with every kind of wave. For instance, if we consider the radio waves produced by an antenna, the frequency of the radio waves is equal to the frequency of the antenna.

In this case, the waves are created by the vibrating bug. The bug is vibrating with a certain frequency $f$ : as a consequence, the frequency $f'$ of the waves produced by the bug will be equal to the frequency of vibration of the bug:

$f'=f$ .

3 0

3 years ago

During a circus performance, a 72-kg humancannonball is shot out of an 18-m-long cannon. If thehuman cannonball spends 0.95 s in

andreyandreev [35.5K]

Answer:

2872.8 N

Explanation:

We have the following information

m =n72kg

Δy = 18m

t = 0.95s.

From here we use the equation

Δy=1/2at2 in order to solve for the acceleration.

So a

=( 2x 18m)/(0.95s²)

= 36/0.9025

= 39.9m/s2.

From there we use the equation

F = ma

F=(72kg) x (39.9)

= 2872.8N.

2872.8N is the average net force exerted on him in the barrel of the cannon.

Thank you!

7 0

3 years ago