A man pushes a 60.8 kg crate across a rough surface with an applied force of 125 N and at a CONSTANT SPEED.

You are raising up a big bucket of water from a 25.9 m deep well. The combined mass of the water and the bucket is 13.9 kg. The

Mars2501 [29]

The total work done is 5980 Joules and the power expended is 57 Watts.

The work done is the work done in the gravitational field as the bucket is raised up Thus work required to remove the bucket Wb;

Wb = 13.9 kg * 25.9 m * 9.8 m/s^2 = 3530 Joules

Height of the center of mass of chain = 25.9 / 2 = 12.95 m

Work done by the chain Wc;

Wc = 12.95 * 19.3 * 9.8 = 2450 Joules

Total work = 3530 + 2450 = 5980 Joules

Power expended = W / t = 5980 J / 105 sec = 57 J/s = 57 Watts

Learn more about work done:brainly.com/question/13662169

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3 0

1 year ago

Can someone help on this I'm really stuck

castortr0y [4]

Here, "Wavelength is same for both waves" it is the distance between two crests or two consecutive troughs, so, it is constant for both of them, you can easily figure it out.

In short, Your Answer would be "Wavelength"

Hope this helps!

5 0

3 years ago

A 0.500 kg rock is whirled in a vertical circle of a radius 0.60 m . the velocity of the rock at the bottom of the swing is 4.0

OLEGan [10]

Explanation:

Centripetal acceleration is:

a = v² / r

a = (4.0 m/s)² / 0.60 m

a = 26.6 m/s²

7 0

3 years ago

The net external force on a golf cart is 411 N north of tha cart has a total mass of 281 kg what is tha cart acceleration

Alisiya [41]

1.46m/s²

Explanation:

Given parameters:

Mass of cart = 281kg

Net external force = 411N

Unknown:

Acceleration of the cart = ?

Solution:

According to newton's second law " the net force on a car is the product of its mass and acceleration";

Force = mass x acceleration

Since acceleration of the cart is unknown;

Acceleration = $\frac{force}{mass}$ = $\frac{411}{281}$

Acceleration = 1.46m/s²

learn more:

Newton's laws brainly.com/question/11411375

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8 0

3 years ago

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

$\sum F = ma$

the resultant of the forces must be zero: $\sum F = 0$ (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

$F$ , the push applied by the worker

$F_f=-\mu mg$ , the force of friction, with $\mu=0.25$ being the coefficient of friction, $m=30.0 kg$ being the mass of the crate, and $g=9.8 m/s^2$ . The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

$F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

$W=Fd cos \theta$

where

F = 73.5 N is the force

d = 4.5 m is the displacement

$\theta=0^{\circ}$ is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

$W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J$

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

$W=F_f d cos \theta$

where

$F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$ is the magnitude of the force of friction

d = 4.5 m is the displacement

$\theta=180^{\circ}$ is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

$W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J$

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

$W=F_f d cos \theta$

We notice that both gravity and normal force are perpendicular to the displacement: therefore, $\theta=90^{circ}$ , and so

$cos \theta=0$

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

$W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0$

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

5 0

2 years ago