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3 years ago

Example sentence for gravitational potential energy

Physics

1 answer:

andreev551 [17]3 years ago

5 0

Answer:

Explanation:

An object falling loses gravitational potential energy and gains kinetic energy. The gravity potential is the gravitational potential energy per unit mass. This energy comes from the gravitational potential energy released when the water falls. ... At 0, all the energy is in gravitational potential energy.

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A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a

Rasek [7]

Answer:

$W=173.48J$

Explanation:

information we know:

Total force: $F=45N$

Weight: $w=100N$

distance: $4m$

vertical component of the force: $F_{y}=12N$

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: $F_{x}=Fcos\theta$

vertical component: $F_{y}=Fsen\theta$

but from the given information we know that $F_{y}=12N$

so, equation these two $F_{y}=Fsen\theta$ and $F_{y}=12N$

$Fsen\theta =12N$

and we know the force $F=45N$ , thus:

$45sen\theta=12$

now we clear for $\theta$

$sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466$

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

$F_{x}=Fcos\theta$

$F_{x}=45cos(15.466)\\F_{x}=43.37N$

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

$W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J$

the work done is W=173.48J

7 0

3 years ago

A 40 g ball rolls around a 30 cm -diameter L-shaped track, shown in the figure, (Figure 1)at 60 rpm . What is the magnitude of t

levacccp [35]

Answer:

0.47 N

Explanation:

Here we have a ball in motion along a circular track.

For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.

This force is called centripetal force, and its magnitude is given by:

$F=m\omega^2 r$

where

m is the mass of the object

$\omega$ is the angular velocity

r is the radius of the circle

For the ball in this problem we have:

m = 40 g = 0.04 kg is the mass of the ball

$\omega =60 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=6.28 rad/s$ is the angular velocity

r = 30 cm = 0.30 m is the radius of the circle

Substituting, we find the force:

$F=(0.040)(6.28)^2(0.30)=0.47 N$

3 0

3 years ago

At t = 0, one toy car is set rolling on a straight track with initial position 13.0 cm, initial velocity -3.6 cm/s, and constant

Klio2033 [76]

Answer:

that's too much to read

8 0

3 years ago

A mass free to vibrate on a level, frictionless surface at the end of a horizontal spring is pulled 35 cm from its equilibrium p

saul85 [17]

Answer:

0.67 s

Explanation:

This is a simple harmonic motion (SHM).

The displacement, $x$ , of an SHM is given by

$x = A\cos(\omega t)$

A is the amplitude and $\omega$ is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or $\frac{\pi}{4}$ radian.