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maria [59]
3 years ago
8

A uniform crate with a mass of 22 kg must be moved up along the 15° incline without tipping. The force P is horizontal. Determin

e the corresponding magnitude of force P.
Physics
1 answer:
Paul [167]3 years ago
6 0

Answer:

F_x=208.25\ N

Explanation:

Given that,

Mass of a crate is 22 kg

It moved up along the 15 degrees incline without tipping.

We need to find the corresponding magnitude of force P. The force P is acting in horizontal direction.

It means that the horizontal component of force is given by :

F_x=F\cos\theta\\\\F_x=mg\cos\theta\\\\F_x=22\times 9.8\times \cos(15)\\\\F_x=208.25\ N

So, the horizontal component of force is 208.25 N.

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a positively charged body makes contact with a body. after a while, the charged body becomes neutralised. state 3 main condition
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The average radius of Mars is 3,397 km. If Mars completes one rotation in 24.6 hours, what is the tangential speed of objects on
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25 times the average speed

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Light with energy equal to three times the work function of a given metal causes the metal to eject photoelectrons. What is the
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<h3 /><h3>What is the photoelectric effect?</h3>

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\rm KE=3 \phi

The ratio of the maximum photoelectron kinetic energy to the work function will be;

R=\frac{E}{\phi} \\\\ R=\frac{3 \phi}{\phi} \\\\ R= 3

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5 0
1 year ago
Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a
blagie [28]

Answer:

The maximum speed at which the car can safety travel around the track is 18.6m/s.

Explanation:

Since the car is in circular motion, there has to be a centripetal force F_c. In this case, the only force that applies for that is the static frictional force f_sbetween the tires and the track. Then, we can write that:

f_s=F_c

And since f_s\leq \mu N and F_c=\frac{mv^{2}}{r}, we have:

\mu N\geq \frac{mv^{2}}{r}

Now, if we write the vertical equation of motion of the car (in which there are only the weight and the normal force), we obtain:

N-mg=0\\\\\\implies N=mg

Substituting this expression for N and solving for v, we get:

\mu mg\geq \frac{mv^{2}}{r}\\\\v\leq \sqrt{\mu gr}

Finally, plugging in the given values for the coefficient of friction and the radius of the track, we have:

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