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1 year ago

The two forces to the right are friction which one would be the force of kinetic friction?

Physics

1 answer:

xeze [42]1 year ago

8 0

Answer:

40 N

Explanation:

Both forces to the right are friction forces. The 40 N force has to be the kinetic friction force because kinetic friction is always less than static friction. So, the 80 N force has to be the static friction force, which is always greater than the kinetic friction.

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A ray of light traveling in air strikes the surface of a liquid. if the angle of incidence is 29.7◦ and the angle of refraction

lana66690 [7]

When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted ray, and it is given by:
$\theta_c = \arcsin ( \frac{n_r}{n_i} )$ (2)
where $n_r$ is the refractive index of the second medium and $n_i$ is the refractive index of the first medium.

We can find the ratio $n_r / n_i$ by using Snell's law:
$n_i \sin \theta_i = n_r \sin \theta_r$ (1)
where
$\theta_i$ is the angle of incidence
$\theta_r$ is the angle of refraction

By using the data of the problem and re-arranging (1), we find
$\frac{n_r}{n_i} = \frac{\sin \theta_i}{\sin \theta_r} = \frac{\sin 16.3^{\circ}}{\sin 29.7^{\circ}} =0.566$

and if we use eq.(2) we can now find the value of the critical angle:
$\theta_c = \arcsin ( \frac{n_r}{n_i} ) = \arcsin (0.566) = 34.5^{\circ}$

3 0

3 years ago

Ten students stand in a circle and are told to make a transverse wave. What best describes the motion of the students? Each stud

Bond [772]

Lifting hands and the down by one student at a time best describe the presentation of the transverse wave by students. Option D is correct.

<h3>What is a Transverse wave?</h3>

The wave in which the oscillation of particles is is perpendicular to the direction of energy transfer.

The students can make a transverse wave by raising their hands up and then down, one student at a time.

The raised hand represents the oscillation of particles while the sequence of the raising hand represents the direction of energy transfer.

Therefore, lifting hands and the down by one student at a time best describe the presentation of the transverse wave by students.

Learn more about Transverse waves:

brainly.com/question/3813804

3 0

2 years ago

Read 2 more answers

A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the

sveticcg [70]

Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

$x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg$

Since the frictional force $f_k$ is equal to $f_k=\mu_k N=\mu_k mg$ , then we have that the acceleration is:

$a=\frac{\mu_k mg}{m}=\mu_k g$

Now, from the definition of acceleration we get:

$a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}$

And, as the final velocity is zero because the box gets to a stop, we have:

$t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}$

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

$t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s$

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

$v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}$

Finally, we calculate the displacement:

$x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm$

This means that the box moves 7.29cm backwards relative to the belt.

4 0

3 years ago

A 7.3 cm diameter loop of wire is initially oriented so that its plane is perpendicular to a magnetic field of 0.61 T pointing u

kenny6666 [7]

Answer:

induced emf = 28.65 mV

Explanation:

given data

diameter = 7.3 cm

magnetic field = 0.61

time period = 0.13 s

to find out

magnitude of the induced emf

solution

we know radius is diameter / 2

radius = 7.3 / 2

radius = 3.65 m

so induced emf is dπ/dt = Adb/dt

induced emf = A × ΔB / Δt

induced emf = πr² × ΔB / Δt

induced emf = π (0..65)² × ( 0.61 - (-0.28)) / 0.13

induced emf = 0.0286538 V

so induced emf = 28.65 mV

3 0

3 years ago

an automobile has a tire pressure of 325 kpa when the temperature is 10°C of the temperature rises to 50°C what is the new press

Finger [1]

Old temperature = 283 K.
New temp = 323 K.

(323/283) x (325 kPa) = 371 kPa.

8 0

3 years ago