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Anon25 [30]
3 years ago
8

After a wave passes through a medium, particles in the medium

Physics
1 answer:
sertanlavr [38]3 years ago
3 0

Answer:

If a particle is affected by a wave, then the particles are displaced. They move along the direction of the wave. Hence, After a wave passes through a medium, particles in the medium are moved along with the wave.

Explanation:

hope this helps you

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The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

5 0
3 years ago
The gravitational attraction between a 20 kg cannonball and a 0.002 kg
Naya [18.7K]

Answer:

2.966\times 10^{-11}\ N

Explanation:

Given:

Mass of the cannonball (M) = 20 kg

Mass of the marble (m) = 0.002 kg

Distance between the cannonball and marble (d) = 0.30 m

Universal gravitational constant (G) = 6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2}

Now, we know that, the gravitational force (F) acting between two bodies of masses (m) and (M) separated by a distance (d) is given as:

F=\dfrac{GMm}{d^2}

Plug in the given values and solve for 'F'. This gives,

F=\frac{(6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2})\times (20\ kg)\times (0.002\ kg)}{(0.30\ m)^2}\\\\F=\frac{6.674\times 20\times 0.002\times 10^{-11}\ m^3 kg^{-1+2} s^{-2}}{0.09\ m^2}\\\\F=2.966\times 10^{-11}\ kg\cdot m\cdot s^{-2}\\\\F=2.966\times 10^{-11}\ N.........(1\ N = 1\ kg\cdot m\cdot s^{-2})

The same force is experienced by both cannonball and marble.

Therefore, the gravitational  force of the marble is 2.966\times 10^{-11}\ N

3 0
3 years ago
Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could als
ankoles [38]

Answer: B. (positive; negative; positive)

5 0
3 years ago
Calculate the Reynolds number for an oil gusher that shoots crude oil 25.0 m into the air through a pipe with a 0.100-m diameter
Alenkasestr [34]

Answer:

Re=1992.24

Explanation:

Given:

vertical height of oil coming out of pipe, h=25\ m

diameter of pipe, d=0.1\ m

length of pipe, l=50\ m

density of oil, \rho = 900\ kg.m^{-3}

viscosity of oil, \mu=1\ Pa.s

Now, since the oil is being shot verically upwards it will have some initial velocity and will have zero final velocity at the top.

<u>Using the equation of motion:</u>

v^2=u^2-2gh

where:

v = final velocity

u = initial velocity

Putting the respective values:

0^2=u^2-2\times 9.8\times 25

u=22.136\ m.s^{-1}

<u>For Reynold's no. we have the relation as:</u>

Re=\frac{\rho.u.d}{\mu}

Re=\frac{900\times 22.136\times 0.1}{1}

Re=1992.24

6 0
3 years ago
Which statement about types of waves and energy is correct?
evablogger [386]

Answer:

I think it's D and good luck

6 0
3 years ago
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