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4 years ago

A setting Sun appears red due to the

Physics

1 answer:

natulia [17]4 years ago

6 0

Answer:

a. scattering of lower frequencies by larger particles in the air.

b. light's longer path through the air at sunset.

Explanation:

Sun light has to travel several distances through the earth's atmosphere during Sun set and Sun rise. This light gets scattered by air molecules present and the shorter wave length blue gets upwards and red comes down. That is why the sun light appears red in our eyes.

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The block on this incline weighs 100 kg and is connected by a cable and pulley to a weight of 10 kg. If the coefficient of frict

blondinia [14]

Answer:

a. 94.54 N

b. 0.356 m/s^2

Explanation:

Given:-

- The mass of the inclined block, M = 100 kg

- The mass of the vertically hanging block, m = 10 kg

- The angle of inclination, θ = 20°

- The coefficient of friction of inclined surface, u = 0.3

Find:-

a) The magnitude of tension in the cable

b) The acceleration of the system

Solution:-

- We will first draw a free body diagram for both the blocks. The vertically hanging block of mass m = 10 kg tends to move "upward" when the system is released.

- The block experiences a tension force ( T ) in the upward direction due the attached cable. The tension in the cable is combated with the weight of the vertically hanging block.

- We will employ the use of Newton's second law of motion to express the dynamics of the vertically hanging block as follows:

$T - m*g = m*a\\\\$ ... Eq 1

Where,

a: The acceleration of the system

- Similarly, we will construct a free body diagram for the inclined block of mass M = 100 kg. The Tension ( T ) pulls onto the block; however, the weight of the block is greater and tends down the slope.

- As the block moves down the slope it experiences frictional force ( F ) that acts up the slope due to the contact force ( N ) between the block and the plane.

- We will employ the static equilibrium of the inclined block in the normal direction and we have:

$N - M*g*cos ( Q )= 0\\\\N = M*g*cos ( Q )$

- The frictional force ( F ) is proportional to the contact force ( N ) as follows:

$F = u*N\\\\F = u*M*g *cos ( Q )$

- Now we will apply the Newton's second law of motion parallel to the plane as follows:

$M*g*sin(Q) - T - F = M*a\\\\M*g*sin(Q) - T -u*M*g*cos(Q) = M*a\\$ .. Eq2

- Add the two equation, Eq 1 and Eq 2:

$M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g = a* ( M + m )\\\\a = \frac{M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g}{M + m} \\\\a = \frac{100*9.81*sin ( 20 ) - 0.3*100*9.81*cos ( 20 ) - 10*9.81}{100 + 10}\\\\a = \frac{-39.12977}{110} = -0.35572 \frac{m}{s^2}$

- The inclined block moves up ( the acceleration is in the opposite direction than assumed ).

- Using equation 1, we determine the tension ( T ) in the cable as follows:

$T = m* ( a + g )\\\\T = 10*( -0.35572 + 9.81 )\\\\T = 94.54 N$

4 0

3 years ago

Which of the following is NOT an example of work?

Gnom [1K]

C. Cause your just standing in line holding a tray.

8 0

4 years ago

A child rides her bike at a rate of 12.0 km/hr down the street. A squirrel suddenly runs in front of her so she applies the brak

laiz [17]

Answer:

C: 16 km/hr/sec

Explanation:

3 0

3 years ago

What happens to a light if the switch is turned off in the circuit?

lukranit [14]

In a parallel circuit nothing happens. All the other lights stay on but if it was a series circuit ALL the lights would turn off at one time.

If there is an ABCD question I can answer that but I wasn't sure so I just said this

Hope it is helpful :)

6 0

3 years ago

When the magnetic flux through a single loop of wire increases by , an average current of 40 A is induced in the wire. Assuming

Zielflug [23.3K]

COMPLETE QUESTION:

When the magnetic flux through a single loop of wire increases by 30 Tm^2 , an average current of 40 A is induced in the wire. Assuming that the wire has a resistance of 2.5 ohms , (a) over what period of time did the flux increase? (b) If the current had been only 20 A, how long would the flux increase have taken?

Answer:

(a). The time period is 0.3s.

(b). The time period is 0.6s.

Explanation:

Faraday's law says that for one loop of wire the emf $\varepsilon$ is

$(1). \: \: \varepsilon = \dfrac{\Delta \Phi_B}{\Delta t }$

and since from Ohm's law

$\varepsilon = IR$ ,

then equation (1) becomes

$(2). \: \:IR= \dfrac{\Delta \Phi_B}{\Delta t }.$

(a).

We are told that the change in magnetic flux is $\Phi_B = 30Tm^2$ , the current induced is $I = 40A$ , and the resistance of the wire is $R = 2.5\Omega$ ; therefore, equation (2) gives

$(40A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

which we solve for $\Delta t$ to get:

$\Delta t = \dfrac{30Tm^2}{(40A)(2.5\Omega)},$

$\boxed{\Delta t = 0.3s}$ ,

which is the period of time over which the magnetic flux increased.

(b).

Now, if the current had been $I =20A$ , then equation (2) would give

$(20A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

$\Delta t = \dfrac{30Tm^2}{(20A)(2.5\Omega)},$

$\boxed{\Delta t = 0.6 s\\}$

which is a longer time interval than what we got in part a, which is understandable because in part a the rate of change of flux $\dfrac{\Delta \Phi_B}{\Delta t}$ is greater than in part b, and therefore , the current in (a) is greater than in (b).

7 0

3 years ago