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Alinara [238K]
3 years ago
9

1. A wall is made up of four elements, as follows:

Physics
1 answer:
balu736 [363]3 years ago
8 0

Part A)

Since all four resistors are made up of different elements and connected in series

SO in series the net resistance is given by

R = R_1 + R_2 + R_3 + R_4

R = 0.81 + 0.62 + 10.9 + 0.45

R = 12.78

Part B)

If insulation is removed then resistance due to three materials

R = R_1 + R_2 + R_3

R = 0.81 + 0.62 + 0.45

R = 1.88

Part c)

Heat will always flow from High temperature to low temperature

So here heat flow is from 90 degree F to 70 degree F

so it will flow from outside to inside

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Explain Archimedes Principle
rusak2 [61]

Answer:

Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces. Archimedes' principle is a law of physics fundamental to fluid mechanics. It was formulated by Archimedes of Syracuse.

Explanation:

6 0
2 years ago
In a single movable pulley, a load of 500 N is lifted by applying 300 N effort. Calculate MA, VR and efficiency.​
galben [10]

Answer:

in pulley there are different kinds.but the most common one are fixed,moveable and compound pulley. in this question we asked about movable pulley.

Explanation:

Given request solutions

L=500N a,M.A=? a,M.A=L/E =5/3

E=300N b,V.R=? b,V.R=2

c,efficiency =? c,£=M.A/V.R=5/6

£=IS NOT THE REAL SYMBOLS OF EFFICIENCY BUT I IS LOOK LIKE THIS

I THINK I HELPED

6 0
2 years ago
PLEASE HELP ME ON THIS ONE ANYONE
andreev551 [17]

Answer:

A & B

Explanation:

A & B Would be the right answer since Morse code cannot be represented through the height of the fire.

6 0
2 years ago
A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.
kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

W=Fd cos \theta

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

\theta=30^{\circ} is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

\Delta K = W

where

\Delta K is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

\Delta K=69.3 J

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, \Delta K, after it has travelled for d=3 m. Using the work-energy theorem again, we find

\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J

And substituting into (1) and re-arrangin the equation, we find

v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s

6 0
3 years ago
Photons of wavelength 65.0 pm are Compton-scattered from a free electron which picks up a kinetic energy of 0.84 keV from the co
ElenaW [278]

Answer:

λ  = 65.6 pm

Explanation:

Given that

λo = 65 pm

The initial energy of the electron

E_o=\dfrac{hC}{\lambda_0 }

Now by putting the values

E_o=\dfrac{hC}{\lambda_0 }

E_o=\dfrac{6.67\times 10^{-34}\times 3\times 10^8}{65\times 10^{-12}}

E_o=3.05\times 10^{-15}\ J

E_o=\dfrac{3.05\times 10^{-15}}{1.6\times 10^{-19}\times 10^3}\ KeV

Eo=19.06 KeV

Given that kinetic energy KE= 0.84 KeV

Therefore the final energy

E= Eo - KE

E = 19.06 - 0.84 KeV

E= 18.22 KeV

The wavelength  λ can be find as

E=\dfrac{hC}{\lambda}

\lambda=\dfrac{hC}{E}

\lambda=\dfrac{6.67\times 10^{-34}\times 3\times 10^8}{19.06 \times 10^3\times 1.6\times 10^{-19}}

λ = 6.56 x 10⁻¹¹ m

λ  = 65.6 pm

3 0
3 years ago
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