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hjlf
2 years ago
14

Look at diagram above. In what direction does the medium move relative to the direction of the wave? Explain.

Physics
1 answer:
Ainat [17]2 years ago
7 0
Answer:
It goes upward and downward
Explanation:
The way the man is moving the string indicates it goes movement of up and sown, I believe the straight line(B) is a distraction.
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Describe the movement of the man when the resultant horizontal force is ON
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Answer:

Explanation:

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3 years ago
What is the relationship between the strength of an
Anastaziya [24]

Answer:

As the number of turns in the coil increases, the strength  of the electromagnet increases.

Explanation:

When current flows through a coil the coil behaves as an electromagnet. The strength of electromagnet depend the amount of current, no of turns of coil and the core of coil.

B=μ₀ N I

μ₀ = permeability of the core

N = Number of turns of the coil

I = Current flowing through the coil

Increasing the current and number of coils increase the strength of electromagnet.

3 0
3 years ago
A person stands at the base of a hill that is a straight incline making an angle φ with the horizontal. For a given initial spee
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Answer:

θ = sin⁻¹\sqrt{2gd}

Explanation:

From one of the equations of motion, v² = u² + 2as.......... equation 1

Since the object thrown was moving against gravity, then the acceleration, a would change to -g and the initial velocity u would change to V₀ sin θ because the object is travelling at angle of θ to the horizontal. By inputting all these parameter into equation 1, we would arrive at:

v² = (u sin θ)² - 2gd

(u sin θ)² = 2gd

d = (u sin θ)²/2g

sin² θ = 2gd

sin θ = \sqrt{2gd}

θ = sin⁻¹ \sqrt{2gd}

4 0
3 years ago
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the less shielding of electrons

6 0
3 years ago
Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 50 W/m · K and
tatuchka [14]

Answer:

solution:

dT/dx =T2-T1/L

&

q_x = -k*(dT/dx)

<u>Case (1)  </u>

dT/dx= (-20-50)/0.35==> -280 K/m

 q_x  =-50*(-280)*10^3==>14 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

 q_x  =-50*(80)*10^3==>-4 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

 q_x  =-50*(80)*10^3==>-4 kW

Case (3)

q_x  =-50*(160)*10^3==>-8 kW

T2=T1+dT/dx*L=70+160*0.25==> 110° C

Case (4)

q_x  =-50*(-80)*10^3==>4 kW

T1=T2-dT/dx*L=40+80*0.25==> 60° C

Case (5)

q_x  =-50*(200)*10^3==>-10 kW

T1=T2-dT/dx*L=30-200*0.25==> -20° C

note:

all graph are attached

6 0
3 years ago
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