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Mila [183]
3 years ago
5

Need help ASAP please Thanks

Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
3 0
Answer: D
Explanation: there is less light at that point.
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Which would you sprinkle around the electromagnet that you produced in the laboratory activity to see its magnet
nadya68 [22]

The Answer are:

1. iron shavings

2. compass

3. galvanometer

Give me Brainliest!!!

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3 years ago
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An elephant pushes with 200 N on a load of trees. it then pushes these trees for 10 N. How much work did the elephant do?
Andreas93 [3]

Answer:

the answer is 2000Nm

Explanation:

wprk done = force × distance moved

w.d = 200N × 10m

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4 years ago
Two long, parallel wires separated by 3.50 cm carry currents in opposite directions. The current in one wire is 1.55 A, and the
vaieri [72.5K]

Answer:

Therefore,

The magnitude of the force per unit length that one wire exerts on the other is

\dfrac{F}{l}=2.79\times 10^{-5}\ N/m

Explanation:

Given:

Two long, parallel wires separated by a distance,

d = 3.50 cm = 0.035 meter

Currents,

I_{1}=1.55\ A\\I_{2}=3.15\ A

To Find:

Magnitude of the force per unit length that one wire exerts on the other,

\dfrac{F}{l}=?

Solution:

Magnitude of the force per unit length on each of @ parallel wires seperated by the distance d and carrying currents I₁ and I₂ is given by,

\dfrac{F}{l}=\dfrac{\mu_{0}\times I_{1}\times I_{2}}{2\pi\times d}

where,

\mu_{0}=permeability\ of\ free\ space =4\pi\times 10^{-7}

Substituting the values we get

\dfrac{F}{l}=\dfrac{4\pi\times 10^{-7}\times 1.55\times 3.15}{2\pi\times 0.035}

\dfrac{F}{l}=2.79\times 10^{-5}\ N/m

Therefore,

The magnitude of the force per unit length that one wire exerts on the other is

\dfrac{F}{l}=2.79\times 10^{-5}\ N/m

7 0
3 years ago
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A large flat block of mass 7 m rests on a level, smooth tabletop (µk ≈ 0). On top of it is a much smaller block of mass m, with
mariarad [96]

Answer:

3. µs g /7

Explanation:

The largest Force appear when the maximal friction Force is required.

Second Newton law for the small block:

F_friction=u_s*N=u_s*(mg)

F-F_friction=ma

F-u_s*(mg)=ma

Second Newton law for the Big Block:

F_friction=7ma

u_s*(mg)=7ma

a=u_s*g/7

8 0
3 years ago
A car has an acceleration of -5 m/s^2. Describe the car’s motion
AlladinOne [14]

Explanation:

because the acceleration is negative, this indicates a deceleration (or slowing down) . Hence we can say that:

The car is decelerating (slowing down), i.e its velocity is decreasing, at a constant rate of 5m/s².

7 0
4 years ago
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