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Alexeev081 [22]
3 years ago
5

A wire carrying a current of 10 A and 2 m in length is placed in a field of flux density 0.15 T. What’s the force on the wire if

it is placed (a) at right angles to the field, (b) at 450 to the field, (c) along the field​
Physics
1 answer:
saul85 [17]3 years ago
7 0

Explanation:

I = 10A

l = 2m

B = 0.15T

F = ?

a) ¶ = 90

F = BILsin¶

F = 0.15×10×2×sin90

F = 3N

b) ¶ = 45 degree

F = BILsin¶

F = 0.15×10×2×sin45

F = 2.12N

c) ¶ = 0 degree

F = BILsin¶

F = 0.15×10×2×sin0

F = 0

Goodluck

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olga nikolaevna [1]

Answer:

B. Marginal cost equals long-run average total cost.

Explanation:

The zero profit condition implies that entry continues until all firms are producing at minimum long run average total cost. Since the marginal cost curve cuts the long run average total cost curve at its minimum point, marginal cost and long run average total cost must be equal in long run equilibrium.

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A skateboarder is moving 5.25 m/s when he starts to roll up a frictionless hill. How much higher is he when his velocity has red
Taya2010 [7]
Mechanical energy E = mgh + 1/2mv²

When he starts, let h = 0 ⇒ E₁ = 1/2mv₁²
When he reaches height h ⇒ E₂ = mgh + 1/2mv₂²

Without friction, energy is conserved at all times.

E₁ = E₂
     ↓
1/2mv₁² = mgh + 1/2mv₂²
     ↓
1/2v₁² = gh + 1/2v₂²
     ↓
gh = 1/2(v₁² - v₂²)
    ↓
h = (v₁² - v₂²) / (2g)


5 0
3 years ago
Read 2 more answers
Part B For this activity, you'll analyze the motion of the marked spot on the skateboarder's leg. For each dropdown labeled “cm”
Anika [276]

Answer:

Explanation:

c. By using the Select Data button and the Select Data Source optionExplanation:A scatter plot is a plot which is used to plot the points of the data on the horizontal and the vertical axis also it depicts how one variable is affected by the another. After preparing the scatter plot to enter the data in the scatter plot we need to use the data button and then data source option so that the data could be entered in the scatter plothence, option c is correct

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2 years ago
PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer this correctly!!!! (40pts)
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2 years ago
A force of 44 N will stretch a rubber band 88 cm ​(0.080.08 ​m). Assuming that​ Hooke's law​ applies, how far will aa 11​-N forc
Setler79 [48]

Answer:

<em>The rubber band will be stretched 0.02 m.</em>

<em>The work done in stretching is 0.11 J.</em>

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = <em>0.02 m</em>

<em>The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch</em>. This is in line with energy conservation.

potential energy stored = \frac{1}{2}ke^{2}

==> \frac{1}{2}* 550* 0.02^{2} = <em>0.11 J</em>

3 0
3 years ago
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