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3 years ago

5. A wire, resistance 23.6 carries a current. 5.9 A. What is the

Physics

1 answer:

posledela3 years ago

3 0

answer : p.d = 23.6 x 5.9 =139.14

Explanation:

p.d = I x r

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In which type of circuit does charge move in only one direction?

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C. A combined circuit
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Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude o

Anna [14]

The magnitude of the tension in the string marked A is 52.5N

Generally, the equation for is mathematically given as

Let's take θ be an angle at A

So, tanθ = 3/8

Let's take α be an angle at B (Below X)

tanα = 5/4

Let's take β be an angle at C (Below x)

tanβ = 1/6

First we take the Horizontal Components

74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)

By solving the equation, we get

A = 78.9 - 0.668B … (1)

Now, we take the vertical components

74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)

By solving the equation, we get

40.07 = 1.015B

B = 39.5N

By substituting the value of B in equation (1)

A = 78.9 - 0.6668× 39.5

A = 52.5N

Hence, the magnitude of the tension in the string marked A is 52.5N

Learn more about Tension here brainly.com/question/2287912

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Energy of motion is known as what energy

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This problem has been solved! See the answer A 21.5 kg person climbs up a uniform 85 N ladder. The upper end of the ladder rests

xenn [34]

Answer:

μ = 0.179

Explanation:

For this exercise we use the rotational equilibrium condition, where we set a reference system at the top of the ladder where it is in contact with the parent, we will assume the counterclockwise rotations as positive

∑ τ = 0

W_stairs x/2 + W_man x’+ fr y - N x = 0

fr y = -W_stairs x/2 - W_man x’ + N x

let's use trigonometry for distances

* the man up the ladder 78%

l ’= 0.78 L

cos 59 = x ’/ l’

x’= 0.78 L cos 59

* the horizontal distance

cos 59 = x / L

x = L cos 59

* vertical distance

sin 59 = Y / L

y = L sin 59

we substitute

fr L sin 59 = -W_stairs Lcos59 / 2 - W_man 0.78 L cos59 + N L cos59

fr sin 59 = - ½ W_stairs cos 59 - 0.78 W_man cos 59 + N cos 59

fr = ctan 59 (N - ½ W_stairs - 0.78 W_man) (1)

Let's write the translational equilibrium equations

Y axis

N -W_stairs - W_man = 0

N = W_stairs + W_man

N = 85 + 21.5 9.8

N = 295.7 N

we substitute in 1

fr = ctg 59 (295.7 - ½ 85 - 0.78 21.5 9.8)

fr = 0.6 (295.7 - 42.5 - 164.346)

fr = 52.89 N

the expression for the friction force is

fr = μ N

μ = fr / N

μ = 52.89 / 295.7

μ = 0.179

8 0

3 years ago