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3 years ago

How much power is generated by a motor that can do 3,000j of work in 75s

Physics

1 answer:

puteri [66]3 years ago

6 0

Answer:

40watts

Explanation:

Given parameters:

Work done = 3000J

Time = 75s

Unknown:

Power generated = ?

Solution:

Power is the rate at which work is done;

Power = $\frac{work done}{time}$

Now, insert the parameters and solve;

Power = $\frac{3000}{75}$ = 40watts

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The correct answer is A: H3PO4

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3 years ago

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

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Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0

3 years ago

A toroid having a square cross section, 5.49 cm on a side, and an inner radius of 18.1 cm has 450 turns and carries a current of

joja [24]

Answer:

(a) 4.27 x 10^-4 Telsa

(b) 3.28 x 10^-4 Telsa

Explanation:

side of square, a = 5.49 cm

inner radius, r = 18.1 cm = 0.181 m

number of turns,N = 450

current, i = 0.859 A

(a)

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(b)

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