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2 years ago

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What is the length of the conductor within the magnetic field 0.01T if a current of 20 A flows in it and acts with a magnetic fo

rce of 0.005N.
F 0.005N
(1 = ———————— = 2.5 cm)
B1 0.01T * 20A

1 answer:

vladimir1956 [14]2 years ago

3 0

Answer:What is the length of the conductor within the magnetic field 0.01T if a current of 20 A flows in it and acts with a magnetic force of 0.005N.

F 0.005N

(1 = ———————— = 2.5 cm)

B1 0.01T * 20A

Explanation:

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2. What is the power rating of an engine capable of lifting a 100 kg object 5 m vertically

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Work=f.d
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3 years ago

A particle moves at a constant speed in a circular path with a radius of r=2.06 cm. If the particle makes four revolutions each

nataly862011 [7]

The centripetal acceleration is $13.0 m/s^2$

Explanation:

For an object in uniform circular motion, the centripetal acceleration is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the circle

The speed of the object is equal to the ratio between the length of the circumference ( $2\pi r$ ) and the period of revolution (T), so it can be rewritten as

$v=\frac{2\pi r}{T}$

Therefore we can rewrite the acceleration as

$a=\frac{4\pi^2 r}{T^2}$

For the particle in this problem,

r = 2.06 cm = 0.0206 m

While it makes 4 revolutions each second, so the period is

$T=\frac{1}{4}s = 0.25 s$

Substituting into the equation, we find the acceleration:

$a=\frac{4\pi^2 (0.0206)}{0.25^2}=13.0 m/s^2$

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2 years ago

Use ohms law to determine the battery voltage you would need to send 2.5 A of current through a light bulb with 3.6 Ω of resista

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2 years ago

Two men standing on the same side of wall and at the same distance from It, such that they are 4oom apart when one fires a gun t

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Answer:

1. 571.43m/s

2. 142.9m and 342.9m

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1.2-0.7=0.7 seconds

Take the distance between them and divide with differnce in time.

400÷0.7=571.43 seconds.

2.Take the time of the two men and divide by two.

0.5÷2= 0.25 secs

1.2÷2= 0.6 secs

multiply each with the velocity.

0.25×571.43=142.9m

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1 year ago

A spring of constant 20 N/m has compressed a distance 8 m by a(n) 0.3 kg mass, then released. It skids over a frictional surface

Fiesta28 [93]

Answer:

$X_2=25.27m$

Explanation:

Here we will call:

1. $E_1$ : The energy when the first spring is compress

2. $E_2$ : The energy after the mass is liberated by the spring

3. $E_3$ : The energy before the second string catch the mass

4. $E_4$ : The energy when the second sping compressed

so, the law of the conservations of energy says that:

1. $E_1 = E_2$

2. $E_2 -E_3= W_f$

3. $E_3 = E_4$

where $W_f$ is the work of the friction.

1. equation 1 is equal to:

$\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2$

where K is the constant of the spring, x is the distance compressed, M is the mass and $V_2$ the velocity, so:

$\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2$

Solving for velocity, we get:

$V_2$ = 65.319 m/s

2. Now, equation 2 is equal to:

$\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd$

where M is the mass, $V_2$ the velocity in the situation 2, $V_3$ is the velocity in the situation 3, $U_k$ is the coefficient of the friction, N the normal force and d the distance, so:

$\frac{1}{2}(0.3)(65.319)^2-\frac{1}{2}(0.3)V_3^2 = (0.16)(0.3*9.8)(2)$

Volving for $V_3$ , we get:

$V_3 = 65.27 m/s$

3. Finally, equation 3 is equal to:

$\frac{1}{2}MV_3^2 = \frac{1}{2}K_2X_2^2$

where $K_2$ is the constant of the second spring and $X_2$ is the compress of the second spring, so:

$\frac{1}{2}(0.3)(65.27)^2 = \frac{1}{2}(2)X_2^2$

solving for $X_2$ , we get:

$X_2=25.27m$

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2 years ago