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3 years ago

12

g An inductor used in a dc power supply has an inductance of 12.0 H and a resistance of It carries a current of 0.300 A. (a) Wha

t is the energy stored in the magnetic field

1 answer:

valkas [14]3 years ago

4 0

Answer:

Explanation:

Energy of an inductor = 1/2 L i²

L is inductance , i is current .

= 1/2 x 12 x .3²

= .54 J

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According to the Big Bang Theory, the density and temperature of the Universe is ___________ now than in the past.

34kurt

Answer:

According to the Big Bang Theory, the density and temperature of the Universe is <u>lower</u> now than in the past.

<em>Hope</em><em> this</em><em> answer</em><em> correct</em><em> </em><em>:</em><em>)</em>

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3 years ago

A motorcycle that is slowing down uniformly. The motorcycle covers 1 ????m=1000 m in 80 sec⁡. The motorcycle then covers the nex

Lynna [10]

Answer:

Part a)

acceleration = -0.042 m/s/s

Part b)

initial speed = 14.17 m/s

final speed = 5.77 m/s

Explanation:

Part a)

Let the initial velocity of the motorcycle is

$v_i = v_o$

now at the end of 80 s let the speed is

$v_f = v_1$

after another 120 s let the speed will be

$v_f' = v_2$

now we know that

$d = \frac{v_i + v_f}{2} (t)$

$d = \frac{v_o + v_1}{2}(80)$

$1000 = 40(v_o + v_1)$

also we know that

$v_1 - v_o = a(80)$

also we have

$1000 = \frac{v_1 + v_2}{2}(120)$

$1000 = 60(v_1 + v_2)$

now we can say

$(v_2 + v_1) - (v_o + v_1) = \frac{50}{3} - \frac{50}{2}$

also we know

$v_2 - v_o = a(120 + 80)$

$-8.33 = 200 a$

$a = -0.042 m/s^2$

Part b)

now we have

$v_1 + v_o = 25$

$v_1 - v_o = (-0.042)(80)$

$v_1 = 10.83 m/s$

so the starting velocity of the trip is

$v_o = 25 - 10.83 = 14.17 m/s$

now speed after t = 200 s is given as

$v_2 = v_o + at$

$v_2 = 14.17 - (0.042)(200)$

$v_2 = 5.77 m/s$

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3 years ago

A firecracker in a coconut blows the coconut into three pieces. two pieces of equal mass fly off south and west, perpendicular t

netineya [11]

Here we have mass that moves at ceratin speed. This means that we have momentum. The law that must be observed is law of conservation of momentum. It states that momentum before certain event is equal to a momentum after that event. Here we have three masses so we can write this as:
$m_{1} v_{1i} + m_{2} v_{2i} + m_{3} v_{3i} = m_{1} v_{1f} + m_{2} v_{2f} + m_{3} v_{3f}$
Before the firecracker blows a coconut does not move, so left side is equal to 0:
$0 = m_{1} v_{1f} + m_{2} v_{2f} + m_{3} v_{3f}$

We know that m1=m2=m and m3=2m. Also we are asked to find v3f so we can rewrite formula:
$v_{3f} = - \frac{m_{1} v_{1f} + m_{2} v_{2f} }{ m_{3} }$

We must take in consideration that two parts with same mass do not move in same direction. The center of mass of these two parts moves between them at angle of 45° with respect to both south and west. The speed of a center of mass is:
$v_{f} = \sqrt{ v_{1f}^{2}+ v_{2f}^{2} } \\ \\ v_{f} = 33.9m/s$
This speed we can insert into formula for v3f:
$v_{3f} = - \frac{m*33.9+m*33.9 }{ 2m } \\ \\ v_{3f} = - \frac{2m*33.9 }{ 2m } \\ \\ v_{3f} = - 33.9m/s$

We can see that part of a coconut with biggest mass has same speed as center of mass of two other parts. Negative sign shows that direction is opposite to direction of two pats. Biggest part moves towards north-east.

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3 years ago

Read 2 more answers

On a very muddy football field, a 110kg linebacker tackles an 85kg halfback. Immediately before the collision, the linebacker is

Alchen [17]

<span>Using conservation of energy and momentum you can solve this question. M_l = mass of linebacker
M_ h = mass of halfback
V_l = velocity of linebacker
V_h = velocity of halfback

So for conservation of momentum,
rho = mv

M_l x V_li + M_h x V_hi = M_l x V_lf + M_h x V_hf

For conservation of energy (kinetic)
E_k = 1/2mv^2/ 1/2mV_li^2 + 1/2mV_{hi}^2 = 1/2mV_{lf}^2 + 1/2mV_{hf}^2

Where i and h stand for initial and final values.
We are already told the masses, \[M_l = 110kg\] \[M_h = 85kg\] and the final velocities \[V_{fi} = 8.5ms^{-1}\] and \[V_{ih} = 7.2ms^{-1} </span>

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3 years ago

THE RELATIVE ANGLE AT THE KNEE CHANGES FROM 0O TO 85O DURING THE KNEE FLEXION PHASE OF A SQUAT EXERCISE. IF 10 COMPLETE SQUATS A

Mice21 [21]

Answer: angular distance = 1700° and 29.7 rad

also the angular displacement = 0

Explanation:

To explain this, i will give a breakdown of this works.

we are asked to find both the angular distance and displacement the knee undergo.

Ok to get the distance of the knee, we would first take note that for one to squat down and get back up, the knee would travel through 85° of flexion to gpo down, and also through another 85° of extension to return standing (upright). So, the actual angular distance of the squat is 170°.

taking ten squats, the knee would have to go through 170° motion times 10 i.e;

10 * 170° = 1700°

Therefore the angulaar distance is 1700°

now converting this distance to radians since we will be required to have our answer in both degree and rad.

Given that 2pi = 360°, it means that one degree will give 57.3°;

∅ (rad) = ∅ (deg) * 2π/360°

∅ (rad) = 1170° * 2π/360° = 29.7 rad

∅ (rad) = 29.7 rad

b. For the other part, let us remember that angular displacement is equal to angular distance divided by time, so the angular displacement displacement of the knee will be zero, because the knee's position at the final third will be the same as the initial position.

cheers i hope this helps!!!!

π

5 0

3 years ago