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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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A box accidentally drops from a truck traveling at a speed of 20.0 m/s and slides along the ground for a distance of 30.0 m Calc
Hoochie [10]

Answer:

a= - 6.667 m/s²

Explanation:

Given that

The initial speed of the box ,u= 20 m/s

The final speed of the box ,v=  0 m/s

The distance cover by box ,s= 30 m

Lets take the acceleration of the box = a

We know that

v²= u ² + 2 a s

Now by putting the values in the above equation we get

0²=20² + 2 a x 30

a=- \dfrac{20^2}{2\times 30} \ m/s^2

a= - 6.667 m/s²

Negative sign indicates that velocity and acceleration are in opposite direction.

Therefore the acceleration of the box will be  - 6.667 m/s² .

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3 years ago
Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an
ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega

Power factor is given by

F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802

The power factor is 0.28802

The average power to the circuit is given by

P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W

The average power to the circuit is 2.57162 W

Power to resistor

P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W

Power to resistor is 14.28 W

Power to inductor

P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W

Power to the inductor is 53.55 W

Power to the capacitor

P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W

The power to the capacitor is 6.07142 W

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3 years ago
A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.
arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

  • Mass of the first puck = m_1\ =\ 5\ kg
  • Mass of the second puck = m_2\ =\ 3\ kg
  • initial velocity of the first puck = u_1\ =\ 3\ m/s.
  • Initial velocity of the second puck = u_2\ =\ -1.5\ m/s.

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = v_2\ =\ 2.31\ m/s.

Let v_1 be the final velocity of first puck after the collision.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = 25\times 10^{-3}\ sec

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

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