How do I answer this question?

A 30 g horizontal metal bar, 13 cm long, is free to slide up and down between two tall, vertical metal rods that are 13 cm apart

natita [175]

Answer:

Terminal speed, v = 6901.07 m/s

Explanation:

It is given that,

Mass of the horizontal bar, m = 30 g = 0.03 kg

Length of the bar, l = 13 cm = 0.13 m

Magnetic field, $B=5.5\times 10^{-2}\ T$

Resistance, R = 1.2 ohms

We need to find the terminal speed oat which the bar falls. When terminal speed is reached,

Force of gravity = magnetic force

$mg=ilB$ ..................(1)

i is the current flowing

l is the length of the rod

Due to the motion in rods, an emf is induced in the coil which is given by :

$E=Blv$ , v is the speed of the bar

$iR=Blv$

$i=\dfrac{Blv}{R}$

Equation (1) becomes,

$mg=\dfrac{B^2l^2v}{R}$

$v=\dfrac{mgR}{B^2l^2}$

$v=\dfrac{0.03\times 9.8\times 1.2}{(5.5\times 10^{-2})^2(0.13)^2}$

v = 6901.07 m/s

So, the terminal speed at which the bar falls is 6901.07 m/s. Hence, this is the required solution.

5 0

3 years ago

Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.8 cm and a current of 12 A . The bigger l

Stolb23 [73]

Answer:

$R_2=6.33cm$

Explanation:

The equation for the magnetic field at the center of a circular loop is:

$B=\frac{\mu_0I}{2R}$

If we name the smaller loop as 1 and the bigger loop as 2 and we impose the same magnitude of the magnetic field at their center, we have:

$B_2=B_1$

$\frac{\mu_0I_2}{2R_2}=\frac{\mu_0I_1}{2R_1}$

$\frac{I_2}{R_2}=\frac{I_1}{R_1}$

$R_2=\frac{I_2}{I_1}R_1$

Which for our values means:

$R_2=\frac{I_2}{I_1}R_1=\frac{(20A)}{(12A)}(3.8cm)=6.33cm$

(Notice that we don't need to convert cm to m, we get our result in cm).

3 0

3 years ago

At its lowest setting a centrifuge rotates with an angular speed of calculate the angular velocity

I am Lyosha [343]

No answer is possible until we know the number that belongs after the words "... angular speed of ".

8 0

3 years ago

John’s car travels a distance from A to B at a speed of 50 km/h for two hours as from B to C at a speed of 40 km/h for two hours

Blababa [14]

Please find attached photograph for your answer. Hope it helps. Please do comment.

7 0

3 years ago

Please help me find the answers!

VashaNatasha [74]

Answer:

1. T₁ is approximately 100.33 N

T₂ is approximately -51.674 N

2. 230°F is 383.15 K

3. Part A

The total torque on the bolt is -4.2 N·m

Part B

Negative anticlockwise

Explanation:

1. The given horizontal force = 86 N

The direction of the given 86 N force = To the left (negative) and along the x-axis

(The magnitude and direction of the 86 N force = -86·i)

The state of the system of forces = In equilibrium

The angle of elevation of the direction of the force T₁ = 31° above the x-axis

The direction of the force T₂ = Downwards, along the y-axis (Perpendicular to the x-axis)

Given that the system is in equilibrium, we have;

At equilibrium, the sum of the horizontal forces = 0

Therefore;

T₁ × cos(31°) - 86 = 0

T₁ = 86/(cos(31°)) ≈ 100.33

T₁ ≈ 100.33 N

Similarly, at equilibrium, the sum of the vertical forces = 0

∴ T₁×sin(31°) + T₂ = 0

Which gives;

100.33 × sin(31°) + T₂ = 0

T₂ = -100.33 × sin(31°) ≈ -51.674

T₂ ≈-51.674 N

2. 230° F to Kelvin

To convert degrees Fahrenheit (°F) to K, we use;

$Degrees \ in \ Kelvin, K = (x^{\circ} F + 459.67) \times \dfrac{5}{9}$

Pluggining in the given temperature value gives;

$Degrees \ in \ Kelvin, K = (230^{\circ} F + 459.67) \times \dfrac{5}{9} = 383.15$

230°F = 383.15 K

3. Part A

Torque = Force × perpendicular distance from the line of action of the force

Therefore, the clockwise torque = 9 N × 0.4 m = 3.6 N·m (clocwise)

The anticlockeisre torque = 13 N × 0.6 m = 7.8 N·m (anticlockwise)

The total torque o the bolt = 3.6 N·m - 7.8 N·m = -4.2 N·m (clockwise) = 4.2 N·m anticlockwise

Part B

The torque is negative anticlockwise.

7 0

3 years ago