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mestny [16]
3 years ago
15

2. A 4.0 kg magnetic toy car traveling at 3.0 m/s east collides and sticks to a 5.0 kg toy magnetic car also traveling at 2.0 m/

s east. Calculate the final speed and direction of the magnetic car (coupled) system?
Physics
1 answer:
Afina-wow [57]3 years ago
4 0

Answer:

2.44 m/s due East

Explanation:

From the question given above, the following data were obtained:

Mass of 1st car (m₁) = 4 Kg

Velocity of 1st car (u₁) = 3 m/s

Mass of 2nd car (m₂) = 5 Kg

Velocity of 2nd car (u₂) = 2 m/s

Final velocity (v) =?

The final velocity can be obtained as follow:

v(m₁ + m₂) = m₁u₁ + m₂u₂

v(4 + 5) = (4×3) + (5×2)

9v = 12 + 10

9v = 22

Divide both side by 9

v = 22/9

v = 2.44 m/s

Thus, the final velocity is 2.44 m/s.

Since both cars was moving due East before collision, and after collision, they stick together, then their direction will be due East.

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Answer:

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Explanation:

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4 0
3 years ago
We wrap a light, nonstretching cable around a 10.0 kg kg solid cylinder with diameter of 38.0 cm cm . The cylinder rotates with
amm1812

Answer: 14.16

Explanation:

Given

d = 38cm

r = d/2 = 38/2 = 19cm = 0.19m

K.E = 510J

m = 10kg

I = 1/2mr²

I = 1/2*10*0.19²

I = 0.18kgm²

When it has 510J of Kinetic Energy then,

510J = 1/2Iω²

ω² = 1020/I

ω² = 1020/0.18

ω² = 5666.67

ω = √5666.67 = 75.28 rad/s

Velocity is the block, v = ωr

V = 75.28 * 0.19

V = 14.30m/s

The "effective mass" M of the system is

M = (14.0 + ½*10.0) kg = 19.0 kg

The motive force would be

F = ma

F = 14 * 9.8

F = 137.2N

so that the acceleration would be

a = F/m

a = 137.2/19

a = 7.22m/s²

Finally, using equation of motion.

V² = u² + 2as

14.3² = 0 + 2*7.22*s

204.49 = 14.44s

s = 204.49/14.44

s = 14.16m

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The answer is convection
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A small circular coil of 5 turns of wire lies in a uniform magnetic field of 0.8 T, so that the normal to the plane of the coil
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Complete question:

A small circular coil of 5 turns of wire lies in a uniform magnetic field of 0.8 T, so that the normal to the plane of the coil makes an angle of 100◦ with the direction of B~ . The radius of the coil is 4 cm, and it carries a current of 1 A.

What is magnitude of the magnetic moment of the coil? Answer in units of A · m2.

Answer:

The magnetic moment of the coil is 0.0252 A.m²

Explanation:

Given;

radius of the coil, r = 4 cm = 0.04 m

number of turns of the coil, N = 5 turns

magnetic field strength B = 0.8 T

current in the coil, I = 1 A

Area of the coil, A = πr² = π(0.04)² = 0.00503 m²

magnetic moment of the coil, μ = NIA

where;

N is the number of turns

I is the current in the coil

A is the area of the coil

magnetic moment of the coil, μ = 5 x 1 x 0.00503 = 0.0252 A.m²

Therefore, the magnetic moment of the coil is 0.0252 A.m²

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