I Need help please with this question..

A roulette wheel with a 1.0m radius reaches a maximum angular speed of 18 rad/s before it stops 35 revolutions ( 220 rad ) after

Alex17521 [72]

Max ang. speed(u) = 18 rad/s
final ang. speed(v) = 0
ang. displacement(s) = 220 rad

ang. acceleration = (v^2 - u^2)/2s = -18^2 / 2*220 = -0.7364 rad/s^2

v = u +at
0 = 18 - 0.7364t
t = 18/0.7364
t = 24.44 seconds

4 0

3 years ago

Consider two identical objects of mass m = 0.250 kg and charge q = 4.00 μC. The first charge is held in place at the origin of a

Gnom [1K]

Answer:

a = 640 m/s²

Explanation:

From work-kinetic energy principles,

The net force acting on the second object is the gravitational force and the electric force due to the first object.

So, the gravitational force on the mass is F₁ = Gm₁m₂/r² since m₁ = m₂ = m, U = -Gm²/r²

Also, the electric force on the charge is F₂ = kq₁q₂/r² since q₁ = q₂ = q, U = kq²/r²

The net Force F = ma

So, -F₁ + F₂ = F (F₁ is negative since it is an attractive force in the negative x -direction and F₂ is positive since it is a repulsive force in the positive x- direction)

-Gm²/r² + kq²/r² = ma

ma = -Gm²/r² + kq²/r²

a = (-Gm²/r² + kq²/r²)/m

a = (-G + kq²/m²)m/r²

Since m = 0.250 kg, q = 4.00 μC = 4.00 × 10⁻⁶ C, r = 3.00 cm = 3.00 × 10⁻² m, G = 6.67 × 10⁻¹¹ Nm²/kg², k = 9 × 10⁹ Nm²/C² and a = acceleration of second mass.

Substituting the variables into the equation, we have

a = (m/r²)(-G + k(q/m)²)]

a = (0.250 kg/{3.00 × 10⁻² m}²)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(4.00 × 10⁻⁶ C/0.250 kg)²)

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(16 × 10⁻⁶ C/kg)²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(256 × 10⁻¹² C²/kg²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 2304 × 10⁻³ Nm²/kg² ]

a = (0.250 kg/9.00 × 10⁻⁴ m)(2.304 Nm²/kg²)

a = 0.576 Nm²/kg /9.00 × 10⁻⁴ m²

a = 0.064 × 10⁴N/kg

a = 64 × 10 N/kg)

a = 640 m/s²

8 0

3 years ago

To measure the acceleration due to gravity on a distant planet, an astronaut hangs a 0.070-kg ball from the end of a wire. The w

mote1985 [20]

Answer:

g = 1.64m/s²

Explanation:

1.5m in 0.078s

V = 15 / 0.078

= 19.23m/s

Tension = mg

μ = 3.10 × 10⁻⁴

T = V²μ

mg = V²μ

g = V²μ / m

g = ((19.23)²(3.10 × 10⁻⁴)) / (0.070)

g = 1.64m/s²

7 0

3 years ago

Does the accuracy of a measurement made always depend on the precision of the instrument used to make the measurement? Why or Wh

mina [271]

Answer:

B

Explanation:

YOUR MUM NOOB SORRY MY MAD THE ANSWER WILL BE B BECAUSE B IS THE BEST YAY.

8 0

3 years ago

A permanent magnet has a magnetic dipole moment of 0.160 A · m^2. The magnet is in the presence of an external uniform magnetic

Elena L [17]

Answer:

the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm

the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J

Explanation:

The torque is given by :

$\bar {N} = \bar {m} * \bar {B}$

where ;

m = 0.160 A.m²

B = 0.0800 T

θ = 35°

So the magnitude of the torque N = mBsinθ

N = (0.160)(0.0800)(sin 35°)

N = 0.007341

N = 7.34×10⁻³ Nm

Hence, the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm

b) The potential energy $\bar{U} = \bar{-m} * \bar{B}$

U = -mBcosθ

U = (- 0.160)(0.0800)(cos 45)

U = -0.010485

U = -1.0485 ×10⁻² J

Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J

6 0

3 years ago