Two identical point charges of +Q coul are separated by a distance of 10 cm.

The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou

Cerrena [4.2K]

Answer:

a) $I = 19.799\,kg\cdot m^{2}$ , b) $T = -3.405\,N\cdot m$ , c) $n_{T} \approx 54.842\,rev$

Explanation:

a) The net torque is:

$T = I\cdot \alpha$

Let assume a constant angular acceleration, which is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}$

$\alpha = 1.793\,\frac{rad}{s^{2}}$

The moment of inertia of the wheel is:

$I = \frac{T}{\alpha}$

$I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }$

$I = 19.799\,kg\cdot m^{2}$

b) The deceleration of the wheel is due to the friction force. The deceleration is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}$

$\alpha = - 0.172\,\frac{rad}{s^{2}}$

The magnitude of the torque due to friction:

$T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )$

$T = -3.405\,N\cdot m$

c) The total angular displacement is:

$\theta_{T} = \theta_{1} + \theta_{2}$

$\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}$

$\theta_{T} = 344.580\,rad$

The total number of revolutions of the wheel is:

$n_{T} = \frac{\theta_{T}}{2\pi}$

$n_{T} = \frac{344.580\,rad}{2\pi}$

$n_{T} \approx 54.842\,rev$

5 0

3 years ago

A 2kg ball is rolled along the floor for 0.8 m at a constant speed of 6 m/s. What is the work done by gravity?

riadik2000 [5.3K]

=F×s×cosa=2×g×0,8×cos90°= 0

3 0

3 years ago

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The electric output of a power plant is 716 MW. Cooling water is the main way heat from the powerplant is rejected, and it flows

Stels [109]

Answer:

(a) 83475 MW

(b) 85.8 %

Explanation:

Output power = 716 MW = 716 x 10^6 W

Amount of water flows, V = 1.35 x 10^8 L = 1.35 x 10^8 x 10^-3 m^3

mass of water, m = Volume x density = 1.35 x 10^8 x 10^-3 x 1000

= 1.35 x 10^8 kg

Time, t = 1 hr = 3600 second

T1 = 25.4° C, T2 = 30.7° C

Specific heat of water, c = 4200 J/kg°C

(a) Total energy, Q = m x c x ΔT

Q = 1.35 x 10^8 x 4200 x (30.7 - 25.4) = 3 x 10^12 J

Power = Energy / time

Power input = $P = \frac{3 \times 10^{12}}{3600}=8.35 \times 10^{8}W$

Power input = 83475 MW

(b) The efficiency of the plant is defined as the ratio of output power to the input power.

$\eta =\frac{Power output}{Power input}$

$\eta =\frac{716}{83475}=0.858$

Thus, the efficiency is 85.8 %.

7 0

3 years ago

In an Atwood machine the two masses (m1 =5.0kg and m2 = 3.0 kg) are released from rest, with m1 at a height of h=0.75 m above th

masya89 [10]

Answer:

Explanation:

m₁ is heavy so it will go down and m₂ will go up. Let common acceleration be a .

For downward acceleration of m₁

m₁g - T₁ = m₁a

For upward acceleration of m₂

T₂- m₂g = m₂a

adding

m₁g - T₁ +T₂- m₂g = (m₁+m₂)a

m₁g- m₂g +(T₂- T₁) = (m₁+m₂)a --------------- ( 1 )

For rotatory motion of disc of mass mp and moment of inertia I

(T₁ -T₂ )R = I α , α is angular acceleration ----------( 2 )

From 1 and 2

m₁g- m₂g - I α / R = (m₁+m₂)a

m₁g- m₂g - I a / R² = (m₁+m₂)a

m₁g- m₂g = (m₁+m₂)a + I a / R²

m₁g- m₂g = [(m₁+m₂) + mp k² / R² ] a ( k is radius of gyration of disk )

a = (m₁- m₂)g / [(m₁+m₂) +mp k² / R² ]

= (m₁- m₂)g / [(m₁+m₂) + .5 mp ] ( for cylinder k² / R² = .5 )

If h be the height by which m₁ falls and its velocity becomes v

v² = 2 a h

a = v² / 2h

v² / 2h = (m₁- m₂)g / [(m₁+m₂) + .5 mp ]

v²[(m₁+m₂) + .5 mp ] = 2(m₁- m₂) gh

- v²[(m₁+m₂) +2(m₁- m₂) gh = .5 mp v²

- v²[(m₁+m₂) +2(m₁- m₂) gh / .5 v ² = mp .

- 1.8² ( 5 + 3 ) + 2 ( 5 - 3 ) 9.8 x .75 / .5 x 1.8² = mp

- 25.92 + 29.4 / 1.62 = mp

2.15 kg = mp .

K.E of masses = 1/2 x 5 x 1.8² + 1/2 x 3 x 1.8²

= 12.96 J

Rotational KE of disc

= 1/2 I ω²

= 1/4 mp r²ω²

= 1/4 x mp x v²

= .25 x 2.15 x 1.8²

= 1.74 J

in percent terms

(1.74 / 12.96) x 100

= 13.5 %

5 0

3 years ago

Which of the following statements are not true about gravity? Check all that apply.

liberstina [14]

C gravity only exists only on earth

3 0

3 years ago

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