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lesya [120]
4 years ago
9

A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.79 rad/s . Its total moment of inertia is 1790

kg⋅m2 . Four people standing on the ground, each of mass 70 kg , suddenly step onto the edge of the merry-go-round.
Physics
1 answer:
Nezavi [6.7K]4 years ago
8 0

Answer:

w_2=0.467rad/s

Explanation:

Four people standing on the ground each of mass and usually this questions have to find the final angular velocity

m_t=4*70kg=280kg

The radius r=4.2/2=2.1m

Angular velocity  w_1=0.79rad/s

The moment of inertia total is I_t=1790 kg/m^2

Momento if inertia

I_1=m_t*r^2

I_1=280kg*(2.1m)^2=1234.8kg*m^2

Angular momentum

I_1*w_1=I_t*w_2

Solve to w2

w_2=\frac{I_1*w_1}{I_t}

w_2=\frac{1790kg*m^2*0.79rad/s}{3024.8kg*m^2}

w_2=0.467rad/s

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You slide a chair across a rough, horizontal surface. The chair's mass is 18.8 kg. The force you exert on the chair is 156 N dir
Luda [366]

Answer:

626.612 J

Explanation:

Work done by friction on the chair is given as

W-ΔEk = W'..................... Equation 1

Where W' = Work done by friction on the chair, W = Work done on the chair by me, Ek = change in Kinetic energy of the chair as a result of the slide.

From the question,

W = FdcosФ.............. Equation 2

ΔEk = 1/2m(v²-u²)................ Equation 3

Where F = Force applied on the chair, d = distance of slide, Ф = angle between the force and the horizontal, m = mass of the chair, v = final velocity of the chair, u = initial velocity of the chair

Substitute equation 2 and equation 3 into equation 1

W' = FdcosФ-1/2m(v²-u²)........................ Equation 4

Given: F = 156, d = 5 m, Ф = 26°, m = 18.8 kg, v = 3.1 m/s, u = 1.3 m/s

Substitute into equation 4

W' = 156×5×cos26°-1/2×18(3.1²-1.3²)

W' = 701.06-74.448

W' = 626.612 J.

Hence the work done by friction = 626.612 J

8 0
4 years ago
A 1 000-V battery, a 3 000-Ω resistor, and a 0.50-μF capacitor are connected in series with a switch. The time constant for such
vampirchik [111]

Answer:

The current in the circuit at a time interval of τ seconds after the switch has been closed is 0.123 A

Explanation:

The time constant for an R and C in series circuit is given by τ = RC.

R = 3000 ohms, C = 0.5 × 10⁻⁶ F = 5.0 × 10⁻⁷ F

τ = 3000 × 5 × 10⁻⁷ = 0.015 s

The voltage across a capacitor as it charges is given be

V(t) = Vs (1 - e⁻ᵏᵗ)

where k = 1/τ

At the point when t = τ, the expassion becomes

V(t = τ) = 1000 (1 - e⁻¹) = 0.632 × 1000 = 632 V

Current flows as a result of potential difference,.

Current in the circuit at this time t =  τ is given by

I = (Vs - Vc)/R

Vs = source voltage = 1000 V

Vc = Voltage across the capacitor = 632 V

R = 3000 ohms

I = (1000 - 632)/3000 = 0.123 A

6 0
4 years ago
The membrane of the axon of a nerve cell is a thin cylindrical shell of radius r = 10-5 m, length L = 0.32 m, and thickness d =
maksim [4K]

Answer:

5.3 x 10⁻⁹ C

Explanation:

r = radius of cylindrical shell = 10⁻⁵ m

L = length = 0.32 m

A = area

Area is given as

A = 2πrL

A = 2 (3.14) (10⁻⁵) (0.32)

A = 20.096 x 10⁻⁶ m²

d = separation = 10⁻⁸ m

k_{appa} = dielectric constant = 4

Capacitance is given as

Q=\frac{k_{appa}\epsilon _{o}A}{d}                               eq-1

V = Potential difference across the membrane = 74 mV = 0.074 Volts

Q = magnitude of charge on each side

Magnitude of charge on each side is given as

Q = CV

using eq-1

Q=\frac{k_{appa} \epsilon _{o}AV}{d}

Inserting the values

Q=\frac{4 (8.85\times 10^{-12})(20.096\times 10^{-6})(0.074)}{10^{-8}}

Q = 5.3 x 10⁻⁹ C

4 0
3 years ago
The acceleration due to gravity on or near the surface of Earth is 32 ft./s/s. Neglecting friction, from what height must a ston
svetoff [14.1K]

Given :

The acceleration due to gravity on or near the surface of Earth is 32 ft/s/s

To Find :

From what height must a stone be dropped on Earth to strike the ground with a velocity of 136 ft/s.

Solution :

Initial velocity of stone, u = 0 ft/s.

Now, by equation of motion :

2as =  v^2 -u^2 \\\\2\times 32 \times s = 136^2 -0^2\\\\s = \dfrac{136^2}{2\times 32}\ ft\\\\s = 289 \ ft

Therefore, height from which stone is thrown is 289 ft.

3 0
3 years ago
A car with mass mc = 1490 kg is traveling west through an intersection at a magnitude of velocity of vc = 9.5 m/s when a truck o
icang [17]

Answer:

v= - 4.507 i - 2.363 j

Explanation:

 Given that

mc= 1490 kg

vc= 9.5 m/s ( - i)

mt=  1650 kg

vt = 6.4 m/s ( -j)

There is any external force so linear momentum will remain conserve.

Lets take final speed is v.

mc .vc + mt . vt = ( mc+mt) v

1490 x 9.5 ( - i) + 1650 x 6.4 ( -j) = ( 1490+1650) v

14,155 ( -i) + 10,560 ( - j) = 3140 v

v= - 4.507 i - 2.363 j

3 0
4 years ago
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