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a_sh-v [17]
3 years ago
10

A 125-g coin is placed 8.0 cm from the axis of rotation of a horizontally rotating turntable as shown. The coefficient of static

friction between the coin and the turntable is μs . The turntable makes exactly 1.0 revolution per second and the coin remains in place without slipping. [a] (8 pts) Draw a free-body force diagram for the coin. [b] (10pts)What is the frictional force acting on the coin? [c] (12pts)When the rotational speed is increased to 1.2 revolutions per second, the coin starts to slip. Calculate the coefficient of static friction μs between the coin and the turntable?
Physics
1 answer:
Marianna [84]3 years ago
5 0

Answer:

zhvshshisvdiscdoscd if vdidg

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A spacecraft in the shape of a long cylinder has a length of 100 m, and its mass with occupants is 1 480 kg. It has strayed too
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Answer:

2352645198509.9604 m/s²

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

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The difference in the gravitational field in this system is given by

\Delta F=\dfrac{GMm}{R_{f}^2}-\dfrac{GMm}{R_{sb}^2}\\\Rightarrow \Delta g=GM\left(\frac{1}{R_f^2}-\frac{1}{R_{sb}^2}\right)\\\Rightarrow g=6.67\times 10^{-11}\times 90\times 1.989\times 10^{30}\left(\frac{1}{(10000+100)^2}-\frac{1}{10000^2}\right)\\\Rightarrow \Delta g=-2352645198509.9604\ m/s^2

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3 years ago
when water boils, it becomes steam .its particles spread out and occupy a greater volume. has its density increased or decreased
Deffense [45]
The density has decreased
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When one bumper car is still and the other one hits it (both cars have the same mass) what happens?
statuscvo [17]

the first bump car will move

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5 0
3 years ago
You would like to store 8.1 J of energy in the magnetic field of a solenoid. The solenoid has 620 circular turns of diameter 6.6
marta [7]

Answer:

(a) The current needed is 56.92 A

(b) The magnitude of the magnetic field inside the solenoid is 0.134 T

(c) The energy density inside the solenoid is 7.144 kJ/m³

Explanation:

Given;

energy stored in the magnetic field of solenoid, E = 8.1 J

number of turns of the solenoid, N = 620 turns

diameter of the solenoid, D = 6.6 cm = 0.066 m

radius of the solenoid, r = D/2 = 0.033 m

length of the solenoid, L = 33 cm = 0.33 m

Inductance of the solenoid is given as;

L= \frac{\mu_o N^2 A}{l}

where;

A is the area of the solenoid = πr² = π (0.033)² = 0.00342 m²

μ₀ is permeability of free space = 4π x 10⁻⁷ H/m

L= \frac{4\pi*10^{-7} *620^2 *0.00342}{0.33} \\\\L = 0.005 \ H

(A). How much current needed

Energy stored in magnetic field of solenoid is given as;

E = \frac{1}{2} LI^2\\\\

Where;

I is the current in the solenoid

E = \frac{1}{2} LI^2\\\\I^2 = \frac{2E}{L}\\\\I = \sqrt{\frac{2*8.1}{0.005}}\\\\ I = 56.92 \ A

(B) The magnitude of the magnetic field inside the solenoid

B = μ₀nI

where;

n is number of turns per unit length

B = μ₀(N/L)I

B = (4π x 10⁻⁷)(620/0.33)(56.92)

B = 0.134 T

(C) The energy density (energy/volume) inside the solenoid

U_B = \frac{B^2}{2\mu_0} \\\\U_B = \frac{(0.134)^2}{2*4\pi*10^{-7}} \\\\U_B = 7143.54 \ J/m^3\\\\U_B = 7.144 \ kJ/m^3

3 0
3 years ago
2 76 Consider a household that uses 14,000 kWh of elec- (tricity per year and 3400 L of fuel oil during a heating sea- son. The
NNADVOKAT [17]

Answer:

The total amount of CO₂ produced will be  = 20680 kg/year

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Explanation:

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Fuel oil used = 3400 L

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Now, the total amount of CO₂ produced will be = (14000 kWh × 0.70 kg/kWh) + (3400 L × 3.2 kg/L)

⇒ The total amount of CO₂ produced will be = 9800 + 10880 = 20680 kg/year

Now,

if the usage of electricity and fuel oil is reduced by 15%, the reduction in the amount of the CO₂ emission will be = 0.15 × 20680 kg/year = 3102 kg/year

6 0
4 years ago
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