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pantera1 [17]
2 years ago
5

A three-way 120 V lamp bulb that contains two filaments is rated for 100-200-300 W. One filament burns out. Afterward, the bulb

operates at the same intensity (dissipates energy at the same rate) on its lowest as on its highest switch positions but does not operate at all on the middle position. What are the (a) larger and (b) smaller values of the filament resistances?
Physics
1 answer:
ololo11 [35]2 years ago
8 0

Answer:

The larger resistance value is 144 ohms and the smaller resistance value is 48 ohms.

Explanation:

Given that Resistance, R, is inversely proportional to Power, the largest value of power will yield the smallest value of resistance and vice versa.

The largest value of resistance will be:

P = V²/R

R = V²/P

R = 120²/100

R = 144 ohms

The smallest value of resistance will be:

R = 120²/300

R = 48 ohms

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A11) A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic fiel
vitfil [10]

Answer:

A

Explanation:

From a Solenoid we know that a magnetic fiel is always inversely proportional to lenght L or BL = constant

B= frac{\mu_0}{2R}

As I is constant

\frac{B2}{B1} = \frac{R1}{R2}

B2 = 2mT*\frac{18}{21}

B2 = 1.714mT

7 0
2 years ago
A 2-kg cart, traveling on a horizontal air track with a speed of 3 m/s, collides with a stationary 4-kg cart. The carts stick to
daser333 [38]

Answer:

Magnitude of impulse, |J| = 4 kg-m/s                                                                                

Explanation:

It is given that,

Mass of cart 1, m_1=2\ kg

Mass of cart 2, m_2=4\ kg  

Initial speed of cart 1, u_1=3\ m/s          

Initial speed of cart 2, u_2=0 (stationary)

The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

m_1u_1+m_2u_2=(m_1+m_2)V

V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}        

V=\dfrac{2\times 3}{(2+4)}

V = 1 m/s

The magnitude of the impulse exerted by one cart on the other is given by:

J=F\times t=m(V-u)

J=m(V-u)

J=2\times (1-3)    

J = -4 kg-m/s

or

|J| = 4 kg-m/s

So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.

8 0
2 years ago
At the circus, a 100-kilogram clown is fired 15 meters per second from a 500-kilogram cannon. What is the recoil speed of the ca
Ahat [919]
Recoil speed= 3m/s, method shown in photo

4 0
2 years ago
The Orion nebula is one of the brightest diffuse nebulae in the sky (look for it in the winter, just below the three bright star
Oxana [17]

Answer:

T=183.21K

Explanation:

We have to take into account that the system is a ideal gas. Hence, we have the expression

PV=nRT

where P is the pressure, V is the volume, n is the number of moles, T is the temperature and R is the ideal gas constant.

Thus, it is necessary to calculate n and V

V is the volume of a sphere

V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (5.9*10^{15}m)^3=8.602*10^{47}m^3

V=8.86*10^{50}L

and for n

n=\frac{(4000M_s)/(2*mH)}{6.022*10^{23}mol^{-1}}=3.95*10^{36}mol

Hence, we have (1 Pa = 9.85*10^{-9}atm)

T=\frac{PV}{nR}=\frac{(6.8*10^{-9}*9.85*10^{-6}atm)(8.86*10^{50}L)}{(0.0820\frac{atm*L}{mol*K})(3.95*10^{36}mol)}\\\\T=183.21K

hope this helps!!

8 0
2 years ago
A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a very light rope that goes over an ideal pulley. If the
AleksAgata [21]

Answer:

acceleration = 2.4525‬ m/s²

Explanation:

Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²

Tension in the rope = T

Sol: m2 > m1

i) for downward motion of m2:

m2 a = m2 g - T

5 a = 5 × 9.81 m/s² - T  

⇒ T = 49.05‬ m/s² - 5 a     Eqn (a)‬

ii) for upward motion of m1

m a = T - m1 g

3 a = T - 3 × 9.8 m/s²

⇒ T =  3 a + 29.43‬ m/s²   Eqn (b)

Equating Eqn (a) and(b)

49.05‬ m/s² - 5 a = T =  3 a + 29.43‬ m/s²

49.05‬ m/s² - 29.43‬ m/s² = 3 a + 5 a

19.62 m/s² = 8 a

⇒ a = 2.4525‬ m/s²

5 0
2 years ago
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