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guapka [62]
3 years ago
13

the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off. the s

imple process in which kind of a change
Physics
1 answer:
Travka [436]3 years ago
3 0
When the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off, this is an example of resistance, which provides light and heat. 
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As a substance melts, heat energy is used to break the connections between molecules, and temperature __________ until all melti
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Temperature stays the same
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2 years ago
A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w
amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁  + m₂u₂ = v(m₁  +  m₂)

10 x 3   + 25 x 0   = v(10  +  25)

30  = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\

the final kinetic energy of the package;

K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J

the fraction of the initial energy lost;

= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71

7 0
3 years ago
Would the forces acting on the sky diver be balanced or unbalanced? Explain your answer.
docker41 [41]

Answer:

your mom is balanced!!!

4 0
2 years ago
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max2010maxim [7]

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3 0
3 years ago
The potential difference between the surface of a 3.0-cm-diameter power line and a point 1.0 m distant is 3.9 kV. Find the line
Korolek [52]

Answer:

The linear charge density is 5.19 X 10⁻⁶ C/m

Explanation:

The potential difference between two cylinders, is given as

V = (λ/2πε)ln(b/a)

where;

λ is the line charge density on the power line.

b is the distance between the power line = 1 m

a is the radius of the wire = 1.5 cm = 0.015 m

ε is the permittivity of free space = 8.9 X 10⁻¹² C

V*2πε = λ* ln(b/a)

3900 *(2π*8.9 x10⁻¹²)= λ *ln(1/0.015)

2.1812 X 10⁻⁷ = 4.1997* λ

λ = 5.19 X 10⁻⁶ C/m

Therefore, the linear charge density is 5.19 X 10⁻⁶ C/m

6 0
2 years ago
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