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Law Incorporation [45]
3 years ago
15

Which simple experiment could be conducted to show an example of how lightning is created? A) Rub a balloon so it picks up extra

electrons from the sweater or your hair and becomes slightly negatively charged. B) Rub a balloon filled with air on a wool sweater or on your hair. Then hold it up to a wall. C) Connect the positive pole of an electrical source to the negative pole, so you create a circuit. D) Walk across a carpet, shuffling your feet and touching something made of metal.
Physics
1 answer:
alexandr1967 [171]3 years ago
7 0

I think the answer is d

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A mass of 2.82 kg is hung from a spring, causing the spring to stretch 0.331 m. If a second mass of 3.09 kg is now added to the
irina1246 [14]

Answer:

0.694 m

Explanation:

Case 1 : When only mass of 2.82 kg is hanged from spring

m = mass hanged from the spring = 2.82 kg

x = stretch caused in the spring = 0.331 m

k = spring constant

Using equilibrium of force in vertical direction

Spring force = weight of the mass

k x = m g

k (0.331) = (2.82) (9.8)

k = 83.5 N/m

Case 2 : When both masses are hanged from spring

m = mass hanged from the spring = 3.09 + 2.82 = 5.91 kg

x = stretch caused in the spring = ?

k = spring constant = 83.5 N/m

Using equilibrium of force in vertical direction

Spring force = weight of the mass

k x = m g

(83.5) x = (5.91) (9.8)

x = 0.694 m

4 0
3 years ago
As you increase current, what happens to the strength of an electromagnet?
Maksim231197 [3]

As you increase current, the strength of an electromagnet intensifies.

3 0
3 years ago
Please help on questions D AND E and please show working out as well so I can understand better thank you only QUESTIONS D AND E
tester [92]

Answer:

d. 87,500 J

e. 49,600 J

Explanation:

The total energy is the heat absorbed by the copper plus the heat absorbed by the water.

d)

E = m₁C₁ΔT + m₂C₂ΔT

E = (1 kg) (390 J/kg/°C) (10 °C) + (2 kg) (4180 J/kg/°C) (10 °C)

E = 87,500 J

e)

E = m₁C₁ΔT + m₂C₂ΔT

E = (2 kg) (390 J/kg/°C) (10 °C) + (1 kg) (4180 J/kg/°C) (10 °C)

E = 49,600 J

5 0
3 years ago
A 52 kg and a 95 kg skydiver jump from an airplane at an altitude of 4750 m, both falling in the pike position. Assume all value
Scilla [17]

Answer: 52 kg skydiver: 9.09 m/s and 522.55 s

              95 kg skydiver: 12.3 m/s and 386.2 s

Explanation: <u>Drag</u> <u>Force</u> is an opposite force when an object is moving in a fluid.

For skydivers, when falling through the air, the forces acting on it are gravitational and drag forces. At a certain point, drag force equals gravitational force, which is constant on any part of the planet, producing a net force that is zero. Since there is no net force, there is no acceleration and, consequently, velocity is constant. When that happens, the person reached the <u>Terminal</u> <u>Velocity</u>.

Drag Force and Velocity are proportional to the squared speed. So, terminal velocity is given by:

F_{G}=F_{D}

mg=\frac{1}{2}C \rho Av_{T}^{2}

v_{T}=\sqrt{\frac{2mg}{\rho CA} }

where

m is mass in kg

g is acceleration due to gravitational force in m/s²

ρ is density of the fluid in kg/m³

C is drag coefficient

A is area of the object in the fluid in m²

Calculating:

The 52kg skydiver has terminal velocity of:

v_{T}=\sqrt{\frac{2(52)(9.8)}{(1.21)(0.7)(0.14)} }

v_{T}= 9.09

The 95kg skydiver's terminal velocity is

v_{T}=\sqrt{\frac{2(95)(9.8)}{(1.21)(0.7)(0.14)} }

v_{T}= 12.3

The 52 kg and 95kg skydivers' terminal velocity are 9.09m/s and 12.3m/s, respectively.

The time each one will reach the floor will be:

52 kg at 9.09 m/s:

t=\frac{4750}{9.09}

t = 522.5

95 kg at 12.3 m/s:

t=\frac{4750}{12.3}

t = 386.2

The 52 kg and 95kg skydivers' time to reach the floor are 522.5 s and 386.2 s, respectively.

3 0
3 years ago
A Carnot engine operates between two heat reservoirs at temperatures THTH and TCTC. An inventor proposes to increase the efficie
Marat540 [252]

Answer:

e_12=1-Tc/Th

This is same as the original Carnot engine.  

Explanation:

For original Carnot engine, its efficiency is given by

e = 1-Tc/Th

For the composite engine, its efficiency is given by

e_12=(W_1+W_2)/Q_H1

where Q_H1 is the heat input to the first engine, W_1 s the work done by the first engine and W_2 is the work done by the second engine.  

But the work done can be written as  

W= Q_H + Q_C with Q_H as the heat input and Q_C as the heat emitted to the cold reservoir. So.  

e_12=(Q_H1+Q_C1+Q_H2+Q_C2)/Q_H1

But Q_H2 = -Q_C1 so the second and third terms in the numerator cancel  

each other.

 e_12=1+Q_C2/Q_H1

but, Q_C2/Q_H2= -T_C/T'

⇒ Q_C2 = -Q_H2(T_C/T')

               = Q_C1(T_C/T')

(T1 is the intermediate temperature)  

But, Q_C1 = -Q_H1(T'/T_H)

so, Q_C2 =  -Q_H1(T'/T_H)(T_C/T') = Q_H1(T_C/T_H) So the efficiency of the composite engine is given by  

e_12=1-Tc/Th

This is same as the original Carnot engine.  

7 0
3 years ago
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