True or False:

Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of +23 pC experiences an electr

Sedbober [7]

Answer:

(A) ratio of electric force to weight will be $23.469\times 10^{-10}$

(b) Electric field will be $E=4.26\times 10^{10}N/C$

Explanation:

We have given mass of bee = 100 mg = $m=100\times 10^{-3}=0.1kg$

Charge on bee $q=23pC=23\times 10^{-12}C$

Electric field E = 100 N/C

Weight of the bee $W=mg=0.1\times 9.8=0.98N$

Electric force on the bee $F=qE=23\times 10^{-12}\times 100=23\times 10^{-10}N$

So the ratio of electric force on the bee and weight is $=\frac{F}{W}=\frac{23\times 10^{-10}}{0.98}=23.469\times 10^{-10}$

(B) To hold the bee in air electric force must be equal to weight of bee

So $mg=qE$

$0.1\times 9.8=23\times 10^{-12}E$

$E=4.26\times 10^{10}N/C$

4 0

3 years ago

Which of these is NOT used to create a simple electromagnet?

timurjin [86]

battery

Explanation:

Electromagnets can be created by wrapping a wire around an iron nail and running current through the wire. The electric field in the wire coil creates a magnetic field around the nail. In some cases, the nail will remain magnetised even when removed from within the wire coil

3 0

2 years ago

Read 2 more answers

What is a chemical substance that cannot be broken down into another substance?

Tema [17]

The answer is C. Elements

Elements cannot be broken down int simpler substances even by chemical means

6 0

3 years ago

Read 2 more answers

Which of the following statements are true?

Virty [35]

Answer:

The true statements are;

a.The decrease in the amplitude of an oscillation caused by dissipative forces is called damping

b.An oscillation that is maintained by a driving force is called forced oscillation

Explanation:

To answer the question, we analyze each of the options as follows;

a.The decrease in the amplitude of an oscillation caused by dissipative forces is called damping.

The above statement is true as damped oscillation is one in which the amplitude is decreased by dissipative forces

b.An oscillation that is maintained by a driving force is called forced oscillation.

When a damped oscillator is allowed to oscillate by itself, it will stop eventually. However when a force is applied to maintain the oscillation the oscillation is then known as a forced oscillation

The above statement is true

c.In a mechanical system, the amplitude of an oscillation diminishes with time unless the lost mechanical energy is replaced.

The above statement is only true in damped oscillation system

d.The increase in amplitude of an oscillation by a driving force is called forced oscillation.

Forced oscillation, as described above, is one the maintains the oscillation of damped oscillator, therefore, the above statement is not correct.

8 0

2 years ago

A 40-cm-diameter, 300 g beach ball is dropped with a 4.0 mg ant riding on the top. The ball experiences air resistance, but the

mars1129 [50]

Answer:

The normal force exerted on the ant is 0.75 N.

Explanation:

Given;

diameter of the ball, D = 40 cm = 0.4m

radius of the ball, r = 0.2m

mass of the beach ball, m₁ = 300 g = 0.3 kg

mass of the ant, m₂ = 4 x 10⁻⁶ kg

speed of the ball, v = 4 m/s

The area of the ball, assuming spherical ball is given by;

A = 4πr²

A = 4π(0.2)² = 0.5027 m²

The drag force (resistance) experienced by the spherical ball is given as;

$F_D = \frac{1}{2}C\rho Av^2$

where;

C is the drag coefficient of the spherical ball = 0.45

ρ is density of air = 1.21 kg/m³

$F_D = \frac{1}{2}C\rho Av^2\\\\F_D = \frac{1}{2}(0.45)(1.21) (0.5027)(4)^2\\\\F_D = 2.19 \ N$

The downward force of the ball due to its weight and that of the ant is given by;

$F_g = mg\\\\F_g =g(m_{ant} + m_{ball})\\\\F_g = g(4*10^{-6} \ kg\ + \ 0.3\ kg)\\\\F_g = g(0.300004 \ kg) \ \ \ (mass \ of \ the \ ant \ is \ insignificant)\\\\F_g = 9.8(0.3)\\\\F_g = 2.94 \ N$

The net downward force experienced by the ball is given by;

$F_{net} = F_g - F_D\\\\F_{net} = 2.94 \ N - 2.19 \ N\\\\F_{net} = 0.75 \ N$

This downward force experienced by the ball is equal to the normal reaction it exerts on the ant.

Thus, the normal force exerted on the ant is 0.75 N.

5 0

2 years ago