Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of 33 pC experiences an electric force in the earth's electric field, which is typically 100 N/C, directed downward.
1. What is the ratio of the electric force on the bee to the bee's weight?
2. What electric field strength would allow the bee to hang suspended in the air?
3. What electric field direction would allow the bee to hang suspended in the air?

Answer:

9.965 nF

Explanation:

The capacitance of the axon C = εA/d where ε = dielectric constant = 24.78 × 10⁻¹² F/m, A = surface area of axon = 2πrL where r = radius of axon = 8 μm = 8 × 10⁻⁶ m and L = length of axon = 8 cm = 8 × 10⁻² m and d = thickness of membrane = 0.01 μm = 0.01 × 10⁻⁶ m

So, C = εA/d

C = ε2πrL/d

Substituting the of the values variables into the equation, we have

C = ε2πrL/d

C =  24.78 × 10⁻¹² F/m × 2π × 8 × 10⁻⁶ m × 8 × 10⁻² m/0.01 × 10⁻⁶ m

C =  9964.63 × 10⁻²⁰ Fm/0.01 × 10⁻⁶ m

C =  996463 × 10⁻¹⁴ F

C = 9.96463 × 10⁻⁹ F

C = 9.96463 nF

C ≅ 9.965 nF

Answer:

Explanation:

Since the x and y components are given

The vectors Magnitude = √32²+(-59)²

=67.12m

Answer:

The weight of the wheelbarrow and the road is 784 N and the force required to lift the wheelbarrow is 784 N.

Explanation:

Given that,

The total mass of the wheelbarrow and the road is 80 kg.

The weight of an object is given by :

W = mg

where

g is acceleration due to gravity

So,

W = 80 × 9.8

= 784 N

So, the force required to lift the wheelbarrow is equal to its weight i.e. 784 N.

Answer:

the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

Explanation:

Given the data in the question;

initial velocity; u = 0 m/s

height; h = 2.5 m

we find the velocity of the ball just before it touches the foam.

using the equation of motion;

v² = u² + 2gh

we know that acceleration due gravity g = 9.81 m/s²

so we substitute

v² = ( 0 )² + ( 2 × 9.81 × 2.5 )

v² = 49.05

v = √49.05

v = 7.00357 m/s

Now as the ball touches the foam

final velocity v₀ = 0 m/s

compresses S = 3 cm = 0.03 m

so

v₀² = v² + 2as

we substitute

( 0 )² = 49.05 + 0.06a

0.06a = -49.05

a = -49.05 / 0.06

a = -817.5 m/s²

Therefore, the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

Answer:

30

Explanation:

since  [tex]\frac{30}{5}[/tex]=6

         [tex]\frac{30}{6}[/tex]=5

then both 5 and 6 are factors of 30

Have a nice day

Answer:

A.

sửa cho tôi nếu tôi sai

Answer:

(a) 81.54 N

(b) 570.75 J

(c) - 570.75 J

(d) 0 J, 0 J

(e) 0 J  

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

(c) Work done by the friction

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J  

Answer:

What is the potential energy? PE= mghPE= hwKE= 1/2mv2

Answer:1960J

Explanation:

Explanation:

1. Distance, d = 5 m

Time, t = 10 s

Speed = distance/time

[tex]v=\dfrac{5}{10}=0.5\ m/s[/tex]

2. Mass, m = 10 kg

Acceleration, a = 3 m/s³

Force, F = mass (m) × acceleration (a)

F = 10 × 3

= 20 N

3. Mass, m = 7 kg

Height, h = 4 m

Potential energy, E = mgh

E = 7 × 9.8 × 4

E = 274.4 J

Hence, this is the required solution.

Answer:

+ 900 J

Explanation:

Since the total energy change ΔE = internal energy change ΔU since there is no change in kinetic and potential energy,

ΔE = ΔU

ΔE = Q - W where Q = heat absorbed by system and W = work done by system

Now since the system absorbs 500 J of heat, Q = + 500 J and work of 400 J is done on the system, W = -400 J

So,  the values of the variables into the equation, we have

ΔE = Q - W

ΔE = + 500 J - (-400 J)

ΔE = + 500 J + 400 J  

ΔE = + 900 J

So, the internal energy change, ΔE = + 900 J

Answer:

ΔV = 2 10¹ V

Explanation:

The calculation of the uncertainty or error in an expression is given by

         ΔV = [tex]\frac{dV}{di}[/tex]  |Δi| + [tex]\frac{dV}{dR}[/tex]  |ΔR |

         V = i R

let's make the derivatives

        [tex]\frac{dV}{di}[/tex] = R

        [tex]\frac{dV}{dR}[/tex] = i

we substitute

         ΔV = R | Δi | + i | ΔR |

in the exercise give the values

         i = (5.9 ± 0.4) A

         R = (42.7 ± 0.6) Ω

we calculate

          ΔV = 42.7  0.4 + 5.9  0.6

          ΔV = 20.6 V

          ΔV = 2 10¹ V

the voltage is

         V = i R

         V = 5.9  42.7

          V = 251.9 V

the result is

         V = (25 ± 2) 10¹ V

Answer:

focal length=12cm

Explanation:

object size is equal to 1.00cm

object distance = 18cm

heigh of image = 2.00cm

image distance = ??

but magnification is given by;

M = 2.00/1.00 = 2

but u/v = M

u/18 = 2

u = 36

1/f = 1/u+1/v

1/f = 1/18+ 1/36

1/f = 1/12

f = 12cm

Answer:

Option B. 3.25×10¯¹⁰ mm.

Explanation:

Measurement (cm) = 0.00325×10⁻⁸ cm

Measurement (mm) =?

The measurement in mm can be obtained as follow:

1 cm = 10 mm

Therefore,

0.00325×10⁻⁸ cm = 0.00325×10⁻⁸ cm × 10 mm / 1 cm

0.00325×10⁻⁸ cm = 3.25×10¯¹⁰ mm

Thus, 0.00325×10⁻⁸ cm is equivalent to 3.25×10¯¹⁰ mm.

The conversion from centimeter to millimeter of the number 0.00325*10^-8cm is 3.25*10^-10mm

The number given is in standard form and can be written as 3.25*10^-11 cm.

To convert this from centimeter to millimeter, we have to multiply this value by 10.

So, let's 3.25*10^-11 by 10 and get our value in mm

[tex]3.25*10^-^1^1 * 10 = 3.25*10^-^1^0[/tex]

From the calculation above, we can see that option B is the right answer since it carries [tex]3.25*10^-^1^0mm[/tex]

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Answer:

vs = 55.84 m/s

Explanation:

In this case, the source (skydiver with buzzer) is moving towards the observer (friend). The formula for this case will be:

[tex]f' = \frac{v}{v-v_s} f[/tex]

where,

f' = shifted frequency = 2150 Hz

f = actual frequency = 1800 Hz

v = speed of sound = 343 m/s

vs = speed of skydiver = ?

Therefore,

[tex]2150\ Hz = \frac{343\ m/s}{343\ m/s - v_s}(1800\ Hz)\\\\343\ m/s - v_s = \frac{343\ m/s}{2150\ Hz} (1800\ Hz)\\\\v_s = 343\ m/s - 287.16\ m/s\\[/tex]

vs = 55.84 m/s

Answer:

ΔT = 25°C

Explanation:

Given that.

The mass of a bullet, m₁ = 30 g = 0.03 kg

The speed of the bullet, v = 900 m/s

Mass of soft iron, m₂ = 1 k

The specific heat of iron, c=490J/kg°C

We need to find the increase in temperature of iron. using the conservation of energy,

Kinetic energy = heat absorbed

[tex]\dfrac{1}{2}m_1v^2=m_2c\Delta T\\\\\Delta T=\dfrac{\dfrac{1}{2}m_1v^2}{m_2c}\\\\\Delta T=\dfrac{\dfrac{1}{2}\times 0.03\times 900^2}{1\times 490}\\\\=24.79^{\circ} C\\\\or\\\\\Delta T=25^{\circ} C[/tex]

So, the correct option is (A).

Answer:

the largest distance we can measure is 10¹⁴ km

Explanation:

Given the data in the question;

Threshold hearing = 10⁻²⁰

smallest distance measured = 1 mm

Largest distance measured will be;

⇒ ( threshold hearing )⁻¹ × smallest distance

= ( 1 / 10⁻²⁰ ) × 1 mm

= 10²⁰ × 1mm

= 10²⁰ mm

we know that; 1000 mm = 10⁶ km

Largest distance = ( 10²⁰ / 10⁶ ) km

= 10¹⁴ km

Therefore, the largest distance we can measure is 10¹⁴ km

Answer:

Dòng điện là một dòng các hạt mang điện, chẳng hạn như electron hoặc ion, di chuyển qua vật dẫn điện hoặc không gian. Nó được đo bằng tốc độ thực của dòng điện tích qua một bề mặt hoặc vào một thể tích điều khiển.

Xin lưu ý rằng tôi đã sử dụng một trình dịch để nhập nội dung này, vì vậy có thể có một số từ không hợp lý.

Explanation:

Answer:

Distance = speed * time

55*5

275 meters.

The train would have covered a distance of 275 m

We can define distance as to how much ground an object has covered despite its starting or ending point.

Distance = speed * time

given

speed= 55 m/s

time = 5 sec

Distance = 55 * 5 = 275 m

The train would have covered a distance of 275 m

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Hey again!

Ok..

Now... The melting Point of this solid is 90°C.

Meaning That as soon as it gets to this temp... It STARTS Melting.

So at that temp... It still has some solid parts in it.

You can say its a Solid Liquid Mixture.

Additional Heat being applied at that point is not raising the temperature;rather its used in breaking the bonds in the solid. This is the Fusion stage.

After Fusion...It'd then Be a Pure Liquid with no solids in it.

So

Q'=MC∆0----- This is the heat needed to take the solid's temp from 30°c - 90°c

Q"=ml ----- This is the heat used in breaking the bonds holding the solids in the solid-liquid phase.

So

Q= Q' + Q"

Q= mc∆0 + ml

∆0 = 90°c - 30°c = 60°c

Q= 2.5(390)(60) + (2.5)(4000)

Q=6.9 x 10⁴Joules

The heat required to change 2.5 kg of the solid at 30.0C to a liquid is 6.9 x 10⁴J.

The specific heat is the amount of heat energy required to change the temperature of 1kg of object by 1°C.

The heat needed to change the solid's temperature from 30°C - 90°C is

Q' = mC∆T

The heat used to change the phase solid-liquid phase .i.e.

Q'' =mL where L =latent heat of fusion

The total heat required is

Q= Q' + Q"

Q= mc∆T + ml

Q= 2.5(390)(90 - 30) + (2.5)(4000)

Q=6.9 x 10⁴Joules

Thus, the heat required to change the solid to liquid is 6.9 x 10⁴J.

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Explanation:

Given that,

Two resistors of 10 ohms and 15 ohms are connected.

In series combination, the equivalent resistance is given by :

[tex]R_s=R_1+R_2\\\\R_s=10+15\\\\R_s=25\ \Omega[/tex]

In parallel combination, the equivalent resistance is given by :

[tex]\dfrac{1}{R_p}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\\dfrac{1}{R_p}=\dfrac{1}{10}+\dfrac{1}{15}\\\\R_p=6\ \Omega[/tex]

Hence, this is the required solution.

Answer:

One

Explanation:

In series combination, the circuit follows one path whereas in parallel it follows two or more than two path

Answer:

The speed of projection is 34 m/s.

Explanation:

Height of building, h = 51 m

horizontal distance, d = 74 m

time, t = 8 s

Let the angle is A and the speed is u.

d = u cos A x t

74 = u cos A x 8

u cos A = 9.25 .... (1)

Use second equation of motion

[tex]h = u sin A t - 0.5 gt^2\\\\-51 = u sinA \times 8 - 0.5\times 9.8\times8\times 8\\\\u sin A = 32.8 .... (2)[/tex]

Squaring and adding both the equations

[tex]u^2 = 9.25^2 + 32.8^2 \\\\u = 34 m/s[/tex]

Answer:

A neutral atom has no net charge and should not be affected by a point charge, and the distance of separation is not a factor.

The force will remain the same and is equal to zero.

We have a point charge and a neutral atom.

We have to investigate if the distance between a neutral atom and a point charge is increased by a factor of six, by what factor does the force on the atom by the point charge change.

The Coulomb's law states that the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. Mathematically -

[tex]$F=\frac{q _{1}\cdot q _{2}}{4\cdot \pi \cdot \varepsilon \cdot \varepsilon _{0}\cdot r^{2}}[/tex]

According to question, we have -

A point charge and a neutral atom.

If initially the distance between the point charge and neutral atom is r meters, then -

q(1) = Q (say)

q(2) = 0   ( Neutral atom has zero charge)

Using Coulomb's law -

[tex]$F=\frac{Q\times 0}{4\cdot \pi \cdot \varepsilon \cdot \varepsilon _{0}\cdot r^{2}}[/tex]

F = 0 Newtons.

Now, if you will increase the distance by the factor of six, still the force will remain zero as there is only one point charge and other is neutral atom. There is no electrostatic force between a charged and uncharged particle.

Hence, the force will remain the same and is equal to zero.

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Answer:

[tex]d=79.9m[/tex]

Explanation:

From the question we are told that:

coefficient of static friction [tex]\mu=0.38[/tex]

Velocity [tex]v=87.9=>24.41667m/s[/tex]

Generally the equation for Conservation of energy is mathematically given by

 [tex]\mu*mgd = 0.5 m v^2[/tex]

 [tex]d=\frac{0.5*24.42^2}{0.38*9.8}[/tex]

 [tex]d=79.9m[/tex]

Answer:

The electric field is given by 4.5 N/C.

Explanation:

Charge density = 80 nC/m3

inner radius, r' = 1 mm

outer radius, r'' = 3 mm

distance,  r = 4 mm

The linear charge density is given by

[tex]\lambda =\rho \times\pi\times (r''^2 - r'^2)\\\\\lambda = 80\times 10^{-9}\times 3.14\times 10^{-6}\times(9-1)\\\\\lambda = 2\times 10^{-12}\\[/tex]

The electric field is given by

[tex]E = \frac{\lambda }{4\pi\varepsilon_or}\\E=\frac{9\times 10^9\times 2 \times 10^{-12}}{0.004}\\\\E=4.5 N/C[/tex]

Answer:

Work done applied = 12 newton-meter

Explanation:

Given examples:

Force applied = 6 newton

Distance of book = 2 meter

Find from the given data:

Work done

Computation:

The equation can be used to compute work.

Work done applied = Force applied x Distance of book

Work done applied = Force x Distance

Work done applied = 6 x 2

Work done applied = 12 newton-meter

Answer: see below explanation, should be straight forward from there? ;)

Explanation: 1 watt = 1 joule per second

Watt is a measure of energy over time

So 10 seconds... u got this :)

Answer:

Option B

Explanation:

A gaseous, highly reactive non-metal

Answer:B

Explanation:I just took the test

It is Potential energy's