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Leto [7]
3 years ago
8

The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the

excess static charge on a cockroach is modeled as point charges located at the end of each antenna, what magnitude of charge q would each antenna possess in order for each antennae to experience a force of magnitude 2.00 μN from the external electric field? Calculate q in units of nanocoulombs (nC) .
Physics
1 answer:
erica [24]3 years ago
4 0

Answer:

0.235 nC

Explanation:

Given:

  • E = the magnitude of electric field = 8.50\ kN/C =8.50\times 10^{3}\ N/C
  • F = the magnitude of electric force on each antenna = 2.00\ \mu N =2.00\times 10^{-6}\ N
  • q = The magnitude of charge on each antenna

Since the electric field is the electric force applied on a charged body of unit charge.

\therefore E = \dfrac{F}{q}\\\Rightarrow q =\dfrac{F}{E}\\\Rightarrow q =\dfrac{2.00\times 10^{-6}\ N}{8.50\times 10^{3}\ N/C}\\\Rightarrow q =0.235\times 10^{-9}\ C\\\Rightarrow q =0.235\ nC

Hence, the value of q is 0.235 nC.

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An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,
insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}

Where

t_{hf} = Temperature at the inside of the pipe = 300 °C

t_{f} = Temperature at the outside of the pipe = 20 °C

r₁ =internal  radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

k_A = 15 W/m·K

k_B = 0.038 W/m·K

h_{hf} = 75 W/m²·K

h_{cf} = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

Q = \frac{t_A-t_B}{R_T} Where R_T for the air film on the pipe outer surface is given by

R_T= \frac{1}{\alpha A}

where A =area of the outside of the pipe

= \frac{1}{10*2\pi*0.07*1 } = 0.227 K/W

Therefore

131.32 W = \frac{t_A-20}{0.227} which gives

t_A = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

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3 years ago
Planet Kling has half the radius and 2 times the mass of the Earth. What is the best estimate for the magnitude of the gravitati
Vlad [161]

Answer:80 m/s^2

Explanation:

6 0
3 years ago
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A discharge lamp rated at 25 W (1 W = 1 J/s) emits yellow light of wavelength 580 nm. How many photons of yellow light does the
Scrat [10]

Answer:

the number of photons of yellow light does the lamp generate in 1.0 s is 7 x 10^{19}

Explanation:

given information:

power, P = 25 W

wavelength. λ - 580 nm = 5.80 x 10^{-7} m

time, t = 1 s

to calculate the number of photon(N), we use the following equation

N = λPt/hc

where

λ = wavelength (m)

P = power (W)

t = time interval (s)

h = Planck's constant (6.23 x 10^{-34} Js)

c = light's velocity (3 x 10^{8} m/s^{2})

So,

N = λPt/hc

   = (5.80 x 10^{-7})(25)(1)/(6.23 x 10^{-34})(3 x 10^{8} m/s^{2})

    = 7 x 10^{19}

3 0
3 years ago
What force is needed to move a barrel 45-m if 3600 J of work are accomplished?​
lesantik [10]

Answer:

<h3>The answer is 80 N</h3>

Explanation:

The force acting on the object can be found by using the formula

f =  \frac{w}{d}  \\

where

d is the distance

w is the work done

We have

f =  \frac{3600}{45}  \\

We have the final answer as

<h3>80 N</h3>

Hope this helps you

8 0
3 years ago
It takes a truck 3.56 seconds to slow down from 112 km/h to 87.4 km/h. What is its average acceleration?
maxonik [38]

Answer:

option B

−1.92 m/s2

Explanation:

Given in the question,

time took by truck to slow down = 3.56 sec

initial speed of truck = 112 km/h

final speed of truck = 87.4 km/h

1 km/h = 0.277778 m/s

112 = 31.1 m/s

87.4 = 24.28 m/s

Formula use to calculate the acceleration

v - u = at

where v is final speed

           u is initial speed

           a is acceleration

           t is time

plug values in the equation

24.28 - 31.1 = a(3.56)

-6.8 = a(3.56)

a = -6.8 / 3.56

a = -1.9 m/s²

3 0
3 years ago
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