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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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The complete queston is The amount of a radioactive element A at time t is given by the formula

A(t) = A₀e^kt

Answer:  A(t) =N e^( -1.2 X 10^-4t)

Explanation:

Given

Half life =  5730 years.

A(t) =A₀e ^kt

such that

A₀/ 2 =A₀e ^kt

Dividing both sides by A₀

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1/2 = e ^k(5730)

1/2 = e^5730K

In 1/2 =  5730K

k = 1n1/2 / 5730

k = 1n0.5 / 5730

K= -0.00012 = 1.2 X 10^-4

So that expressing   N in terms of t, we have

A(t) =A₀e ^kt

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A(t) =N e^ -1.2 X 10^-4t

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Mercury is in the 80th position in the periodic table. How many protons does it have?
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Mercury has 80 protons.
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8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s
allsm [11]

Answer:

a) \omega = 50\,\frac{rad}{s}, b) \omega = 0\,\frac{rad}{s}

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

\tau = F \cdot r

\tau = m\cdot a \cdot r

\tau = m \cdot \alpha \cdot r^{2}

Where \alpha is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}

\alpha = -3\,\frac{rad}{s^{2}}

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)

\omega = 50\,\frac{rad}{s}

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}

t = 66.667\,s

Since t > 66.667\,s, then the angular velocity is equal to zero. Therefore:

\omega = 0\,\frac{rad}{s}

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3 years ago
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Answer:

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Unique and methodical activity performed to improve the reservoir of knowledge is known as the research. . A research project might be a continuation of previous work in the topic.

By the following way, the test is conducted;

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Athletes require oxygen to give energy and to eliminate carbon dioxide, which is a waste product produced when energy is produced.Because it is a reliable indicator of ventilatory efficiency in cardiac patients.

Ventilatory threshold performance is used to determine the lactate threshold. The maximal rate of oxygen consumption recorded during activity is 5.VO2 max. It is higher in trained people than in unskilled people.

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