Ask question

Ask question

All categories

3 years ago

7

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

ic force in the earth’s electric field, which is typically 100 N/C, directed downward.
a) What is the ratio of the electric force on the bee to the bee's weight?
b) What electric field strength would allow the bee to hang suspended in the air? Express your answer with the appropriate units.

1 answer:

Sedbober [7]3 years ago

4 0

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 $mg=qE$

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

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

You might be interested in

Why didn't the astronauts land on the moon math worksheet answers?

Ghella [55]

Answer and Explanation:

Here's the answer!

4 0

3 years ago

The lowest-pitch tone to resonate in a pipe of length L that is closed at one end and open at the other end is 200 Hz. Which one

ozzi

Answer:

e. 400 Hz

Explanation:

In closed organ pipe, only odd harmonics of fundamental note is possible .

The fundamental frequency is 200 Hz . Then other overtones will be having following frequencies .

200 x 3 , 200 x 5 , 200 x 7 , 200 x 9 etc

600 Hz , 1000 Hz , 1400 Hz , 1800 Hz .

Frequency not possible is 400 Hz .

7 0

3 years ago

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

4 0

3 years ago

A person is lifting a heavy box using a lever. What is the purpose of the lever in this situation?

ruslelena [56]

Answer:

to reduce the <em>force</em> needed to lift the box and <em>change</em> the direction of the force

Explanation:

1. "A lever consists of a rigid bar that is able to pivot at one point. This point of rotation is known as the fulcrum. A force is applied at some point away from the fulcrum (typically called the effort)."

By this definition, we know that force is needed to lift an object using a lever.

2.<u> "When the input and output forces are on opposite sides of the fulcrum, </u><u>the lever changes the direction of the applied force.</u> This occurs only with first-class levers. When both the input and output forces are on the same side of the fulcrum, the direction of the applied force does not change"

For example, on a sew saw, if a force is applied on one end, you on the other side/end would go up, meaning <u>a change in direction</u>.

3. Lastly, we know <u><em>a lever is typically used to reduce work</em></u>, in other words, the force needed to move something.

Basically, if we were to put a lever into an equation:

reduced force + change in direction = lever

(<em>the expection</em>) <u>unless load and force are on the same side</u>, there will be <u>no change in direction. </u>

For example, if you and your friend sit on the same side of a sew saw, the sew saw would not go up or down, meaning no change in direction.

So if not stated otherwise you can assume the load and force are on opposite sides. The purpose of a lever in that situation would be to reduce the force needed to lift the box and change the direction of the force.

*While reading my explanation, it may be helpful to look up a diagram containing a lever, with a load, fulcrum, and applied force.

6 0

2 years ago

The flow of electrons through a wire or any conductor is called____.

Karo-lina-s [1.5K]

Electricity. Ruler. 69. N.

5 0

3 years ago