A postman needs to deliver a parcel to Peter's house 5

Honey bees can acquire a small net charge on the order of 1 pC as they fly through the air and interact with plants. Estimate th

murzikaleks [220]

Answer:

$3.35\cdot 10^{-16}N$

Explanation:

The force exerted on a charged particle due to a magnetic field is given by:

$F=qvB sin \theta$

where

q is the charge of the particle

v is the velocity of the particle

B is the magnetic field strength

$\theta$ is the angle between the directions of B and v

In this problem we have:

$q=1 pC=1\cdot 10^{-12} C$ is the charge of one honey bee

$v=15 mi/h = 6.7 m/s$ is the velocity of the bee

$B=5.0\cdot 10^{-5} T$ is the average strength of the Earth's magnetic field

$\theta=90^{\circ}$ , because the bee flies east to west while the Earth's magnetic field is south to north

Substituting into the equation, we find:

$F=(1\cdot 10^{-12}C)(6.7 m/s)(5.0\cdot 10^{-5} T)(sin 90^{\circ})=3.35\cdot 10^{-16}N$

7 0

3 years ago

if the current through a resistor is increased by a factor of 4, how does this affect the power dissipated?

lbvjy [14]

The new current will be 4I. The power dissipated by the resistor will increase by a factor of 16.

<h3>What is a resistor?</h3>

Using electrical resistance as a circuit element, a resistor is a passive electrical component with two terminals. In electrical circuits, resistors are used, among other things, to limit current flow, modify signal levels, divide voltages, bias active devices, and terminate transmission lines.

As test loads for generators, power distribution systems, and motor controls, high-power resistors that can create many watts of heat rather than just electrical energy can be used.

Variable resistors can be used as sensors for force, heat, light, humidity, humidity, and chemical activity or for adjusting circuit components.

Electrical networks and electronic circuits frequently contain resistors, which are found everywhere in electronic equipment. Practical resistors are discrete components that come in a wide range of materials and forms.

To learn more about resistor refer :

brainly.com/question/24858512

#SPJ4

4 0

1 year ago

Which element is the main component of<br> steel and the most widely used of all metals

Blizzard [7]

Answer:

iron

Explanation:

8 0

1 year ago

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Four charges with equal magnitudes of 10.6 × 10-12 C are placed at the corners of a rectangle. The lengths of the sides of the r

cricket20 [7]

Answer:

Figure a. E_net = 99.518 N/C

Figure b. E_net = 177.151 N / C

Explanation:

Given:

- Attachment for figures missing in the question.

- The dimensions for rectangle are = 7.79 x 3.99 cm

- All four charges have equal magnitude Q = 10.6*10^-12 C

Find:

Find the magnitude of the electric field at the center of the rectangle in Figures a and b.

Solution:

- The Electric field generated by an charged particle Q at a distance r is given by:

E = k*Q / r^2

- Where, k is the coulomb's constant = 8.99 * 10^9

Part a)

- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:

E_1 + E_3 = 0

- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:

E_net = E_2 + E_4

E_2 = E_4

E_net = 2*E = 2*k*Q / r^2

- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:

r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )

r = sqrt ( 1.9151*10^-3 ) = 0.043762 m

- Plug the values in the E_net expression developed above:

E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3

E_net = 99.518 N/C

Part b)

- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).

- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.

Hence,

E_net = 2*E_part(a)*cos(Q)

- Where, Q is the angle between resultant, vertical in direction, and each of the electric field. We can calculate Q using trigonometry as follows:

Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.

- Now, compute the net electric field E_net:

E_net = 2*(99.518)*cos(27.12)

E_net = 177.151 N / C

5 0

2 years ago

An object that covers more distance in the same amount of time has a higher speed.

DiKsa [7]

Answer:

true

Explanation:

5 0

2 years ago

Read 2 more answers