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3 years ago

WILL MARK BRAINLIEST!!!!!!!

2 answers:

7 0

In a series circuit, all devices are constrained to a single flow of current. There can only be a single value for the current for all devices, otherwise this would violate the conservation of charge. Therefore the current must be the same across each resistor.

<h3>The answer is D.</h3>

Shkiper50 [21]3 years ago

3 0

Answer:

option D.

Explanation:

In a series circuit, two or more resistors are connected in a line. The equivalent resistance of the circuit is greater than any individual resistor. As more resistors are added, the current decreases. The voltage across each resistor is different. The current remains same across the circuit.

Thus, option D is correct.

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Through which element would someone most likely receive a shock

Elenna [48]

Gold or copper than silver

7 0

3 years ago

Read 2 more answers

A copper rod of cross-sectional area 11.6 cm2 has one end immersed in boiling water and the other in an ice-water mixture, which

julia-pushkina [17]

Answer:

0.686 g of ice melts each second.

Solution:

As per the question:

Cross-sectional Area of the Copper Rod, A = $11.6\ cm^{2} = 11.6\times 10^{- 4}\ m^{2}$

Length of the rod, L = 19.6 cm = 0.196 m

Thermal conductivity of Copper, K = $390\ W/m.^{\circ}C$

Conduction of heat from the rod per second is given by:

$q = \frac{KA\Delta T}{L}$

where

$\Delta T = 100^{\circ} - 0^{\circ} = 100^{\circ}C$ = temperature difference between the two ends of the rod.

Thus

$q = \frac{390\times 11.6\times 10^{- 4}\times 100}{0.196} = 228.48\ J/s$

Now,

To calculate the mass, M of the ice melted per sec:

$M = \frac{q}{L_{w}}$

where

$L_{w}$ = Latent heat of fusion of water = 333 kJ/kg

$M = \frac{228.48}{333\times 10^{3}} = 6.86\times 10^{- 4}\ kg = 0.686\ g$

5 0

3 years ago

The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh

Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

$v^2=v^2_o-2ax$

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

$\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}$

Next, consider that the formula for the force is:

$F=ma$

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

$F=(820kg)(0.66\frac{m}{s^2})=542.20N$

Hence, the required force to stop the car is 542.20N

4 0

1 year ago

What is the best use of an atomic model to explain the charge of the particles in thomson's beams

hoa [83]

The best use of an atomic model to explain the charge of the particles in Thomson's beams is:

<u>An atom's smaller negative particles are at a distance from the central positive particles, so the negative particles are easier to remove.</u>

<u>Explanation:</u>

In Thomson's model, an atom comprises of electrons that are surrounded by a group of positive particles to equal the electron's negative particles, like negatively charged “plums” that are surrounded by positively charged “pudding”.

Atoms are composed of a nucleus that consists of protons and neutrons . Electron was discovered by Sir J.J.Thomson. Atoms are neutral overall, therefore in Thomson’s ‘plum pudding model’:

atoms are spheres of positive charge
electrons are dotted around inside

Thomson's conclusions made him to propose the Rutherford model of the atom where the atom had a concentrated nucleus of positive charge and also large mass.

6 0

2 years ago

When is your kinetic energy the least when swinging on a park swing?

Afina-wow [57]

Answer:

An active pendulum has the most kinetic energy at the lowest point of its swing when the weight is moving fastest.

Explanation:

SO YOU HAVE THE LEAST KINETIC ENERGY AT THE HIGHEST POINT OF THE SWING WHEN IT'S NOT ACTIVE

7 0

2 years ago