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3 years ago

Which contribution added to Rutherford's discovery of a positive nucleus?

Physics

2 answers:

SCORPION-xisa [38]3 years ago

7 0

Its B. because he did discovered that electrons are scatterd in a atom

OverLord2011 [107]3 years ago

7 0

B) Electrons are scattered in an atom.

is the answer

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Dolphins communicate using compression waves (longitudinal waves). Some of the sounds dolphins make are outside the range of hum

lana66690 [7]

Answer: send the message underwater because a more dense medium would make the sound travel faster.

Explanation:

Dolphins communicate using compression waves - longitudinal waves. Longitudinal waves requires a medium to travel. A longitudinal wave transfers energy by the vibration of medium particles in the direction of the wave motion. Compression are the regions where density of the medium is higher and rarefaction is a low density region.

A longitudinal wave travels faster in a denser medium. It has maximum speed in solid and minimum in gas. Thus, to transfer message quickly to dolphin B., dolphin A should send the message underwater and not in air. This is because water has higher density than air. Molecules collide more quickly in water than in air and it takes less time for signal to travel.

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3 years ago

Can someone help me please

TEA [102]

Answer:

Explanation:

Because it is impossible for it to show the real depth of the ocean and how deep it is

8 0

3 years ago

Physics students study a piano being pulled across a room on a rug. They know that when it is at rest, it experiences a gravitat

Vanyuwa [196]

The static frictional force is greater than the kinetic frictional force, so the static frictional force is greater than 1200 N.

5 0

3 years ago

Read 2 more answers

Two parallel plates are 1 cm apart and are connected to a 500 V source. What force will be exerted on a single electron half way

VladimirAG [237]

Answer: 8*10^-15 N

Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that, F=q*E

The electric field between the plates is given by:

E = ΔV/d = 500 V/0.01 m=5*10^3 N/C

the force applied to the electron is: F=e*E=8*10^-15 N

3 0

2 years ago

You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene

uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is $150\,^{\circ}C$ . According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

$\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T)$ (1)

Where:

$\rho_{k}$ - Density of kerosene, measured in kilograms per cubic meter.

$V_{k}$ - Volume of kerosene, measured in cubic meters.

$c_{k}$ , $c_{t}$ - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

$T$ - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

$V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}$

If we know that $m_{t} = 1.83\,kg$ , $c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{t,o} = 88\,^{\circ}C$ , $T_{k,o} = 24.0\,^{\circ}C$ , $T = 57\,^{\circ}C$ , $c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C}$ and $\rho_{k} = 820\,\frac{kg}{m^{3}}$ , then the volume of the liquid needed to accomplish this task without boiling is:

$V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}$

$V_{k} = 2.273\times 10^{-4}\,m^{3}$

$V_{k} = 0.273\,L$

0.273 liters are needed to accomplish this task without boiling.

3 0

3 years ago