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2 years ago

Define the word logrolling in your own words?

Physics

2 answers:

Elan Coil [88]2 years ago

3 0

A crazy sport thats kinda dangerous

Doss [256]2 years ago

3 0

Answer:

Logrolling is the practice of legislators exchanging favors, or quid pro quo, such as vote trading, in order to enact legislation that is of interest to each of them.

<u>OAmalOHopeO</u>

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What’s the voltage of a battery in a circuit with resistance of 3 ohms and current of 5 amps?

alexira [117]

Answer: The correct answer is-15 Volts.

Explanation-

Voltage of a battery can be defined as the difference in electric potential that lies between the positive and negative terminals of a battery.

It can be calculated using Ohm's law, which states that the electric potential difference between two points on a circuit is equal to the product of the current that flows between the two points (I) and the total resistance that sis present between the two points. It can be mathematically depicted as-

ΔV = I • R

Putting the value of 'I' and 'R', we get-

ΔV = 5 X 3

= 15 V

8 0

2 years ago

What is the velocity of a proton that has been accelerated by a potential difference of 15 kV? (i)\:\:\:\:\:9.5\:\times\:10^5\:m

andre [41]

Answer:

Velocity of a proton, $v=1.7\times 10^6\ m/s$

Explanation:

It is given that,

Potential difference, $V=15\ kV=15\times 10^3\ V$

Let v is the velocity of a proton that has been accelerated by a potential difference of 15 kV.

Using the conservation of energy as :

$\dfrac{1}{2}mv^2=qV$

q is the charge of proton

m is the mass of proton

$v=\sqrt{\dfrac{2qV}{m}}$

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 15\times 10^3\ V}{1.67\times 10^{-27}\ kg}$

$v=1695361.75\ m/s$

$v=1.69\times 10^6\ m/s$

$v=1.7\times 10^6\ m/s$

So, the velocity of a proton is $1.7\times 10^6\ m/s$ . Hence, this is the required solution.

8 0

2 years ago

A speed-time graph is shown below:

VladimirAG [237]

Answer: $0.5 m/s^{2}$

Explanation:

Average acceleration $a_{ave}$ is the variation of velocity $\Delta V$ over a specified period of time $\Delta t$ :

$a_{ave}=\frac{\Delta V}{\Delta t}}$

Where:

$\Delta V=V_{f}-V_{o}$ being $V_{o}=0 cm/s$ the initial velocity and $V_{f}=4 cm/s$ the final velocity (according to the information given from the described graph)