All categories

3 years ago

If the force of an object doubles and the mass also doubles, by what factor does the acceleration change?

1 answer:

3 0

Starting with the equation for newtons second law of motion: $F = ma \rightarrow 2F = 2ma \rightarrow \frac{2F}{2m} = a \rightarrow \frac{F}{m} = a$

You might be interested in

What is one property of iodine

mamaluj [8]

Answer: I2

Explanation:

6 0

3 years ago

The difference in potential between the cathode and anode of a spark plug is 14700 v. what energy does the electron give up as i

Verdich [7]

The energy the electron gives up passing between the electrodes is equal to the product between its charge and the potential difference between the electrodes:
$\Delta U = e \Delta V$
where
e is the electron charge
$\Delta V$ is the potential difference

Plugging numbers into the equation, we find that the electron gives up is
$\Delta U = (1.6 \cdot 10^{-19} C)(14700 V)=2.35 \cdot 10^{-15} J$

8 0

3 years ago

What is one of the earliest ways that ancient peoples produced light?

VARVARA [1.3K]

Answer:

Candles

Explanation:

The answer is the first option or "candles." In ancient times the people would lit candles and place them in difference places in their homes, temples, etc... to produce light. It's not option B because light emitting diodes are used to light up toys and such which weren't invented in ancient times. It's not option C because fluorescent bulbs are long light bulbs used to light up huge spaces without as many light bulbs which were again not invented at the time, and it's also not option D because incandescent bulbs are you usual light bulbs which weren't invented at the time.

Hope this helps.

4 0

3 years ago

Please help ASAP.

Elden [556K]

Answer:

1st question c part

2nd question c part

8 0

3 years ago

Determine the energy required to accelerate an electron from (a) 0.500c to 0.900c and (b) 0.900c to 0.990c

Lera25 [3.4K]

Answer:

0.582 MeV

2.45 MeV

Explanation:

$E_r=0.511\ MeV$ =Electron rest energy

(a) 0.500c to 0.900c

$W=\left(\frac{1}{\sqrt{1-\frac{v_f^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}}\right)E_r\\\Rightarrow W=\left(\frac{1}{\sqrt{1-0.9^2}}-\frac{1}{\sqrt{1-0.5^2}}\right)0.511\\\Rightarrow W=0.582\ MeV$

Energy required is 0.582 MeV

(b) 0.900c to 0.990c

$W=\left(\frac{1}{\sqrt{1-\frac{v_f^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}}\right)E_r\\\Rightarrow W=\left(\frac{1}{\sqrt{1-0.99^2}}-\frac{1}{\sqrt{1-0.9^2}}\right)0.511\\\Rightarrow W=2.45\ MeV$

Energy required is 2.45 MeV

6 0

3 years ago