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3 years ago

Q2. Fill in the spaces with the correct form of the verb in future progressive tense.

Physics

1 answer:

-BARSIC- [3]3 years ago

8 0

Answer:

will you fed the dogs while mom is in the hospital?

Explanation:

hope it may help

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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

3 years ago

Which has the greater acceleration, a person going from 0 m/s to 10 m/s in 10 seconds or an ant going from 0 m/s to 0.25 m/s in

monitta

Person-1

Initial velocity=u=0m/s
Final velocity=v=10m/s
Time=10s=t

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{10-0}{10}$

$\\ \sf\longmapsto Acceleration=\dfrac{10}{10}$

$\\ \sf\longmapsto Acceleration=1m/s^2$

Person-2

initial velocity=0m/s=u
Final velocity=v=0.25m/s
Time=t=2s

$\\ \sf\longmapsto Acceleration=\dfrac{0.25-0}{2}$

$\\ \sf\longmapsto Acceleration=\dfrac{0.25}{2}$

$\\ \sf\longmapsto Acceleration=0.125m/s^2$

Person-1 is accelerating faster.

4 0

3 years ago

Where do the electrons that form the auroras enter the magnetosphere? a. Through holes c. At the equator b. Between the magnetic

marta [7]

I think the answer is d. In the magnetotail. I hope this helps! :)

7 0

3 years ago

Read 2 more answers

If a piece of an object is dropped down vertically is the moment of inertia gonna be 0? And why?

Degger [83]

Answer:

The object is dropped, we know the initial velocity is zero. Once the object has left contact with whatever held or threw it, the object is in free-fall. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g.

No so sure

Explanation:

Hope it helps

5 0

2 years ago

Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a

Sergio039 [100]

Answer:

Yes, it is reasonable to neglect it.

Explanation:

Hello,

In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):

$m=1molec*\frac{1mol}{6.022x10^{23}molec} *\frac{32g}{1mol}*\frac{1kg}{1000g}=5.31x10^{-26}kg$

After that, we compute the potential energy 1.00 m above the reference point:

$U=mhg=5.31x10^{-26}kg*1.00m*9.8\frac{m}{s^2}=5.2x10^{-25}J$

Then, we compute the average kinetic energy at the specified temperature:

$K=\frac{3}{2}\frac{R}{Na}T$

Whereas $N_A$ stands for the Avogadro's number for which we have:

$K=\frac{3}{2} \frac{8.314\frac{J}{mol*K}}{6.022x10^{23}/mol}*(23+273)K\\ \\K=6.13x10^{-21}J$

In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.

Regards.

8 0

3 years ago