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coldgirl [10]
3 years ago
6

A business letter is informal correspondence and is sent to people within

Physics
1 answer:
jarptica [38.1K]3 years ago
5 0

Answer:

FALSE!!!

Explanation:

Business letters are formal and normally given to someone of higher importance.

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Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900
Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron m_{e}=9.11\times10^{-31}\ kg

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}

E=0.582\ Mev

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}

E=0.350\ Mev

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0
3 years ago
Why do you think it is a driving force of weather around the world?
Finger [1]

Answer:

The sun.

Explanation:

The sun provides energy for living organisms, and it drives our planet’s weather and climate patterns.

Remember, Earth is spherical and the energy from the sun does not reach all areas with equal intensity. Areas exposed to the sun are directly on the sun’s rays (i.e. those nearest to the equator) and hence, receive greater solar input. In contrast, those in higher latitudes receive sunlight that is spread over a larger area and that has taken a longer path through the atmosphere. As a result, these higher latitudes receive less solar energy.

Also, ocean circulation and precipitation are all factors of weather

4 0
2 years ago
B. If we drop a ball from the roof, its falls<br>downward​
frosja888 [35]

Answer:

yes ,what's the question

and if the question is why

then it's for gravity

4 0
2 years ago
Read 2 more answers
Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?
Alexandra [31]

11) 1.04\cdot 10^7 J

12) 1.04\cdot 10^7 J

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

PE=mgh

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

g=9.8 m/s^2 (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

PE=(8325)(9.8)(127)=1.04\cdot 10^7 J

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

KE_t + PE_t = KE_b + PE_b

where

KE_t is the kinetic energy at the top

PE_t is the potential energy at the top

KE_b is the kinetic energy at the bottom

PE_b is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

KE_t=0

Also, at the bottom the height is zero, so the potential energy is zero

PE_b=0

Therefore, we find:

KE_b=PE_t=1.04\cdot 10^7 J

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

KE=1.04\cdot 10^7 J

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

KE_1 + PE_1 = KE_2 + PE_2

where

KE_1 is the kinetic energy at the top of the 1st hill

PE_1 is the potential energy at the top of the 1st hill

KE_2 is the kinetic energy at the top of the 2nd hill

PE_2 is the potential energy at the top of the 2nd hill

From the previous questions, we know that

KE_1=0

and

PE_1=1.04\cdot 10^7 J

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J

So, the kinetic energy at the second hill is

KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J

And so, the speed is

v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s

4 0
2 years ago
I will Mark Brainliest 1. ) Allena and Charissa were discussing whether Cl2 is an element or a compound. Allena said that it is
WITCHER [35]
The answer to question one is A.
The answer to question two is A.
The answer to question three is D.

7 0
3 years ago
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