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sveta [45]
3 years ago
13

A train departs from its station at a constant acceleration of 5 m/s. What is the speed of the train at the end of 20s?

Physics
2 answers:
shutvik [7]3 years ago
7 0

Answer:

Velocity, v = 100 m/s

Explanation:

It is given that,

Initial speed of the train, u = 0

Acceleration of the train, a=5\ m/s^2

Let v is the velocity of the train at the end of 20 seconds, t = 20 s

Using the equation of kinematics as :

v = u + a × t

v = a × t

v = 5 × 20

v = 100 m/s

So, the velocity of the train after 20 s is 100 meters. Hence, this is the required solution.

morpeh [17]3 years ago
4 0
Acceleration = (final velocity-initial velocity)/time
5 = (v-0)/20
v = 100m/s
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How much work must be done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners
Natasha_Volkova [10]

Answer:

Potential\ Energy=Work \ Done=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

Explanation:

The potential energy is given by:

U=Q*V

where:

Q is the charge

V is the potential difference

Potential Difference=V=\frac{kq}{r}

So,

Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}

Where:

k is Coulomb Constant=8.99*10^9 Nm^2/C^2

q is the charge on electron=-1.6*10^-19 C

r is the distance=3.0*10^{-10}m

For 3 Electrons Potential Energy or work Done is:

Potential\ Energy=3*\frac{kq^2}{r}

Potential\ Energy=3*\frac{(8.99*10^9)(-1.6*10^{-19})^2}{3*10^{-10}}\\Potential\ Energy=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

7 0
3 years ago
Two motorcycles are traveling in opposite directions at the same speed when one of the cyclists blasts her horn, which has frequ
Galina-37 [17]

Answer:

6ms^-1

Explanation:

Given that the frequency difference is

( 563- 544) = 19

So alsoThe wavelength of each wave is = v/f = 344 /544

and there are 19 of this waves

So it is assumed that each motorcycle has moved 0.5 of this distance

in one second thus the speed of the motorcycles will be

=> 19/2 x 344/544 = 6.0 m/s

5 0
3 years ago
In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s
sladkih [1.3K]

Answer:

  • quality factor (Q) = 69.99
  • inductor = 1.591 x 10⁻⁴ H
  • capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R =  10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW)  = \frac{R}{2\pi L}

BW = \frac{R}{2\pi L}

make L (inductor) the subject of the formula

L = \frac{R}{2\pi *BW}  =  \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH

F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}

make C (capacitor)  the subject of the formula

C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F

quality factor (Q) = \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99

quality factor (Q) =  69.99

5 0
3 years ago
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vazorg [7]

Answer:

The number of valence electrons increases to 8 or, the atom gives up loosely held valence electrons.

Explanation:

Please mark brainliest and have a great day!

3 0
3 years ago
Which of these activities increases the amount of carbon in the atmosphere?
Llana [10]

What are the choices

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3 years ago
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