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zheka24 [161]
3 years ago
13

What are 5 aspects of resistors that are connected in parallel ?

Physics
1 answer:
amid [387]3 years ago
6 0

Answer:

Voltage.

current.

Resistance.

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If a girl is running along a straight road with a uniform velocity 1.5 m/s, find her
Dafna1 [17]

Answer:

Explanation:

The definition of acceleration is the change in velocity over a period of time. If the girl's velocity is constant, that means it's not changing. Therefore, acceleration is 0 m/s/s

7 0
3 years ago
The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de
spin [16.1K]

Answer:

the spring constant k = 5.409*10^4 \ N/m

the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

Explanation:

From Hooke's Law

F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m

Thus; the spring constant k = 5.409*10^4 \ N/m

The amplitude is decreasing 37% during one period of the motion

e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}

b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) }    }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s

Therefore; the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

5 0
3 years ago
* A ball is projected horizontally from the top of
bagirrra123 [75]

Explanation:

Given

Ball is projected horizontally from a building of height h=19.6\ m

time taken to reach ground is given by

\text{Cosidering vertical motion}\\\Rightarrow h=ut+0.5at^2\\\Rightarrow 19.6=0+0.5\times 9.8t^2\\\Rightarrow t^2=4\\\Rightarrow t=2\ s

(b) Line joining the point of projection and the point where it hits the ground makes an angle of 45^{\circ}

From the figure, it can be written

\Rightarrow \tan 45^{\circ}=\dfrac{h}{x}\\\\\Rightarrow x=h\cdot 1\\\Rightarrow x=19.6

Considering horizontal motion

\Rightarrow x=u_xt\\\Rightarrow 19.6=u_x\times 4\\\Rightarrow u_x=4.9\ m/s

(c) The vertical velocity with which it strikes the ground is given by

\Rightarrow v^2-u_y^2=2as\\\Rightarrow v^2-0=2\times 9.8\times 19.6\\\Rightarrow v=\sqrt{384.16}\\\Rightarrow v=19.6\ m/s

Thus, the ball strikes with a vertical velocity of 19.6\ m/s

6 0
3 years ago
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Calculate the electric field at the center of a square 46.4 cm on a side, if one corner is occupied by a +42.0 µc charge and the
liraira [26]

centre of square disrance to each corner found by Pythagoras' theorem.

coulombs law used to clculate field of each charge at centre

fields added vectorially for res

8 0
3 years ago
A ball initially at rest rolls down a hill and has an acceleration of 3.3 m/s2. If it accelerates for 7.5 s, how many meters wil
ollegr [7]

The distance the ball rolls is given by

<em>x</em> = 1/2 <em>a</em> <em>t</em>²

so that

<em>x</em> = 1/2 * (3.3 m/s²) * (7.5 s)² = 92.8125 m ≈ 93 m

8 0
4 years ago
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