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2 years ago

14

Physics

1 answer:

Ksivusya [100]2 years ago

4 0

Answer:

ok so youll tell me when you have problems

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I need question 5 please and correct any others if they’re wrong

Kryger [21]

The resistance would increase.

8 0

3 years ago

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You are standing 30 m from a window that is 30 m off the ground. Your friend tries to toss you something, but only throws it str

pochemuha

Answer:

minimum initial velocity is 21.35 m/s

Explanation:

given data

distance S = 30 m

height h = 30 m

maximum acceleration a = 2 m/s²

to find out

minimum initial velocity that your friend could have thrown the object to enable you to catch

solution

first we get here time with the help of second equation of motion

time = $\sqrt{\frac{2h}{a} }$ ..................1

put her value we get

time = $\sqrt{\frac{2*30}{2} }$

time = 5.477 second

and that is time which tossed object must be take so we apply here again second equation of motion that is

-S = ut - 0.5 × gt² .......................2

-30 = u× 5.477 - 0.5 ×9.8×5.477²

solve it we get

u = 21.35 m/s

so minimum initial velocity is 21.35 m/s

3 0

3 years ago

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why does a spark jump from your hand to the doorknob as you reach for the door after walking across a carpeted floor?

Anna11 [10]

<span>because electrical charge builds up otherwise know as static electricity</span>

8 0

3 years ago

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The force that pulls away from a curve is known as __________ .

tatyana61 [14]

Answer: <u>Centrifugal Force </u>

Explanation:

The force that pulls out from the center on a body in circular motion is called centrifugal force and it increases with acceleration. Centrifugal force results in strong outward pull on your vehicle.

4 0

2 years ago

A series RLC circuit consists of a 52.0 Ω resistor, a 4.80 mH inductor, and a 330 nF capacitor. It is connected to an oscillator

Tju [1.3M]

Answer:

(G) 75.11 ohm

(H) 0.08 A

(I) 46.2 degree

Explanation:

R = 52 ohm

L = 4.8 m H = 4.8 x 106-3 H

C = 330 nF = 330 x 10^-9 F

Vo = 6 V

(G)

f = 5000 Hz

Let the impedance is Z.

$X_{L}= 2 \pi fL = 2 \times 3.14\times 5000\times 4.8\times 10^{-3}=150.72 ohm$

$X_{c}= \frac{1}{2 \pi fC}=\frac{1}{2\times 3.14\times 5000\times 330\times 10^{-9}}=96.51 ohm$

$Z=\sqrt{R^{2}+\left ( X_{L}-X_{c} \right )^{2}}$

$Z=\sqrt{52^{2}+\left (150.72-96.51)^{2}}=75.11 ohm$

(H) Let Io be the peak current

$I_{0}=\frac{V_{0}}{Z}=\frac{6}{75.11}=0.0798 A = 0.08 A$

(I) Let Ф be the phase angle

$tan\phi = \frac{X_{L}-X_{C}}{R}$

$tan\phi =\frac{150.72-96.51}}{52}=1.0425$

Ф = 46.2 degree

4 0

3 years ago