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3 years ago

HELP IM TIMED. ILL MARK BRAINLIEST PLSSSSSS.

Physics

1 answer:

Alex17521 [72]3 years ago

5 0

Answer:

According to Newton's Second Law of Motion :

Where,

F = Force Applied

m = Mass of the object

a = Acceleration

Now, we will use this law to solve this question.

Given :

Acceleration or a = 15.3 m/s²

Force = 44 N

Mass = ?

Substitute, the given values in the formula.

F = ma

⇒ m = F/a

m = 44/15.3

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A series of pulses, each of amplitude 0.1m , is sent down a string that is attached to a post at one end. The pulses are reflect

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The net displacement at a point on the string where the pulses cross is 0.2 m.

The term "displacement" refers to a shift in an object's position. It has a magnitude and a direction, making it a vector quantity. An arrow pointing from the starting point to the finishing point serves as its symbol.

A string that is connected to a post at one end is used to transmit a sequence of pulses, each measuring 0.1 meters in amplitude.

At the post, the pulses are reflected and return along the string without losing any of their amplitude.

Now, let's say the ends are free.

There is no inversion on reflection if the end is free. The amplitude at their intersection is 2A.

Now, since A = 0.1 m

Then, 2A = 2(0.1) = 0.2 m

As a result, the net displacement at the string's intersection of two pulses is 0.2 m.

The correct option is (c).

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1 year ago

A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subject

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Answer:

The deflection of the spring is 34.56 mm.

Explanation:

Given that,

Diameter = 10 mm

Number of turns = 10

$Radius_{mean} = 60\ mm$

$Diameter_{mean} = 120\ mm$

Load = 200 N

We need to calculate the deflection

Using formula of deflection

$\delta=\dfrac{8pD^3n}{Cd^4}$

Put the value into the formula

$\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}$

$\delta =34.56\ mm$

Hence, The deflection of the spring is 34.56 mm.

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3 years ago

A spring is hung from the ceiling. A 0.442-kg block is then attached to the free end of the spring. When released from rest, the

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Answer:

$K=58.8N/m$

Explanation:

From the question we are told that:

Mass $M=0.442$

Drop distance $d=0.150$

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$K=\frac{2mg}{x}$

$K=\frac{2*0.442*9.8}{1.150}$

$K=58.8N/m$

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3 years ago

Which type of energy is released when a bond between atoms is broken

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Answer:

Explanation:

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3 years ago

a tank circuit contains a capacitor and an inductor that produce 30 of reactance at the resonant frequency. the inductor has a q

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The total circuit current at the resonant frequency is 0.61 amps

What is a LC Circuit?

A capacitor and an inductor, denoted by the letters "C" and "L," respectively, make up an LC circuit, also referred to as a tank circuit, a tuned circuit, or a resonant circuit.
These circuits are used to create signals at particular frequencies or to receive signals from more complicated signals at particular frequencies.

Q =15 = (wL)/R

wL = 30 ohms = Xl

R = 2 ohms

Zs = R + jXl = 2 +j30 ohms where Zs is the series LR impedance

| Zs | = 30.07 <86.2° ohms

Xc = 1/(wC) = 30 ohms

The impedance of the LC circuit is found from:

Zp = (Zs)(-jXc)/( Zs -jXc)

Zp = (2+j30)(-j30)/(2 + j30-j30) = (900 -j60)2 = 450 -j30 = 451 < -3.81°

I capacitor = 277/-j30 = j9.23 amps

I Zs = 277/(2 +j30) = (554 - j8,310)/904 = 0.61 - j9.19 amps

I net = I cap + I Zs = 0.61 + j0.04 amps = 0.61 < 3.75° amps

Hence, the total circuit current at the resonant frequency is 0.61 amps

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1 year ago