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krek1111 [17]
3 years ago
13

I need help on 2/3 please asap thx ♥️

Physics
1 answer:
ch4aika [34]3 years ago
5 0

<u>questions 2</u>

F=m

therefore

a=F/m

a=408/68

=6m/s^2.

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An ideal spring with spring constant k is hung from the ceiling. The initial length of the spring, with nothing attached to the
hram777 [196]

The mass m of the object = 5.25 kg

<h3>Further explanation</h3>

Given

k = spring constant = 3.5 N/cm

Δx= 30 cm - 15 cm = 15 cm

Required

the mass m

Solution

F=m.g

Hooke's Law

F = k.Δx

\tt m.g=k.\Delta x\\\\m.10=3.5\times 15\\\\m=5.25~kg

7 0
2 years ago
As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i
11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

Explanation:

Given that,

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Bounced height = 1.48 m

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Velocity of the ball bearing just before hitting the steel plate

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mgh=\dfrac{1}{2}mv^2

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v=\sqrt{2\times9.8\times1.94}

v=6.166\ m/s

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

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mgh=\dfrac{1}{2}mv^2'

Put the value into the formula

9.8\times1.48=\dfrac{1}{2}\times v^2'

v'=\sqrt{2\times9.8\times1.48}

v'=5.38\ m/s

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

a=\dfrac{v-v'}{t}

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a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}

a=776.98\ m/s^2

a=0.77\times10^{3}\ m/s^2

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6 0
3 years ago
How many items are present in the compound (NH4)2Cr2O7
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In 2Cr2O7 there’s 2 items; Cr and O
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3 years ago
A straight wire of length 4.5 cm moves at a constant speed of 5.2m/s perpendicular to its length and a uniform magnetic field. I
liraira [26]
Well you have to minus the 4.5 to 5.2 and the answer to that would be -11.5 and calculated that to be 4.5
3 0
3 years ago
Charge: A piece of plastic has a net charge of +2.00 μC. How many more protons than electrons does this piece of plastic have? (
snow_lady [41]

Answer

given,

net charge = +2.00 μC

we know,

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1 micro coulomb  charge =  6.28 x 10¹⁸ x 10⁻⁶ electron

                                         = 6.28 x 10¹² electrons

2.00 μC = 2 x 6.28 x 10¹² electrons

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since net charge is positive.

The number of protons should be 1.256 x 10¹³ more than electrons.

hence, +2.00 μC have 1.256 x 10¹³ more protons than electrons.

6 0
3 years ago
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