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3 years ago

7

helpppp I just saw my crush when I was walking out of a supermarket and like I was looking a mess so now I'm just wondering how

it would went when we saw each other if I told him I like him (I think he knows but untill then not gonna confess)

1 answer:

UkoKoshka [18]3 years ago

3 0

Answer:

okay

Explanation:

(wish i could help but im just answering for points)

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oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student

puteri [66]

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

3 0

2 years ago

Please help I need help this is so confusing to me

spayn [35]

C. increased stirring i think

7 0

3 years ago

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How many days sun rises in a year

Lana71 [14]

Answer:

365

Explanation:

there are 365 days in a year

8 0

3 years ago

Copper has a modulus of elasticity of 110 GPa. A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm is pull

bearhunter [10]

Answer:

$strain = 1.4 \times 10^{-3}$

Explanation:

As we know by the formula of elasticity that

$E = \frac{stress}{strain}$

now we have

$E = 110 GPA$

$E = 110 \times 10^9 Pa$

Area = 15.2 mm x 19.1 mm

$A = 290.3 \times 10^{-6}$

now we also know that force is given as

$F = 44500 N$

here we have

stress = Force / Area

$stress = \frac{44500}{290.3 \times 10^{-6}}$

$stress = 1.53 \times 10^8 N/m^2$

now from above formula we have

$strain = \frac{stress}{E}$

$strain = \frac{1.53 \times 10^8}{110 \times 10^9}$

$strain = 1.4 \times 10^{-3}$

7 0

3 years ago

A 3.0 N force and a 5.0 N force are acting in opposite directions on the same 5.0 kg ball. What is the magnitude of the accelera

Nataly [62]

F(net)=ma

(5-3)=5a
2=5a
a=0.4

Hope I can help u

4 0

3 years ago

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