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goblinko [34]
3 years ago
15

On a separate sheet of paper, tell why scientists in different countries can easily compare the amount of matter in similar obje

cts in their countries
Physics
1 answer:
Harrizon [31]3 years ago
8 0

Answer: no u

Explanation: no u

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Work is done when spring is extended or compressed. Elastic potential energy is stored in the spring. Provided inelastic deformation has not happened, the work done is equal to the elastic potential energy stored.

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2 years ago
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Which statements describe the motion of the object represented by the graph? Check all that apply.
Sav [38]

Answer:a, d, e

Explanation:

6 0
3 years ago
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g Suppose Howard is pulling a bucket of bricks up along the side of a building with a rope. The bricks have a mass of 20 kg and
cupoosta [38]

Answer:

= 236N

Explanation:

tension T = mg + ma

Given that,

m = 20kg

g = 9.8 m/s²

a = 2.0 m/s²

T = m(g + a)

T = 20( 9.8 + 2.0)

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4 0
3 years ago
What is the intensity of 60dB sound?​
polet [3.4K]

Answer:

The intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².

Explanation:

Given;

intensity of the sound level, dB = 60 dB

The intensity of the sound in W/m² is calculated as;

dB = 10 Log[\frac{I}{I_o} ]\\\\

where;

I₀ is threshold of hearing = 1 x 10⁻¹² W/m²

I is intensity of the sound in W/m²

Substitute the given values and for I;

dB = 10 Log[\frac{I}{I_o} ]\\\\60 = 10 Log[\frac{I}{I_o} ]\\\\6 =  Log[\frac{I}{I_o} ]\\\\10^6 = \frac{I}{I_o} \\\\I = 10^6 \ \times \ I_o\\\\I = 10^6 \ \times \ 1^{-12} \ W/m^2 \\\\I = 1\ \times \ 10^{-6} \ W/m^2

Therefore, the intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².

5 0
3 years ago
A point charge A of charge +4micro coloumb and another B of -1 micro coloumb are placed at a distance in air 1m apart then the d
andrew11 [14]

Answer:

Explanation:

Given that,

A point charge is placed between two charges

Q1 = 4 μC

Q2 = -1 μC

Distance between the two charges is 1m

We want to find the point when the electric field will be zero.

Electric field can be calculated using

E = kQ/r²

Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.

Then, the magnitude of the electric at point x is zero.

E = kQ1 / r² + kQ2 / r²

0 = kQ1 / x²  - kQ2 / (1-x)²

kQ1 / x² = kQ2 / (1-x)²

Divide through by k

Q1 / x² = Q2 / (1-x)²

4μ / x² = 1μ / (1 - x)²

Divide through by μ

4 / x² = 1 / (1-x)²

Cross multiply

4(1-x)² = x²

4(1-2x+x²) = x²

4 - 8x + 4x² = x²

4x² - 8x + 4 - x² = 0

3x² - 8x + 4 = 0

Check attachment for solution of quadratic equation

We found that,

x = 2m or x = ⅔m

So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.

5 0
3 years ago
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