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2 years ago

12

I have no clue what to do to find the problem.

1 answer:

oksano4ka [1.4K]2 years ago

7 0

kess dense material comes up. glycerin dencity= 37.8 / 30 syrup density = 80 / 60 so glycerin is less dense and comes up

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If you apply a 4 N force to a 2 kg object, how fast will it accelerate? (Your

zubka84 [21]

Answer:2n

Explanation:

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2 years ago

Explain why the light bulb isn't lighting up in the circuit<br> pictured on the right.

FromTheMoon [43]

The wires should be attached to each screw on the light not only one

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2 years ago

What is the wavelength of the wave

sladkih [1.3K]

Explanation:

it is equal to the speed (v) of a wave train in a medium divided by its frequency (f): λ = v/f. Waves of different wavelengths.

8 0

2 years ago

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A copper cylinder is initially at 21.1 ∘C . At what temperature will its volume be 0.163 % larger than it is at 21.1 ∘C?

fomenos

To solve this problem we will apply the concepts related to the final volume of a body after undergoing a thermal expansion. To determine the temperature, we will use the given relationship as well as the theoretical value of the volumetric coefficient of thermal expansion of copper. This is, for example to the initial volume defined as $V_1$ , the relation with the final volume as

$V_2 = V_1 +0.163\% V_1$

$V_2 = V_1 +0.00163V_1$

$V_2 = 1.00163V_1$

Initial temperature = $21.1\°C$

Let T be the temperature after expanding by the formula of volume expansion

we have,

$V_2 = V_1 (1+\gamma \Delta t)$

Where $\gamma$ is the volume coefficient of copper $5.1*10^{-5}/C$

$1.00163V_1 = V_1(1+\gamma(T-21.1\°))$

$1.00163 = 1+5.1*10^{-5}(T-21.1\°)$

$0.00163 = 0.000051T-0.0010761$

$T = 53.0608\°C$

Therefore the temperature is 53.06°C

7 0

2 years ago

Your answer should be precise to 0.1 m/s. Use a gravitational acceleration of 10 m/s/s. At it lowest point, a pendulum is moving

saw5 [17]

Explanation:

It is given that,

Speed, v₁ = 7.7 m/s

We need to find the velocity after it has risen 1 meter above the lowest point. Let it is given by v₂. Using the conservation of energy as :

$\dfrac{1}{2}mv_1^2=\dfrac{1}{2}mv_2^2+mgh$

$v_2^2=v_1^2-2gh$

$v_2^2=(7.7)^2-2\times 10\times 1$

$v_2=6.26\ m/s$

So, the velocity after it has risen 1 meter above the lowest point is 6.26 m/s. Hence, this is the required solution.

4 0

3 years ago

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