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Taya2010 [7]
3 years ago
12

PLEASE ANSWER QUICK (7th grade science measurment of volume)

Physics
1 answer:
lubasha [3.4K]3 years ago
7 0

Answer : The volume of water in graduated cylinder is 15.5 mL.

Explanation :

As we know that for the measurement of the volume of liquid in graduated cylinder are shown by placing the graduated cylinder on the flat surface and then view the height of the liquid in the graduated cylinder with the naked eyes directly level with the liquid.

The liquid will tend to curve downward that means this curve is known as the meniscus.

In the case of colored liquid, we are always read the upper meniscus of the liquid for the measurement.

In the case of colorless liquid, we are always read the lower meniscus of the liquid for the measurement.

In the given image, there are 5 larger and 5 smaller division between the 15 and 20 and the solution is colored. The value of larger division is 1 mL and smaller division is 0.5 mL.

So, we will read the upper meniscus of the liquid for the measurement.

Hence, the volume of water in graduated cylinder is 15.5 mL.

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A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s aft
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Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

m_1 v_1 = m_2 v_2

m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s

Using the equation to find v_1,

v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475

Using the conservation of energy equation, we have,

KE= \frac{1}{2}m*v^2

KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J

KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J

Total KE= 92077.75+13425530=92708.9J

Now this energy over the cannonball

KE=\frac{1}{2}m*v_2^2

92708.9=\frac{1}{2}15.5v_2^2

V_2 = 109.37m/s

The gain in velocity is 0.37m/s

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Stars combine Hydrogen to make Helium during nuclear fusion. Living things are made of heavier elements like Carbon, Oxygen, Iro
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it takes 90 j of work to stretch a spring 0.2 m from its equilibrium position. How muc work is needed to stretch it an additiona
Vinvika [58]

Work needed: 720 J

Explanation:

The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the stretching of the spring from the equilibrium position

In this problem, we have

E = 90 J (work done to stretch the spring)

x = 0.2 m (stretching)

Therefore, the spring constant is

k=\frac{2E}{x^2}=\frac{2(90)}{(0.2)^2}=4500 N/m

Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of

x = 0.2 + 0.4 = 0.6 m

Substituting,

E'=\frac{1}{2}kx^2=\frac{1}{2}(4500)(0.6)^2=810 J

Therefore, the additional work needed is

\Delta E=E'-E=810-90=720 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

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3 years ago
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The number of electrons in a neutral atom is equal to the number of protons. The mass number of the atom (M) is equal to the sum of the number of protons and neutrons in the nucleus. The number of neutrons is equal to the difference between the mass number of the atom (M) and the atomic number. Therefore, it is true!

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3 years ago
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