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3 years ago

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The diagram shows a person holding a bow and arrow. Which will most likely increase the kinetic energy in this system? a not rel

easing the string b increasing the mass of the arrow c releasing the string d pulling back less on the string

1 answer:

Fantom [35]3 years ago

7 0

Answer:

b. Increasing the mass of the arrow.

Explanation:

The formula is K=1/2mv^2. Increasing the mass also increases the kinetic energy.

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A mercury barometer reads 745.0 mm on the roof of a building and 760.0 mm on the ground. Assuming a constant value of 1.29 kg/m3

Ipatiy [6.2K]

Answer:

The height of the building is 158.140 meters.

Explanation:

A barometer is system that helps measuring atmospheric pressure. Manometric pressure is the difference between total and atmospheric pressures. Manometric pressure difference is directly proportional to fluid density and height difference. That is:

$\Delta P \propto \rho \cdot \Delta h$

$\Delta P = k \cdot \rho \cdot \Delta h$

Where:

$\Delta P$ - Manometric pressure difference, measured in kilopascals.

$\rho$ - Fluid density, measured in kilograms per cubic meter.

$\Delta h$ - Height difference, measured in meters.

Now, an equivalent height difference with a different fluid can be found by eliminating manometric pressure and proportionality constant:

$\rho_{air} \cdot \Delta h_{air} = \rho_{Hg} \cdot \Delta h_{Hg}$

$\Delta h_{air} = \frac{\rho_{Hg}}{\rho_{air}} \cdot \Delta h_{Hg}$

Where:

$\Delta h_{air}$ - Height difference of the air column, measured in meters.

$\Delta h_{Hg}$ - Height difference of the mercury column, measured in meters.

$\rho_{air}$ - Density of air, measured in kilograms per cubic meter.

$\rho_{Hg}$ - Density of mercury, measured in kilograms per cubic meter.

If $\Delta h_{Hg} = 0.015\,m$ , $\rho_{air} = 1.29\,\frac{kg}{m^{3}}$ and $\rho_{Hg} = 13600\,\frac{kg}{m^{3}}$ , the height difference of the air column is:

$\Delta h_{air} = \frac{13600\,\frac{kg}{m^{3}} }{1.29\,\frac{kg}{m^{3}} }\times (0.015\,m)$

$\Delta h_{air} = 158.140\,m$

The height of the building is 158.140 meters.

5 0

3 years ago

Read 2 more answers

Suppose a piano tuner stretches a steel piano wire 7.5 mm. The wire was originally 0.975 mm in diameter, 1.45 m long, and has a

Tasya [4]

The forces a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

<h3>What is force?</h3>

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.

The application of a force may be used to describe the steel as an elastic element within a certain range of applied force.

Given data;

Young modulus, E=2.10 × 10¹¹ N/m²

Cross-sectional area,A

Final length, $\rm L_f = 1.45 m = 1450 \ mm$

Initial length, $\rm L_i = 7.5 mm$

$\rm F = \frac{(L_f-L_i)(E)(A)}{L_1} \\\\ \rm F = \frac{(1450 - 7.5)(2.0 \times 0^{11})(0.746)}{1450} \\\\ F = 1.4 \times 10^{11} \ N$

Hence, the force a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

To learn more about the force refer to the link;

brainly.com/question/26115859

#SPJ1

8 0

2 years ago

You used a telescope and other mathematics to discover that Jupiter is 5.20 au from the sun. Use the equation to find its orbita

meriva

Answer:

11.9 years

Explanation:

We can find the orbital period by using Kepler's third law, which states that the ratio between the square of the orbital period and the cube of the average distance of a planet from the Sun is constant for every planet orbiting aroudn the Sun:

$\frac{T^2}{r^3}=const.$

Using the Earth as reference, we can re-write the law as

$\frac{T_e^2}{r_e^2}=\frac{T_j^2}{r_j^3}$

where

Te = 1 year is the orbital period of the Earth

re = 1 AU is the average distance of the Earth from the Sun

Tj = ? is the orbital period of Jupiter

rj = 5.20 AU is the average distance of Jupiter from the Sun

Substituting the numbers and re-arranging the equation, we find:

$T_j=\sqrt{\frac{T_e^2 r_j^3}{T_j^2}}=\sqrt{\frac{(1 y)^2 (5.2 AU)^3}{(1 AU)^3}}=11.9 y$

4 0

3 years ago

Read 2 more answers

3) Ęplain why muddy water is<br> Heterogeneous<br> mixture

nataly862011 [7]

Answer:

muddy water is a heterogeneous mixture, which is Suspension.

7 0

3 years ago

Why does paper feel different after it has been wet and dried?

dsp73

"in the paper making process, the pulp is strained to create the initial textured/brittle paper, but the process doesn't end there. The textured paper must be put through high pressured steam rollers to create the smooth paper you've come to know. This setting process causes all the fibers to be pressed flat and packed together evenly on the surface. When paper gets wet, it "unsets" itself, breaking up the tension that held the fibers together, again causing the uneven texture it initially had." Got this from reddit

7 0

4 years ago