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2 years ago

When using a physical balance on what pan we place the object and why

Physics

1 answer:

kondaur [170]2 years ago

4 0

Answer:

wood

Explanation:

Wooden pans are usually used
Wood has less density than water so it will not sink in liquid substances derived from water.
So during physical balance it won't affect the balance which we need

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A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is:

$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

$W_{loss} = 6574.75\,J$

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0

2 years ago

An object that travels around another object in space is called a(n)

HACTEHA [7]

Satellite. think of the moon.

Hope this helps!
Vote me Brainliest!

3 0

3 years ago

Read 2 more answers

The human body has an average density of 979 kg/m3 , what fraction of a person is submerged when floating gently in fresh water?

FromTheMoon [43]

A person is submerged of about 97.9%.

The average density of the human body is given as 979 kg/ m³.

<h3>Define Law of floatation.</h3>

Law of floatation can be defined as the volume of the liquid displaced when a body floats on the liquid surface is equal to the body submerged in the water.

As body has the stable equilibrium state, the buoyancy of the fluid will be equal to the weight.

Weight of the body floating = Weight of the body immersed in fluid

Law of floatation = Density of the floating object / density of fluid

As fluid is the freshwater here, the density of fluid will be 1000 kg/ m³.

= (979 kg/ m³) / ( 1000 kg/ m³)

= 97.9 %

A person is submerged when floating gently in fresh water about 97.9%.

Learn more about law of floatation,

brainly.com/question/17032479

#SPJ4

4 0

10 months ago

If the initial velocity of a ball is sent straight upward at 10.5m/s from the ground what will its final velocity be when it hit

Shalnov [3]

Answer: -10.08 m/s

Explanation:

Here we only need to analyze the vertical problem.

When the ball is in the air, the only force acting on it will be the gravitational force, this means that the acceleration of the ball, is equal to the gravitational acceleration, then:

a(t) = -9.8m/s^2

Where the negative sign is because gravity pulls the ball down.

To get the velocity equation we need to integrate over time, we get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the initial vertical velocity, here it is v0 = 10.5 m/s

Then the velocity equation is:

v(t) = (-9.8m/s^2)*t + 10.5 m/s

To get the position equation, we need to integrate again over time, we get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t + p0

Where p0 is the initial position, we know that the ball is sent upward from the ground, so p0 = 0m

Then the position equation is:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

Now we need to find the value of t such that the position is equal to zero (this means that the ball hits the ground again).

Then we need to solve:

p(t) = 0 = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

If we divide both sides by t, we get:

0 = (1/2)*(-9.8m/s^2)*t + (10.5 m/s)

Now we can solve it:

(1/2)*(9.8m/s^2)*t = 10.5 m/s

t = (10.5 m/s)/((1/2)*(9.8m/s^2)) = 2.14 s

This means that after 2.14 seconds, the ball will hit the ground again.

The velocity of the ball when it hits the ground is equal to:

v(2.14s) = (-9.8m/s^2)*2.14s + 10.5 m/s = -10.08 m/s

3 0

3 years ago

Students in a science class were provided with this image and asked to identify which color of light would refract the most if w

djverab [1.8K]

Violet would refract the most as it has the shortest wavelength amongst the visible spectrum.

8 0

2 years ago

When using a physical balance on what pan we place the object and why​

When using a physical balance on what pan we place the object and why