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3 years ago

Give three examples of how gravity is at work in our solar system.

2 answers:

6 0

Well.....
Gravity from the sun pulls the planets torward it while inertia pulls it outward....but I guess that would be why it orbits sorry if this doesn't help but uh

mariarad [96]3 years ago

3 0

Gravity is the force that keeps all of the planets, dwarf planets, moons, asteroids and comets orbiting the Sun.<span> The force of gravity is inversely proportional to the distance of any two objects. This means that objects that are close to the Sun experience a stronger gravitational pull, and therefore, orbit the Sun at a greater speed</span>

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1. In a biology experiment the number of yeast cells is determined after 24 hours of growth at

Anna11 [10]

Answer:

In a biology experiment the number of yeast cells is determined after 24 hours of culture at different temperatures. a) Identify the independent and dependent variables in the experiment. b) Draw the graph of the

Explanation:

5 0

3 years ago

A hockey puck wont slide very far on concrete or asphalt, but it will slide for a very long time on ice. Why is this?

Ilya [14]

When hockey players push the puck along the ice it slides causing heat which melts the ice causing the friction against the ice to be less.

7 0

3 years ago

Read 2 more answers

When air resistance balances the weight of an object that is falling, the velocity _____.

miskamm [114]

Decreases. Air resistance will slow a falling object to its terminal velocity, placing a limit on its acceleration.

5 0

3 years ago

An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the

zzz [600]

Answer:

The magnitude of the electric flux is $3.53\ N-m^2/C$

Explanation:

Given that,

Electric field = 2.35 V/m

Angle = 25.0°

Area $A= 1.65 m^2$

We need to calculate the flux

Using formula of the magnetic flux

$\phi=E\cdot A$

$\phi = EA\cos\theta$

Where,

A = area

E = electric field

Put the value into the formula

$\phi=2.35\times1.65\times\cos 25^{\circ}$

$\phi=2.35\times1.65\times0.91$

$\phi=3.53\ N-m^2/C$

Hence, The magnitude of the electric flux is $3.53\ N-m^2/C$

8 0

3 years ago

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

3 years ago