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3 years ago

Give two ways that vegetable oil is different from fat

Physics

1 answer:

Paladinen [302]3 years ago

7 0

Answer:

In simple terms, fats are animal fats whereas oils are vegetable oils. The other difference is fats tend to be solids at room temperature; on the other hand, oils tend to be liquid at room temperature.

Explanation:

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011 10.0 points

Sedbober [7]

<h2>The temperature of the air is 66.8° C</h2>

Explanation:

From the Newton's velocity of sound relationship , the velocity of sound is directly proportional to the square root of temperature .

In this case The velocity of sound = frequency x wavelength

= 798 x 0.48 = 383 m/sec

Suppose the temperature at this time = T K

Thus 383 ∝ $\sqrt{T}$ I

The velocity of sound is 329 m/s at 273 K ( given )

Thus 329 ∝ $\sqrt{273}$ II

Dividing I by II , we have

$\frac{383}{329}$ = $\sqrt{\frac{T}{273} }$

or $\frac{T}{273}$ = 1.25

and T = 339.8 K = 66.8° C

4 0

4 years ago

A satellite is in a circular orbit around Mars, which has a mass M = 6.40 × 1023 kg and radius R = 3.40 ×106 m.

Pepsi [2]

Answer:

a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately $\displaystyle \sqrt\frac{4.27 \times 10^{13}}{R}\; \rm m \cdot s^{-1}$ .

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately $\displaystyle 9.62 \times 10^{-7}\, R^{3/2}\; \rm s$ .

c) The orbital radius required would be approximately $\rm 2.04 \times 10^7\; m$ .

d) The escape velocity from the surface of that planet would be approximately $\rm 5.01\times 10^3\; m \cdot s^{-1}$ .

Explanation:

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}}$ ,

where

$G$ is the constant of universal gravitation.
$M$ is the mass of the first mass. (In this case, let $M$ be the mass of the planet.)
$m$ is the mass of the second mass. (In this case, let $m$ be the mass of the satellite.)
$r$ is the distance between the center of mass of these two objects.

On the other hand, the net force on an object in a centripetal motion should be:

$\displaystyle \frac{m \cdot v^{2}}{r}$ ,

where

$m$ is the mass of the object (in this case, that's the mass of the satellite.)
$v$ is the orbital speed of the satellite.
$r$ is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for $v$ :

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}} = \frac{m \cdot v^{2}}{r}$ .

$\displaystyle v^2 = \frac{G \cdot M}{r}$ .

$\displaystyle v = \sqrt{\frac{G \cdot M}{r}}$ .

Take $G \approx 6.67 \times \rm 10^{-11} \; m^3 \cdot kg^{-1} \cdot s^{-2}$ , Simplify the expression $v$ :

$\begin{aligned} v &= \sqrt{\frac{G \cdot M}{r}} \cr &= \sqrt{\frac{6.67 \times \rm 10^{-11} \times 6.40 \times 10^{23}}{r}} \cr &\approx \sqrt{\frac{4.27 \times 10^{13}}{r}} \; \rm m \cdot s^{-1} \end{aligned}$ .

Since the orbit is a circle of radius $R$ , the distance traveled in one period would be equal to the circumference of that circle, $2 \pi R$ .

Divide distance with speed to find the time required.

$\begin{aligned} t &= \frac{s}{v} \cr &= 2 \pi R}\left/\sqrt{\frac{G \cdot M}{R}} \; \rm m \cdot s^{-1}\right. \cr &= \frac{2\pi R^{3/2}}{\sqrt{G \cdot M}} \cr &\approx 9.62 \times 10^{-7}\, R^{3/2}\; \rm s\end{aligned}$ .

Convert $24.6\; \rm \text{hours}$ to seconds:

$24.6 \times 3600 = 88560\; \rm s$

Solve the equation for $R$ :

$9.62 \times 10^{-7}\, R^{3/2}= 88560$ .

$R \approx 2.04 \times 10^7\; \rm m$ .

If an object is at its escape speed, its kinetic energy (KE) plus its gravitational potential energy (GPE) should be equal to zero.

$\displaystyle \text{GPE} = -\frac{G \cdot M \cdot m}{r}$ (Note the minus sign in front of the fraction. GPE should always be negative or zero.)

$\displaystyle \text{KE} = \frac{1}{2} \, m \cdot v^{2}$ .

Solve for $v$ . The value of $m$ shouldn't matter, for it would be eliminated from both sides of the equation.

$\displaystyle -\frac{G \cdot M \cdot m}{r} + \frac{1}{2} \, m \cdot v^{2}= 0$ .

$\displaystyle v = \sqrt{\frac{2\, G \cdot M}{R}} \approx 5.01\times 10^{3}\; \rm m\cdot s^{-1}$ .

5 0

4 years ago

Where is a proton located within an atom?

Westkost [7]

Answer:

in the nucleus of the atom

Explanation:

a p 3 x

8 0

3 years ago

Brandon pushes an object on a ramp as shown in the diagram.

stiks02 [169]

The force which has the greatest effect on causing this object to slow while it remains in contact with the ramp is: B. a frictional force.

<h3>What is a force?</h3>

A force can be defined as a push or pull of an object or physical body, which typically results in a change of motion (acceleration), especially due to the interaction of the object with another.

<h3>The types of force.</h3>

In Science, there are different types of force and these include the following:

Gravitational force

Tension force.

Electrical force

Normal Force.

Magnetic force

Air resistance force

Applied force

Frictional force.

<h3>What is a frictional force?</h3>

Friction force can be defined as a type of force that resists and slows the relative motion of two physical objects when there surfaces come in contact. This ultimately implies that, a frictional force prevents two surfaces from easily sliding over or slipping across one another.

In this context, we can infer and logically deduce that the force which has the greatest effect on causing this object to slow while it remains in contact with the ramp is a frictional force.

Read more on frictional force here: brainly.com/question/25253774

#SPJ1

Complete Question:

Brandon pushes an object on a ramp as shown in the diagram.

While Brandon pushes the object and it remains in contact the ramp, which force has the greatest effect on causing it to slow?

A. the applied force

B. a frictional force

C. the force due to gravity

D. a force of air resistance

5 0

2 years ago

A car can accelerate from rest to 28 m/s and travels 280meters. How long does this acceleration take?

Reika [66]

Answer:

$a=1.4m/s^2$

Explanation:

From the question we are told that

Velocity v=28m/s

Distance d=280

Generally the Newtons equation for motion is mathematically given as

Acceleration

$v^2=2as$

$a=\frac{V^2}{2s}$

$a=\frac{28^2}{2*280}$

$a=1.4m/s^2$

4 0

3 years ago

Give two ways that vegetable oil is different from fat​

Give two ways that vegetable oil is different from fat