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2 years ago

If you have 10 ounces of meat, what percentage of the calories comes from fat if the meat is 90% lean?

Physics

1 answer:

Lady bird [3.3K]2 years ago

4 0

Answer:

There are 602 calories in 10 ounces of Ground Beef (85% Lean / 15% Fat). serving sizes of Ground Beef

Explanation:

hope this helps if not please let me now

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2. In contrast to Venus, the coldest temperature yet measured on the surface of

garik1379 [7]

Answer:

Explanation:

Temperature Scales

There are three temperature scales in the modern sciences: Fahrenheit, Celsius, and Kelvin. Fahrenheit temperature scale assigns the value 32 for the freezing point of water and 212 for the boiling point of water and divides that interval into 180 parts. Celsius scale has a similar reference, giving 0 to the freezing point of water and 100 for the boiling point of water. The conversion between them is as follows

$\displaystyle F=\frac{9}{5}C+32$

$K=C+273.15$

The coldest temperature yet measured on the surface of any body in the solar system is -235°C. Converting to Fahrenheit

$\displaystyle F=\frac{9}{5}(-235)+32=-391^oF$

$K=-235+273.15=38.15^oK$

4 0

3 years ago

????????????????what’s the answer

kompoz [17]

Answer:

I don't know what do u. mean

6 0

3 years ago

A student that jumps a vertical height of 50 cm during the hang time activity.

muminat

The hang time of the student is 0.64 seconds, and he must leave the ground with a speed of 3.13 m/s

Why?

To solve the problem, we must consider the vertical height reached by the student as max height.

We can use the following equations to solve the problem:

Initial speed calculations:

$v_{f}^{2}=v_{o}^{2}+2*a*d$

At max height, the speed tends to zero.

So, calculating, we have:

 $v_{f}^{2}=v_{o}^{2}+2*a*d\\\\0=v_{o}^{2}+2*(-9.81\frac{m}{s^{2}})*0.5m\\\\v_{o}^{2}=9.81\frac{m^{2} }{s^{2}}\\\\v_{o}=\sqrt{9.81\frac{m^{2} }{s^{2}}}=3.13\frac{m}{s}$ 

Hang time calculations:

We must remember that the total hang time is equal to the time going up plus the time going down, and both of them are equal,so, calculating the time going down, we have have:

$y-yo=vo.t+\frac{1}{2}*9.81\frac{m}{s^{2}}*t^{2} \\\\0.5m-0=0*t+4.95*t^{2}\\\\t^{2}=\frac{0.5}{4.95}\\\\t=\sqrt{0.101}=0.318s$

Then, for the total hang time, we have:

$TotalHangTime=2*0.318seconds=0.64seconds$

Have a nice day!

3 0

4 years ago

When driving at slower speeds you need to use what type of steering

OLga [1]

Answer:

slower speeds = larger and faster steering wheel movements

faster speeds = small and slow steering wheel movements

Explanation:

When driving at slower speeds you need to use larger and faster steering wheel movements. This is because at slow speeds the car does not have enough momentum to make certain maneuvers with small steering wheel movements in a given amount of time, therefore making large and faster steering wheel movements gives the car enough time with the momentum it has to make the desired maneuver. At faster speeds only small and slow steering wheel movements are needed and while cause the car to quickly change to the desired direction due to the increased momentum of the car.

8 0

4 years ago

Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and the

MissTica

Answer:

V = 10581.59 V

Explanation:

To solve this problem, we have to take into the relation between the kinetic energy of an electron and the energy generated by the potential difference of an electric field. This relation is given by

$E_{k} = E_{V}$

$\frac{1}{2}mv^{2} = eV$

where m is the mass of the electron, v is the velocity of the electron, e is the electric charge and V is the potential difference that accelerated the electron.

By taking V from the last expression we have:

$V = \frac{mv^{2}}{2e} = \frac{(9.1*10^{-31} )(6.1*10^{7})^{2}}{2*(-1.6*10^{-19})}} = 10581.59 V$

where we have taken that

me = 9.1*10^(-31) Kg

e = -1.6*10^(-19) C

v = 6.1*10^(7)

I hope this is useful for you

Regards

6 0

4 years ago