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kipiarov [429]
3 years ago
7

Here is something to blow your mind. In one slice of "everything" pizza there are 650 Calories (1000 chemistry calories). That e

quates to 2,700,000 J of energy. How many slices of pizza did you and your 131 other classmates "burn up" in those 50 minutes. Round up to two digits.
Physics
1 answer:
amid [387]3 years ago
6 0

Answer:

Number of slices of pizza is 993.

Explanation:

It is given that,  in one slice of "everything" pizza there are 650 Calories. The conversion factor from calories to joules is :

1 calorie = 4.184 joules

650 calories = 2719.6 joules

Total energy in the pizza, E = 2700000 J

Let there are n number of slices of pizza. It is given by :

n=\dfrac{2700000}{2719.6}

n = 992.79

or

n = 993

So, there are 993 slices of pizza. Hence, this is the required solution.

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Please send me solution of the question pls​
ELEN [110]

Answer:

20m

Explanation:

P.E=mgh

2000=10×10×h

2000=100h

Divide both side by 100

2000/100=20

4 0
3 years ago
The weight of the atmosphere above 1 m- of
mars1129 [50]

Answer:

1.09 kg.m

Explanation:

no need

4 0
2 years ago
The total amount of energy contained in an object
Elenna [48]

The total amount of energy stored in the particles of an object is called its internal energy. The internal energy of an object is made up of the kinetic energy due to the random motion of the particles and the potential energy due to the interactive forces among the particles.

7 0
3 years ago
The size of a balloon increases when the pressure inside it increases. The balloon gets bigger when it is left in the heat from
Masteriza [31]

Explanation:

This happens because the gas inside tend to expand because its temperature gets higher.

This is why the balloon that is put in a freezer for too long tend to gets smaller, because the gas temperature that is inside the balloon decreases.

(you can try it at home)

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5 0
3 years ago
The objective lens and the eyepiece of a telescope are spaced 85 cm apart. If the eyepiece is123 D what is the total magnificati
dedylja [7]

Answer:

The magnification would be "103.55". A further explanation is given below.

Explanation:

The given values are:

Distance between lens and eyepiece,

L = 85 cm

Eyepiece is,

= 123 D

Now,

The refractive power of eye piece will be:

⇒ \frac{1}{f_e}=123D

   f_e=\frac{1}{123D}

   f_e=0.813 \ cm

The length of the telescope will be:

⇒ L=f_0+f_e

⇒ f_0=L-f_e

On substituting the values, we get

⇒     =85-0.813

⇒     =84.187 \ cm

Now,

The magnification of the telescope will be:

⇒ M=\frac{f_0}{f_e}

⇒      =\frac{84.187}{0.813}

⇒      =103.55

5 0
2 years ago
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