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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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<span>The oven will transform electrical energy into heat energy</span>
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Three small balls of the same size but different masses are hung side-by-side in parallel on the strings of same length. They to
andrey2020 [161]

Answer:

m1/6 ( c )

Explanation:

since all the balls starts having the same momentum after the two collisions we will apply the principal of conservation of energy

After first collision

m1v = m1v1 + m2v2 --- ( 1 )

After second collision

m2v2 = m2v2 + m3v3   ---- ( 2 )

combining equations 1 and 2

m1v = m1v1 + m2v2 + m3v3  ----- ( 3 )

All balls moving at the same momentum ( p ) = m1v1 = m2v2 = m3v3

note ; 3p = m1v ∴ m3 = \frac{m1v}{3v3}  -----  ( 4 )

applying conservation of energy

3v = v1 + v2 + v3 ------- ( 5 )

also 3m1v1 = m1v = v1 = v/3 =

v2 + v3 = 8/3 v ----- ( 6 )

next eliminate V3 for equation 6 by applying conservation of energy and momentum

m1 =  2m2 ------ ( 7 )

now using p1 = p2 = m1v1 = 1/2 m1v1  hence v2 = 2v1  where v1 = 1/3 v

hence ; v2 = 2/3 v ------- ( 8 )

solving with equation 6 and 8

v3 = 2v ------ ( 9 ) ∴  v/v3 = 1/2 ---- ( 10 )

solving with equation 9 and 10

m3 = m1/3 * 1/2 = m1/6

8 0
3 years ago
A hunter stood at about 60 m away from a tree. He used the bow to release the arrow in order to shoot a coconut held by a monkey
Marianna [84]

Answer:

Time taken to reach the tree

they have given the maximum height which is 25m

therefore...

v=u+at (upward)

0 = 45sin(20)-9.81t

t = 1.56s

3 0
3 years ago
5. A 905 kg test car travels around a 3.04 km circular track. If the magnitude of the centripetal force is 2100 N. What is the c
balu736 [363]

Answer:

Explanation:

The equation for centripetal force is

F=\frac{mv^2}{r}. We have all the values we need except for the radius. We have the circumference of the circle, though, so we will solve for the radius using that and the fact that C = 2πr:

3.04 = 2(3.1415)r and

r = .484 m, to the correct number of sig fig's.

Now that we have everything we need and isolating the v NOT squared:

v=\sqrt{\frac{rF}{m} } and filling in:

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1.1 m/sec

8 0
3 years ago
Solve these problems on charges by finding out Q1 and Q2. Q1Q2= 4x10^6, Q1 + Q2 =8.00×10^-6​
inna [77]

Answer:

Answer:

Q_1 = 7Q

1

=7

Q_2 = 10Q

2

=10

Q_3 = 13.5Q

3

=13.5

Step-by-step explanation:

Given

5, 7, 7, 8, 10, 11, 12, 15, 17.

Required

Determine Q1, Q2 and Q3

The number of data is 9

Calculating Q1:

Q1 is calculated as:

Q_1 = \frac{1}{4}(N + 1)Q

1

=

4

1

(N+1)

Substitute 9 for N

Q_1 = \frac{1}{4}(9 + 1)Q

1

=

4

1

(9+1)

Q_1 = \frac{1}{4}*10Q

1

=

4

1

∗10

Q_1 = 2.5th\ itemQ

1

=2.5th item

This means that the Q1 is the mean of the 2nd and 3rd data.

So:

Q_1 = \frac{1}{2}(7+7)Q

1

=

2

1

(7+7)

Q_1 = \frac{1}{2}*14Q

1

=

2

1

∗14

Q_1 = 7Q

1

=7

Calculating Q2:

Q2 is calculated as:

Q_2 = \frac{1}{2}(N + 1)Q

2

=

2

1

(N+1)

Substitute 9 for N

Q_2 = \frac{1}{2}(9 + 1)Q

2

=

2

1

(9+1)

Q_2 = \frac{1}{2}*10Q

2

=

2

1

∗10

Q_2 = 5th\ itemQ

2

=5th item

Q_2 = 10Q

2

=10

Calculating Q3:

Q3 is calculated as:

Q_3 = \frac{3}{4}(N + 1)Q

3

=

4

3

(N+1)

Substitute 9 for N

Q_3 = \frac{3}{4}(9 + 1)Q

3

=

4

3

(9+1)

Q_3 = \frac{3}{4}*10Q

3

=

4

3

∗10

Q_3 = 7.5th\ itemQ

3

=7.5th item

This means that the Q3 is the mean of the 7th and 8th data.

So:

Q_3 = \frac{1}{2}(12+15)Q

3

=

2

1

(12+15)

Q_3 = \frac{1}{2}*27Q

3

=

2

1

∗27

Q_3 = 13.5Q

3

=13.5

4 0
2 years ago
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