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Andre45 [30]
3 years ago
9

9. A 2 liter bottle of Coke weighs about 2 kilograms. If that bottle were combined with an equal amount of anti-Coke, how many J

oules of energy would be released (note that you get units Joules if you calculate the energy using units of kilograms and meters)
Physics
1 answer:
givi [52]3 years ago
3 0

Answer:

The amount of energy that would be released is equal to 4182 Joules.

Explanation:

Total amount of coke = 2 kg = 2000 g

1 calorie per gram is equal to 4.184 Joules of energy

4.184 J/gC*2000g = 8368 J

1 food calorie is roughly equal to 4186 J

8368 - 4186

Therefore, the amount of energy that would be released is equal to 4182 Joules.

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In this problem, you will answer several questions that will help you better understand the moment of inertia, its properties, a
scoundrel [369]

Answer:

a)  Total mass form, density and axis of rotation location are  True

b)   I = m r²

Explanation:

a) The moment of inertia is the inertia of the rotational movement is defined as

       I = ∫ r² dm

Where r is the distance from the pivot point and m the difference in body mass

In general, mass is expressed through density

        ρ = m / V

        dm = ρ dV

From these two equations we can see that the moment of inertia depends on mass, density and distance

Let's examine the statements, the moment of inertia depends on

- Linear speed       False

- Acceleration angular False

-  Total mass form True

-  density True

- axis of rotation location   True

b) we calculate the moment of inertia of a particle

For a particle the mass is at a point whereby the integral is immediate, where the moment of inertia is

          I = m r²

4 0
3 years ago
A risk-free, zero-coupon bond with a $5000 face value has ten years to maturity. The bond currently trades at $3650. What is the
PtichkaEL [24]

Answer:

The Yield to Maturity of the bond is YTM = 3.20%

Explanation:

Mathematically the Yield to Maturity of the bond  YTM is as follows

           YTM = \frac{C + \frac{F-P}{n} }{\frac{F+P}{2} }

Where C is the amount of payment to be made = $0

           P is the price i.e the present value =$3650

           F is the face value of the bond=$5000

          n is the year of maturity of the bond = 10 years

            YTM = \frac{0+\frac{5000-3650}{10} }{\frac{5000+3650}{2} } * \frac{100}{1}

                      =3.20%

                     

                     

     

   

5 0
3 years ago
Which material is very strong and tough but shows very little elongation as it absorbs energy? A. spider silk B.rubber C.Kevlar®
Furkat [3]

C. Kevlar

:) ++ * ++ !

5 0
2 years ago
Read 2 more answers
A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught
meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2} (Eq. 1)

Where:

y_{o} - Initial height of the softball, measured in meters.

y - Final height of the softball, measured in meters.

v_{o} - Initial velocity of the softball, measured in meters per second.

t - Time, measured in seconds.

g - Gravitational acceleration, measured in meters per square second.

If we know that y = y_{o}, t = 3.56\,s and g = -9.807\,\frac{m}{s^{2}}, the initial velocity of the softball is:

v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0

3\cdot v_{o} -44.132\,m= 0

v_{o} = 14.711\,\frac{m}{s}

The initial velocity of the softball is 14.711 meters per second.

8 0
3 years ago
How much heat is needed to melt 13.74 kg of silver that is initially at 20°C?
salantis [7]

Answer:

Q = 4.63 \times 10^6 J

Explanation:

As we know that melting point of silver is

T = 961.8 degree C

Latent heat of fusion of silver is given as

L = 111 kJ/kg

specific heat capacity of silver is given as

s = 240 J/kg C

now we will have

Q = ms\Delta T + mL

\Delta T = 961.8 - 20

\Delta T = 941.8 degree C

now from above equation

Q = (13.74)(240)(941.8) + (13.74)(111 \times 10^3)

Q = 3.1 \times 10^6 + 1.53 \times 10^6

Q = 4.63 \times 10^6 J

3 0
2 years ago
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