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shutvik [7]
3 years ago
5

Develop a hypothesis for why one of the two types of soup should indeed be rolling down faster than the other. This hypothesis s

hould directly relate to the moment of inertia and should involve predictions about how the two types of soup behave while the cans are rolling downhill. Hint: Soup is a liquid.
Physics
1 answer:
Eva8 [605]3 years ago
7 0

Answer:

Assume two identical cans filled with two types of soup having same mass are rolling down on an inclined plane in same conditions. In terms of inertia different types of soup will indicate different viscosity. The higher viscosity fillings indicates more part of the soup mass is rotating together with the can’s body. This means that for the can with lower viscosity soup has a lower moment of inertia and the can with higher viscosity has higher moment of inertia while the same gravity makes them to roll.

incline angle = θ ; can's mass =  m ; Radius of the can's = R , Angular acceleration for Can 1 = α1 ; Angular acceleration for Can 2 = α2

T1 = Inertia of Can with high viscosity soup

T2 = Inertia of Can with low viscosity soup

M1 rolling moment of Can 1

M2 rolling moment of Can 2

equation is given by

T1*α1 = M1   - (a)

T2*α2 = M2 - (b)

M1 = M2 = m*g*R*sin(θ). (c)

as assumed T1 > T2

from the three equation (a), (b) & (c)

the α2 > α1

Angular acceleration of Can 2 is higher than Can 1. Already stated that Can 1 has more viscous soup as compared to Can 2.

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madreJ [45]

A) A concave mirror forming a larger, virtual image

Explanation:

The figure is missing; see attachment.

There are two types of mirror:

  • Concave (converging) mirrors: a concave mirror is a mirror that reflects the light inward
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The image formed by a mirror can also be of two types:

  • Real image: it is formed on the same side of the object, with respect to the mirror
  • Virtual image: it is formed on the opposite side of the object, with respect to the mirror

In the figure of this problem (see attachment), we see that:

- The mirror reflects the light from the object inward --> so it is a concave mirror

- The image is formed on the other side of the mirror --> it is a virtual image

So the correct option is

A) A concave mirror forming a larger, virtual image

Learn more about mirrors:

brainly.com/question/8737441

#LearnwithBrainly

7 0
3 years ago
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What is kinematics?<br>explain!!~​<br><br>thankyou ~
Hatshy [7]

Answer:

<u>Kinematics:-</u>

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  • Features of motion in an object
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Hope it helps!

6 0
2 years ago
Read 2 more answers
A coil with an inductance of 2.3 H and a resistance of 14 Ω is suddenly connected to an ideal battery with ε = 100 V. At 0.13 s
klemol [59]

Given Information:

Resistance = R = 14 Ω

Inductance = L = 2.3 H

voltage = V = 100 V

time = t = 0.13 s

Required Information:

(a) energy is being stored in the magnetic field

(b) thermal energy is appearing in the resistance

(c) energy is being delivered by the battery?

Answer:

(a) energy is being stored in the magnetic field ≈ 219 watts

(b) thermal energy is appearing in the resistance ≈ 267 watts

(c) energy is being delivered by the battery ≈ 481 watts

Explanation:

The energy stored in the inductor is given by

U = \frac{1}{2} Li^{2}

The rate at which the energy is being stored in the inductor is given by

\frac{dU}{dt} = Li\frac{di}{dt} \: \: \: \: eq. 1

The current through the RL circuit is given by

i = \frac{V}{R} (1-e^{-\frac{t}{ \tau} })

Where τ is the the time constant and is given by

\tau = \frac{L}{R}\\ \tau = \frac{2.3}{14}\\ \tau = 0.16

i = \frac{110}{14} (1-e^{-\frac{t}{ 0.16} })\\i = 7.86(1-e^{-6.25t})\\\frac{di}{dt} = 49.125e^{-6.25t}

Therefore, eq. 1 becomes

\frac{dU}{dt} = (2.3)(7.86(1-e^{-6.25t}))(49.125e^{-6.25t})

At t = 0.13 seconds

\frac{dU}{dt} = (2.3) (4.37) (21.8)\\\frac{dU}{dt} = 219.11 \: watts

(b) thermal energy is appearing in the resistance

The thermal energy is given by

P = i^{2}R\\P = (7.86(1-e^{-6.25t}))^{2} \cdot 14\\P = (4.37)^{2}\cdot 14\\P = 267.35 \: watts

(c) energy is being delivered by the battery?

The energy delivered by battery is

P = Vi\\P = 110\cdot 4.37\\P = 481 \: watts

4 0
3 years ago
Carlos was camping and getting cold as the sun went down. He wanted to light a fire for warmth and light. However, he discovered
storchak [24]
1. The chemical reaction produced by Carlo's fire is exergonic because energy is "going out". As the reaction proceeds, entropy increases as the energy stored in the dry wood and leaves are used up as fuel to create the fire which produces low quality light and warmth.  

2. This reaction is a classic example of an exothermic reaction. Exothermic reactions are characterized with the presence of heat and light in the products. Combustion reactions are always exothermic in nature.

3. Catalyst are substances that are used to speed up reactions by lowering the activation requirement. Catalysts aren't consumed in the reaction and can still be chemically retrieved afterwards. In this situation, the leaves cannot be retrieved after the reaction ends. The leaves speed up the heating of the wood but it does not behave as a catalyst. 
6 0
3 years ago
In Example 2.7, we investigated a jet landing on an aircraft carrier. In a later maneuver, the jet comes in for a landing on sol
igor_vitrenko [27]

Answer:

a) 20 seconds

b) No.

Explanation:

t = Time taken for jet to stop

u = Initial velocity = 100 m/s (given in the question)

v = Final velocity = 0 (because the jet will stop at the end)

s = Displacement of the jet (Distance between the moment the jet touches the ground to the point the point it stops)

a = Acceleration = -5.00 m/s² (slowing down, so it is negative)

a) Equation of motion

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-100}{-5}\\\Rightarrow t=20\ s

The time required for the plane to slow down from the moment it touches the ground is 20 seconds.

s=ut+\frac{1}{2}at^2\\\Rightarrow s=100\times 20+\frac{1}{2}\times -5\times 20^2\\\Rightarrow s=1000\ m

The distance it requires for the jet to stop is 1000 m so in a small tropical island airport where the runway is 0.800 km long the plane would not be able to land. The runway needs to be atleast 1000 m long here the runway on the island is 1000-800 = 200 m short.

5 0
3 years ago
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