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3 years ago

A 2-kg wheel rolls down the road with a linear speed of 15m/s. Find its trwansitional and rotational kinetic energies.

Physics

1 answer:

Elza [17]3 years ago

6 0

Answer:

The translational kinetic energy is 225 J

The rotational kinetic energy is 225 J

Explanation:

Given;

mass of the wheel, m = 2-kg

linear speed of the wheel, v = 15 m/s

Transnational kinetic energy is calculated as;

E = ¹/₂MV²

where;

M is mass of the moving object

V is the velocity of the object

E = ¹/₂ x 2 x (15)²

E = 225 J

Rotational kinetic energy is calculated as;

E = ¹/₂Iω²

where;

I is moment of inertia

ω is angular velocity

$E = \frac{1}{2} I \omega^2\\\\E = \frac{1}{2} *mr^2*(\frac{v}{r})^2\\\\E = \frac{1}{2} *mr^2*\frac{v^2}{r^2} \\\\E = \frac{1}{2}mv^2$

E = ¹/₂ x 2 x (15)²

E = 225 J

Thus, the translational kinetic energy is equal to rotational kinetic energy

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A weather balloon has a volume of 90.0 l when it is released from sea level. What

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The final atmospheric pressure is $5.19\cdot 10^4 Pa$

Explanation:

Assuming that the temperature of the air does not change, we can use Boyle's law, which states that for a gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume. In formula,

$pV=const.$

where

p is the gas pressure

V is the volume

The equation can also be rewritten as

$p_1 V_1 = p_2 V_2$

where in our problem we have:

$p_1= 1.03\cdot 10^5 Pa$ is the initial pressure (the atmospheric pressure at sea level)

$V_1 = 90.0 L$ is the initial volume

$p_2$ is the final pressure

$V_2 = 175.0 L$ is the final volume

Solving the equation for p2, we find the final pressure:

$p_2 = \frac{p_1 V_1}{V_2}=\frac{(1.01\cdot 10^5)(90.0)}{175.0}=5.19\cdot 10^4 Pa$

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3 years ago

Problem 2: (15 pts) A 10-m high cylindrical container is half-filled at the bottom with water of density =1000 kg/m3 while the t

Step2247 [10]

Answer:

$\Delta p = 90.7 kPa$

Explanation:

specific gravity of oil is $= \frac{\rho_{oil}}{\rho_w}$

$\rho_{oil} = 0.85*1000 = 850 kg/m3$

we know that

change in pressure for oil is given as

$\Delta p = \rho gh$

here density and h is for oil

$\Delta p = 850*5 *9.81 = 41,692.5 kPa$

change in pressure for WATER is given as

$\Delta p = \rho gh$

here density is for water and h is for water

$\Delta p = 1000*5 *9.81 = 49,050 kPa$

pressure change due to both is given as

$\Delta p = 41692.3 + 49050 = 90742.5 N/m2$

$\Delta p = 90.7 kPa$

8 0

3 years ago

Read 2 more answers

A satellite orbits earth with a mean altitude of 361 km. If the orbit is circular, what are the satellite's time period and spee

Advocard [28]

Answer:

v = 7.69 x 10³ m/s = 7690 m/s

T = 5500 s = 91.67 min = 1.53 h

Explanation:

In order for the satellite to orbit the earth, the force of gravitation on satellite must be equal to the centripetal force acting on it:

$F_{gravitation}= F_{centripetal}\\\\\frac{GM_{s} M_{E}}{r^2} = \frac{M_{s} v^2}{r}\\\\\frac{GM_{E}}{r} = v^2\\\\v = \sqrt{\frac{GM_{E}}{r} } \\\\$

where,

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Me = Mass of Earth = 5.97 x 10²⁴ kg

r = distance between the center of Earth and Satellite = Radius of Earth + Altitude = 6.371 x 10⁶ m + 0.361 x 10⁶ m = 6.732 x 10⁶ m

v = orbital speed = ?

Therefore,

$v = \sqrt{\frac{(6.67 x 10^{-11}N.m^2/kg^2)(5.97 x 10^{24} kg)}{6.732 x 10^6 m} }\\\\$

v = 7.69 x 10³ m/s

For time period satellite completes one revolution around the earth. It means that the distance covered by satellite is equal to circumference of circle at the given altitude.

So, its orbital speed can be given as:

$v = \frac{Circumference of Circle at Given Altitude}{T}\\\\v = \frac{2\pi r}{T}\\\\$

where,

T = Time Period of Satellite = ?

Therefore,

$T = \frac{2\pi r}{v}\\\\T = \frac{(2\pi )(6.732 x 10^6 m}{7.69 x 10^3 m/s}\\\\$

T = 5500 s = 91.67 min = 1.53 h

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3 years ago

What is the voltage drop across the 10.0 2 resistor?

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Answer:

the answer is equal to 2.00v

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3 years ago

Differences between freezing point and melting point (Atleast 5 differences)

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Answer:

What is freezing point?

A liquid's freezing point is determined at which it turns into a solid. Corresponding to the melting point, the freezing point often rises with increasing pressure. In the case of combinations and for some organic substances, such as lipids, the freezing point is lower than the melting point. The first solid which develops when a combination freezes often differs in composition from the liquid, and the development of the solid alters the composition of the remaining liquid, typically lowering the freezing point gradually. Utilizing successive melting and freezing to gradually separate the components, this approach is used to purify mixtures.

What is melting point?

The temperature at which a purified substance's solid and liquid phases may coexist in equilibrium is referred to as the melting point. A solid's temperature goes up when heat is added to it until the melting point is achieved. The solid will then turn into a liquid with further heating without changing temperature. Additional heat will raise the temperature of the liquid once all of the solid has melted. It is possible to recognize pure compounds and elements by their distinctive melting temperature, which is a characteristic number.

The difference between freezing point and melting point:

While a substance's melting point develops when it transforms from a solid to a liquid, a substance's freezing point happens when a liquid transforms into a solid when the heat from the substance is removed.
When the temperature rises, the melting point can be seen, and when the temperature falls, the freezing point can be seen.
When a solid reaches its melting point, its volume increases; meanwhile, when a liquid reaches its freezing point, its volume decreases.
While a substance's freezing point is not thought of as a distinctive attribute, its melting point is.
While external pressure is a significant component in freezing point, atmospheric pressure is a significant element in melting point.
Heat must be supplied from an outside source in order to reach the melting point for such a state shift. When a material is at its freezing point, heat is needed to remove it from the substance in order to alter its condition.

Reference: Berry, R. Stephen. "When the melting and freezing points are not the same." Scientific American 263.2 (1990): 68-75.

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2 years ago

A 2-kg wheel rolls down the road with a linear speed of 15m/s. Find its trwansitional and rotational kinetic energies.​

A 2-kg wheel rolls down the road with a linear speed of 15m/s. Find its trwansitional and rotational kinetic energies.