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3 years ago

At what point of thermometric scale does Kelvin scale reading coincide with Fahrenheit scale

Physics

1 answer:

amid [387]3 years ago

3 0

Answer:

<em>At 574.59 Kelvin, the Fahrenheit temperature will be 574.59 °F.</em>

Explanation:

We first need to find a relation between the Kelvin scale and the Fahranheit scale. We'll use the Celsius scale to relate them.

The Kelvin and Celsius scales are related by the formula:

K = °C + 273.15

Solving for °C:

°C = K - 273.15

Besides, the Kelvin and Celsius scales are related by:

°C = 5 ⁄ 9(°F - 32)

Now we find a temperature, say X, where both scales coincide. Equating both formulas:

X - 273.15=5 ⁄ 9(X - 32)

Multiply by 9:

9X - 2,458.35 = 5X - 160

Simplifying:

4X = 2,458.35 - 160=2,298.35

Solving:

X =2,298.35 / 4 = 574.59

At 574.59 Kelvin, the Fahrenheit temperature will be 574.59 °F.

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A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it i

kaheart [24]

Answer:

The moment of inertia of the system decreases and the angular speed increases.

Explanation:

This very concept might not seem to be interesting at first, but in combination with the law of the conservation of angular momentum, it can be used to describe many fascinating physical phenomena and predict motion in a wide range of situations.

In other words, the moment of inertia for an object describes its resistance to angular acceleration, accounting for the distribution of mass around its axis of rotation.

Therefore, in the course of this action, it is said that the moment of inertia of the system decreases and the angular speed increases.

5 0

3 years ago

A 5.30 g bullet moving at 963 m/s strikes a 610 g wooden block at rest on a frictionless surface. The bullet emerges, traveling

Tems11 [23]

Answer:

a) $V_{wf}$ = 4.67m/s

b) V = 8.29 m/s

Explanation:

Givens:

The bullet is 5.30g moving at 963m/s and its speed reduced to 426m/s. The wooden block is 610g.

a) From conservation of linear momentum

Pi = Pf

$m_{b}V{b_{i} } + V_{wi} = m_{w} V_{wf} + m_{b}V_{bf}$

where $m_{b},V{b_{i}$ are the mass and the initial velocity of the bullet, $m_{w}$ and $V_{wi}$ are the mass and the initial velocity of the wooden block, and $V_{wf}$ and $V_{bf}$ are the final velocities of the wooden block and the bullet

The wooden block is initial at rest $(V_{wi} = 0)$ this yields

$m_{b}V{b_{i} } = m_{w} V_{wf} + m_{b}V_{bf}$

By solving for $V_{wf}$ adn substitute the givens

$V_{wf}$ = $\frac{m_{b}V_{bi} - m_{b} V_{bf} }{m_{W} }$

= $\frac{5.3(g)(963(m/s)-426(m/s) }{610(g)}$

$V_{wf}$ = 4.67m/s

b) The center of mass speed is defined as

$V = \frac{m_{b} }{m_{b}+m_{w} } V_{bi}$

substituting:

$V = \frac{5.3(g)}{5.3(g)+610(g)} X 963(m/s)$

V = 8.29 m/s

7 0

3 years ago

Calculate the Schwarzschild radius (in kilometers) for each of the following.1.) A 1 ×108MSun black hole in the center of a quas

Westkost [7]

Answer:

(I). The Schwarzschild radius is $2.94\times10^{8}\ km$

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is $1.1\times10^{-7}\ km$

(IV). The Schwarzschild radius is $7.4\times10^{-29}\ km$

Explanation:

Given that,

Mass of black hole $m= 1\times10^{8} M_{sun}$

(I). We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Where, G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times1\times10^{8}\times1.989\times10^{30}}{(3\times10^{8})^2}$

$R_{g}=2.94\times10^{8}\ km$

(II). Mass of block hole $m= 6 M_{sun}$

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times6\times1.989\times10^{30}}{(3\times10^{8})^2}$

$R_{g}=17.7\ km$

(III). Mass of block hole m= mass of moon

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times7.35\times10^{22}}{(3\times10^{8})^2}$

$R_{g}=1.1\times10^{-7}\ km$

(IV). Mass = 50 kg

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times50}{(3\times10^{8})^2}$

$R_{g}=7.4\times10^{-29}\ km$

Hence, (I). The Schwarzschild radius is $2.94\times10^{8}\ km$

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is $1.1\times10^{-7}\ km$

(IV). The Schwarzschild radius is $7.4\times10^{-29}\ km$

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3 years ago

Janet jumps off a high diving platform with a horizontal velocity of 2.06 m/s and lands in the water 1.8 s later. How high is th

Archy [21]

Initial height of platform is 15.87m or 16m.

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3 years ago

Approximately how much air is in a column 1-cm2 in cross section that extends from sea level to the top of the atmosphere?

laila [671]

Approximately 101 N air is in a column 1-cm2 in cross-section that extends from sea level to the top of the atmosphere

The basic level for determining height and depth on Earth is the sea level. The ocean's surface tends to seek the same level since it is one continuous body of water. However, the sea level is never fully level due to winds, currents, river discharges, and changes in gravity and temperature.

At the equator, the radius of the Earth at sea level is 6378.137 km (3963.191 mi). At the poles, it is 6,356.752 km (3,949.903 km), and on average, it is 6,371.001 km (3,958.756 mi). The elevation of the shoreline—the boundary between the ocean and the land—is referred to as sea level. Land that is higher than this altitude is above sea level, and land that is lower is below sea level.

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8 0

2 years ago