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3 years ago

Given a temperature of 300 Kelvin, what is the approximate temperature in degrees Celsius?

Physics

2 answers:

kramer3 years ago

7 0

26.85 degrees celsius is the answer

S_A_V [24]3 years ago

3 0

<span>Jun 16, 2012 - Given a temperature of 300 Kelvin, what is the approximate temperature in degrees Celsius? –73°C 27°C 327°C 673°C.</span><span>
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Which of the following biomes has the highest number of different species of plants and animals?

Nataliya [291]

The answer is tropical rainforest

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3 years ago

suppose the ball has the smallest possible frequency that allows it to go all the way around the circle. what tension in the str

tatyana61 [14]

The complete question is missing, so i have attached the complete question.

Answer:

A) FBD is attached.

B) The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).

Explanation:

A) I've attached the image of the free body diagram.

B) The formula for the net force is given as;

F_net = mv²/r

We know that angular velocity;ω = v/r

Thus;

F_net = mω²r

Now, the minimum downward force is the weight and so;

mg = m(ω_min)²r

m will cancel out to give;

g = (ω_min)²r

(ω_min)² = g/r

ω_min = √(g/r)

The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).

5 0

2 years ago

Fill in the correct answers. A big wheel with twice the circumference of a small wheel will rotate with ____ the force and ___ t

Lina20 [59]

Answer:

Explanation:

""""""" Twice,Half """"""

3 0

3 years ago

Read 2 more answers

What is the magnitude of the electric field strength between them, if the potential 7.95 cm from the zero volt plate (and 2.05 c

Dmitriy789 [7]

Answer:

ΔVab = Ed

ΔVab = Va-Vb = Va-V0 = Va

E = Va/ d

= 413V / 0.0795 m

= 5194.97 V/M

Explanation:

the potential difference between two uniform plates is calculated by the formula of electric field.

6 0

3 years ago

A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s

strojnjashka [21]

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

$m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v$

Where:

$m_{M}$ , $m_{S}$ - Masses of the small meteorite and the communication satellite, measured in kilograms.

$v_{M}$ , $v_{S}$ - Speeds of the small meteorite and the communication satellite, measured in meters per second.

$v$ - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

$v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}$

If $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ and $v_{S} = 0\,\frac{m}{s}$ , the final speed is now calculated:

$v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}$

$v = 0.1\,\frac{m}{s}$

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

$K_{S} + K_{M} - K - Q_{disp} = 0$

$Q_{disp} = K_{S}+K_{M}-K$

Where:

$K_{S}$ , $K_{M}$ - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

$K$ - Kinetic energy of the satellite-meteorite system, measured in joules.

$Q_{disp}$ - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

$Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]$

Given that $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ , $v_{S} = 0\,\frac{m}{s}$ and $v = 0.1\,\frac{m}{s}$ , the dissipated heat is:

$Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]$ $Q_{disp} = 200000\,J$

$Q_{disp} = 200\,kJ$

The energy coverted to heat is 200 kilojoules.

4 0

3 years ago