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3 years ago

The standard unit of work in the metric system is named after the scientist _____.

Physics

2 answers:

Lena [83]3 years ago

7 0

James Prescott Joule (1818 - 1889)

Harlamova29_29 [7]3 years ago

4 0

Answer: Hello there! The standard unit of work in the metric system, the joule, is named after the James Prescott Joule (1819 -1889), where a joule is $j = \frac{kg*m}{s^{2} }$ . He was a physicist who did a lot of work with thermodynamics, for example, he worked on the Joule-Thomson effect.

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A 5.0 kg block hangs from the ceiling by a mass-less rope. A Second block with a mass of 10.0 kg is attached to the first block

gayaneshka [121]

The tension in the first and second rope are; 147 Newton and 98 Newton respectively.

Given the data in the question

Mass of first block; $m_1 = 5.0kg$

Mass of second block, $m_2 =10kg$

Tension on first rope; $T_1 =\ ?$

Tension on second rope; $T_2 =\ ?$

To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:

$F = m\ *\ a$

Where F is the force, m is the mass of the object and a is the acceleration ( In this case the block is under gravity. Hence ''a" becomes acceleration due to gravity $g = 9.8m/s^2$ )

For the First Rope

Total mass hanging on it; $m_T = m_1 + m_2 = 5.0kg + 10.0kg = 15.0kg$

So Tension of the rope;

$F = m\ * \ g\\\\F = 15.0kg \ * 9.8m/s^2\\\\F = 147 kg.m/s^2\\\\F = 147N$

Therefore, the tension in the first rope is 147 Newton

For the Second Rope

Since only the block of mass 10kg is hang from the second, the tension in the second rope will be;

$F = m\ * \ g\\\\F = 10.0kg \ * 9.8m/s^2\\\\F = 98 kg.m/s^2\\\\F = 98N$

Therefore, the tension in the second rope is 98 Newton

Learn More, brainly.com/question/18288215

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2 years ago

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9. Ripening is a _______________ and spoiling is a ______________.

IrinaK [193]

Answer:

Explanation:

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2 years ago

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You push a coin across a table. The coin stops. How does this motion relate to balanced and unbalanced forces?

Snezhnost [94]

If you are pushing the coin across the table at a constant rate, the friction of the table and the horizontal force of your hand pushing are equal, and the coin itself moves at a constant rate. If you push a coin and let it go, there is no horizontal force keeping the coin going. Friction slows the coin to a stop. In both cases, the gravitational downward pull of Earth is equally but oppositely resisted by the upward push of table on the coin.

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3 years ago

the ratio of the energy per second radiated by the filament of a lamp at 250k to that radiated at 2000k, assuming the filament i

Naily [24]

Answer:

(a) $\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}$

(b) P = 0.816 Watt

Explanation:

(a)

The power radiated from a black body is given by Stefan Boltzman Law:

$P = \sigma AT^4$

where,

P = Energy Radiated per Second = ?

σ = stefan boltzman constant = 5.67 x 10⁻⁸ W/m².K⁴

T = Absolute Temperature

So the ratio of power at 250 K to the power at 2000 K is given as:

$\frac{P_{250k}}{P_{2000k}}=\frac{\sigma A(250)^4}{\sigma A(2000)^4}\\\\\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}$

(b)

Now, for 90% radiator blackbody at 2000 K:

$P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4$

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3 years ago

A diver makes 1.0 revolutions on the way from a 9.2-m-high platform to the water. Assuming zero initial vertical velocity, find

____ [38]

Answer:

Average angular velocity ≈ 4.59 rad/s

Explanation:

Using the equation of motion,

H = ut + (1/2)t² ............................ equation 1.

Where H= height, u = initial velocity(m/s), g = acceleration due to gravity(m/s²), t = time(s) u= 0 ∴ ut =0

H =(1/2)gt².................................... equation 2.

making t² the subject of the relation in equation 2,

∴ t² = 2H/g

Where H = 9.2 m, g= 9.8 m/s

∴ t² = ( 2×9.2)/9.8

t = √(2 × 9.2/9.8) = √(18.4/9.8)

t = 1.37 s.

The average angular velocity = θ/t

Where θ = is the number of revolution that the diver makes, t = time

θ = 1 rev.

Since 1 rev = 2π (rad)

t = 1.37 s

Average angular velocity = 2π/t

π = 3.143

Average angular velocity = (2×3.143)/1.37 = 6.286/1.37

Average angular velocity ≈ 4.59 rad/s

8 0

3 years ago