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3 years ago

You put a bottle of soft drink in a refrigerator and leave it until its temperature has dropped 12.0K .

Physics

1 answer:

Bingel [31]3 years ago

3 0

Answer: a) - 437.8° F, b) - 261°c.

Explanation: a) the kelvin and Fahrenheit temperature scale are related by the formulae below.

5 (°F - 32) = 9 (k - 273)

Where °F = temperature in Fahrenheit and k = temperature in kelvin.

For question A, k = 12.0, by substituting to have the value for °F, we have

5(°F - 32) = 9 ( 12 - 273)

5(°F - 32) = 9(-261)

5(°F - 32) = - 2349

°F - 32 = - 2349/5

°F - 32 = - 469.8

°F = - 469.8 + 32

°F = - 437.8

Question B

The centigrade and kelvin scale are related by the formulae below

°c = k - 273

Where °c = temperature in centigrade and k = temperature in kelvin =12

°c = 12 - 273

°c = - 261

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While in sketch mode, the number displayed to the right of the cursor defines the sketch element`s exact size. Therefore, there

galina1969 [7]

Answer:

B) False.

Explanation:

Most of the times, the numbers that appears right to the cursor are representing the coordinates (x & y) of the cursor. So as the cursor moves around the screen those number changes accordingly.Therefore its wrong to say that there is no need to add a dimension to set the dimensional value.

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3 years ago

At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a pri

viva [34]

Answer:

$v_{f2}$ =6.5% $v_{i1}$

Explanation:

Mass of the ball: $m_{1} =0.12kg$ ]

Initial velocity of the ball: $v_{i1}$

final velocity of the ball: $v_{f1}$ which is -30/100 of $v_{i1}$ = $-0.3v_{i1}$

Mass of the bottle: $m_{2} =2.4kg$

Initial velocity of the bottle: $v_{i2}=0m/s$

final velocity of the bottle: $v_{f2}$ is unknown (to find)

by using conservation momentum, which stated that the initial momentum is equal to the final momentum.

$m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}$

so since the bottle is at rest firstly, therefore $v_{i2} =0$

$m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}$

$m_{1} v_{i1} =m_{1} v_{f1} +m_{2} v_{f2}$ equation 1

so now substitute $v_{f1}$ into equation 1

$m_{1} v_{i1} =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}$

$m_{1} v_{i1} = -0.3m_{1}v_{i1} +m_{2} v_{f2}$

collect the like terms

$m_{1} v_{i1} +0.3m_{1}v_{i1} =m_{2} v_{f2}$

$1.3m_{1} v_{i1} =m_{2} v_{f2}$

divide both side by $m_{2}$

$v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }$

Now substitute

$v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2} =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}$

$v_{f2} =$ 6.5% $v_{i1}$

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3 years ago

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Give a example of scalar quantity

Serhud [2]

speed, volume, mass, temperature and power

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3 years ago

Read 2 more answers

The 0.15kg baseball has a speed of v=30 m/s just before it is struck by the bat. It then travels along the trajectory shown befo

notka56 [123]

The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N

The mass of the baseball, m = 0.15 kg

The speed at which it moves, v = 30 m/s

Time at which the baseball was in contact with the bat, t = 0.75 ms

t = 0.75/1000 s

t = 0.00075 s

The impulsive force is given by the formula:

$F=\frac{mv}{t}$

Substitute m = 0.15 kg, v = 30, and t = 0.00075s into the formula above:

$F=\frac{0.15 \times 30}{0.00075} \\\\F=6000N$

The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N

Learn more here: brainly.com/question/25892144

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2 years ago

A piece of styrofoam has a charge of 0.002 mC and is placed 0.5 m from a grain of salt with a charge of 0.03 nC. How much electr

aleksklad [387]

Answer:

2.16×10⁻⁶ N

Explanation:

Applying,

F = kqq'/r² (coulomb's Law)....................... Equation 1

Where F = electrostatic force, k = coulomb's constant, q = charge on the styrofoam, q' = charge on the grain of salt, r = distance between the charges.

From the question,

Given: q = 0.002 mC = 2.0×10⁻⁶ C, q' = 0.03 nC = 3.0×10⁻¹¹ C, r = 0.5 m

Constant: k = 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = (2.0×10⁻⁶)(3.0×10⁻¹¹)(8.99×10⁹)/0.5²

F = 2.16×10⁻⁶ N

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3 years ago