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irakobra [83]
2 years ago
10

2. On January 21 in 1918, Granville, North Dakota, had a surprising change in temperature. Within 12 hours, the temperature chan

ged from 237 K to 283 K. What is this change in temperature in the Celsius and Fahrenheit scales?
Physics
1 answer:
wolverine [178]2 years ago
7 0

Answer:

The temperature change in Celsius is 46°C.

The temperature change in Fahrenheit is 82.8°F.

Explanation:

A degree of Celsius scale is equal to that of kelvin scale; therefore,

\Delta C = \Delta K  = 283K-237K\\\\ \boxed{\Delta C = 46^oC}

A degree in Fahrenheit is 1.8 times the Celsius degree; therefore

\Delta F  = 1.8(46^o)

\boxed{\Delta F = 82.8^oF}

Hence, the temperature change in Celsius is 46°C, and the temperature change in Fahrenheit is 82.8°F.

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A motorist inflates the tires of her car to a pressure of 180 kPa on a day when the temperature is -8.0° C. When she arrives at
GuDViN [60]

Apply Gay-Lussac's law:

P/T = const.

P = pressure, T = temperature, the quotient of P/T must stay constant.

Initial P and T values:

P = 180kPa, T = -8.0°C = 265.15K

Final P and T values:

P = 245kPa, T = ?

Set the initial and final P/T values equal to each other and solve for the final T:

180/265.15 = 245/T

T = 361K

4 0
2 years ago
A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi
Vlad [161]

Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block m = 6.60 kg

Initial speed of block v _{i} = 1.56 \frac{m}{s}

Final speed of block v_{f} = 0 \frac{m}{s}

Coefficient of kinetic friction \mu _{k} = 0.62

Ramp inclined at angle \theta = 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

  \mu _{k} mgd \cos 28.4 =  \Delta K - \Delta U

Where \Delta U = mg d\sin 28.4

\mu _{k} mgd \cos 28.4 =  \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4

\mu _{k} mgd \cos 28.4 +mgd\sin 28.4  =  \frac{1}{2}mv_{i} ^{2}

d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}

 d = 0.122 m

Therefore, the distance travel by block before coming to rest is 0.122 m

7 0
2 years ago
If the gas in a container absorbs 275 Joules of heat, has 125 Joules of work done on it, then does 50 Joules of work, what is th
cluponka [151]

Answer:

    The increase in the internal energy = 350 J

Explanation:

Given that

Q= 275  J

W= - 125 J

W' = 50 J

W(net)= -125  + 50 = -75 J

Sign -

1.Heat rejected by system - negative

2.Heat gain by system - Positive

3.Work done by system = Positive

4.Work done on the system-Negative

Lets take change in the  internal energy =ΔU

We know that

Q= ΔU + W(net)

275 = ΔU -75

ΔU= 275 + 75 J

ΔU=350 J

The increase in the internal energy = 350 J

7 0
2 years ago
The ultimate source of energy that powers the Sun is__________.
dolphi86 [110]

Answer:C

Explanation:

Mass energy of hydrogen fusing into helium

5 0
1 year ago
Read 2 more answers
A cylindrical wire has a resistance R and resistivity ρ. If its length and diameter are BOTH cut in half, what will be its resis
snow_lady [41]

Answer:

The resistance will be 2×R

Explanation:

We note that the resistivity of a cylindrical wire is given by the following relation;

\rho = \frac{RA}{L}

Where:

ρ = Resistivity of the wire

R = The wire resistance

A = Cross sectional area of the wire = π·D²/4

L = Length  of the wire

Rearranging, we have;

R= \frac{\rho L}{A}

If the length and the diameter are both cut in half, we have;

L₂ = L/2

A₂ =π·D₂²/4 = \pi \cdot \left (\frac{D}{2}   \right )^{2} \times \frac{1}{4}  = \pi \cdot \frac{D^{2}}{16} = A/4

Therefore, the new resistance, R₂ can be expressed as follows;

R_2= \frac{\rho \frac{L}{2} }{\frac{A}{4} } = \rho \frac{L}{2} \times \frac{4}{A} = 2 \times  \frac{\rho L}{A}

Hence, the new resistance R₂ =  2×R, that is the resistance will be doubled.

8 0
2 years ago
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