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AveGali [126]
3 years ago
13

Any temperature on the Kelvin scale can be changed to Celsius degrees by?

Physics
1 answer:
igor_vitrenko [27]3 years ago
5 0
           0° C  =  273.15 K

Notice that the K temp is 273.15 more than the C temp.

So, to get any Celsius temp, subtract 273.15 from the Kelvin temp.
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- A wooden crate that measures 2.0 m long and 0.40 m wide rests on
devlian [24]

Answer:

P = F/A = 600.0 / (2.0(0.40)) = 750 N/m² = 750 Pa

Explanation:

8 0
3 years ago
3. A very brittle piece of iron metal (C=.44 J/gC) is heated to 200 C and placed
lubasha [3.4K]

Answer:The temperature of a piece of Metal X with a mass m of 95.4g increases from 25.0°C to 48.0°C as the metal absorbs 849 J of heat.

Explanation:

4 0
2 years ago
4. Jimmy dropped a 10 kg bowling ball from a building that is 25 meters high.
KIM [24]

Answer:

The K.E of the bowling ball right before it hits the ground, K.E = 2450 J            

Explanation:

Given data,

The mass of the bowling ball, m = 10 kg

The height of the building, h = 25 m

The total mechanical energy of the body is given by,

                                     E = P.E + K.E

At height 'h' the P.E is maximum and the K.E is zero,

According to the law of conservation of energy, the K.E at the ground before hitting the ground is equal to the P.E at 'h'

Therefore, P.E at 'h'

                                  P.E = mgh

                                         = 10 x 9.8 x 25

                                         =  2450 J

Hence, the K.E of the bowling ball right before it hits the ground, K.E = 2450 J                                                                      

3 0
3 years ago
một động cơ nhiệt khi hoạt động nhiệt lượng mà nó nhận vào gấp bốn lần công mà nó thực hiện. Hiệu suất của động cơ là?
OlgaM077 [116]

\frac{1}{4}  = 0.25 = 25\%

6 0
3 years ago
A proton in a particular electric field experiences an electric force whose magnitude is equal to the weight of the proton. Dete
gavmur [86]

Answer:

Electric field, E=10^{-7}\ N/C

Explanation:

It is given that, a proton in a particular electric field experiences an electric force whose magnitude is equal to the weight of the proton such that,

qE=mg

E=\dfrac{mg}{q}

E=\dfrac{1.67\times 10^{-27}\times 9.8}{1.6\times 10^{-19}}

E=1.02\times 10^{-7}\ N/C

or

E=10^{-7}\ N/C

So, the magnitude of electric field at the location of the proton is 10^{-7}\ N/C. Hence, this is the required solution.

8 0
4 years ago
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