Gravitational Force

Un'automobile di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

sesenic [268]

Answer:

e di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiute di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiut

Explanation:

e di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiute di 1200 kg rallenta,passando da una velocità di 130 km/h a una velocità di 70 km/h .

Quale è stato il lavoro compiut

3 0

2 years ago

3 An un calibrated mercury in glass thermometer immersed in melting ice. The length of the mercury thread is 25 mm when the ther

sammy [17]

Answer:

25 mm = 0 deg C

200 mm = 100 deg C

200 - 25 = 175 = change in thread per 100 deg C

95 - 25 = 70 mm - change in thread from 0 deg C

70 / 175 * 100 = 40 deg C final temperature at 95 mm

5 0

3 years ago

On a straight, level, two-lane road, two cars moving in opposite directions approach and pass each other. Car A is in the eastbo

ludmilkaskok [199]

Answer:

a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).

Explanation:

a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:

$\vec v_{A/B} = \vec v_{A} - \vec v_{B}\\\vec v_{A/B} = 11 \frac{m}{s} \cdot i + 31 \frac{m}{s} \cdot i\\\vec v_{A/B} = 42 \frac{m}{s} \cdot i$

42 m/s, positive direction (to the east).

b) The relative velocity of Car B as seen by the drive of Car A is:

$\vec v_{B/A} = \vec v_{B} - \vec v_{A}\\\vec v_{B/A} = -31 \frac{m}{s} \cdot i - 11 \frac{m}{s} \cdot i\\\vec v_{B/A} = - 42 \frac{m}{s} \cdot i$

42 m/s, negative direction (to the west).

5 0

3 years ago

What quantity of heat is needed to convert 1 kg of ice at -13 degrees C to steam at 100 degrees C?

Effectus [21]

Answer:

Heat energy needed = 3036.17 kJ

Explanation:

We have

heat of fusion of water = 334 J/g

heat of vaporization of water = 2257 J/g

specific heat of ice = 2.09 J/g·°C

specific heat of water = 4.18 J/g·°C

specific heat of steam = 2.09 J/g·°C

Here wee need to convert 1 kg ice from -13°C to vapor at 100°C

First the ice changes to -13°C from 0°C , then it changes to water, then its temperature increases from 0°C to 100°C, then it changes to steam.

Mass of water = 1000 g

Heat energy required to change ice temperature from -13°C to 0°C

H₁ = mcΔT = 1000 x 2.09 x 13 = 27.17 kJ

Heat energy required to change ice from 0°C to water at 0°C

H₂ = mL = 1000 x 334 = 334 kJ

Heat energy required to change water temperature from 0°C to 100°C

H₃ = mcΔT = 1000 x 4.18 x 100 = 418 kJ

Heat energy required to change water from 100°C to steam at 100°C

H₄ = mL = 1000 x 2257 = 2257 kJ

Total heat energy required

H = H₁ + H₂ + H₃ + H₄ = 27.17 + 334 + 418 +2257 = 3036.17 kJ

Heat energy needed = 3036.17 kJ

5 0

3 years ago

Suppose for your cookout you need to make 100 hamburgers. You know that 2.00 pounds will make 9 hamburgers. How many pounds will

VMariaS [17]

2 pounds = 9 burgers figure out ow many 9's you can get out of 100: 100/9=11 but that only makes 99 you need 100 so we would add another one making 12. now multiply 12 by 2: 12·2=24. You would need 24 pounds of meet :)

8 0

3 years ago