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dmitriy555 [2]
3 years ago
15

How much work must be done to raise a 1100kg car 2m above the ground?​

Physics
2 answers:
N76 [4]3 years ago
6 0
21560J? i dont know
Xelga [282]3 years ago
4 0

Answer:

21560 J

Explanation:

Work = mg*h = 1100*9.8*2 = 21560 J

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A building made with a steel structure is 565 m high on a winter day when the temperature is 0◦F. How much taller is the buildin
anyanavicka [17]

To solve this problem we apply the thermodynamic equations of linear expansion in bodies.

Mathematically the change in the length of a body is subject to the mathematical expression

\Delta L = L_0 \alpha \Delta T

Where,

L_0 = Initial Length

\alpha = Thermal expansion coefficient

\Delta T = Change in temperature

Since we have values in different units we proceed to transform the temperature to degrees Celsius so

0\°F \Rightarrow (0-32)*\frac{5}{9} = -17.77\°C

103\°F \Rightarrow (103-32)*(\frac{5}{9})= 39.44\°C

The coefficient of thermal expansion given is

\alpha = 1.1*10^{-5}/\°C

The initial length would be,

L_0 = 565m

Replacing we have to,

\Delta L = L_0 \alpha \Delta T

\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))

\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))

\Delta L = 0.355m

This means that the building will be 35.5cm taller

3 0
3 years ago
A 0.50-kg block slides across a tabletop with an initial velocity of 20 cm/s and comes to rest in a distance of 70 cm. Find the
mars1129 [50]

Answer:

FAE= 0.014 N

Explanation:

The KE of block is decreased because of the slowing action of the friction force .

Change in KE of block = work done on block by friction ƒ

⠀ ➪ ½mu²ƒ - ½mu²i = Fƒs cos θ

Because the friction force on the block is opposite in direction to the displacement , cos θ = -1

➢ Using Uƒ = 0 , Vƒ = 0.20 m/s , and s = 0.70 m

✒ We find ,

➪½mu²ƒ - ½mu²i = Fƒs cos θ

➪0-½ (0.50 kg) (0.20 m/s)² = (Fƒ) (0.70 m) (-1)

➪ Fƒ = 0.014 N

Hope this helped, can i pls have brainliest

6 0
3 years ago
Read 2 more answers
Tính lực tương tác giữa 2 điện tích q1=4.10^-6C và q2=-3.10^-6C cách nhau một 3cm đặt trong dầu hoả = 2
olga2289 [7]

Explanation:

decimal diamond ml dias and ka n t u t a n mo papa mo

4 0
2 years ago
Which statement correctly compares ultraviolet light to visible light? Ultraviolet light has both a lower frequency and longer w
Effectus [21]

The correct statement is

Ultraviolet light has both a higher frequency and a higher radiant energy than visible light.

because ultraviolet light has wavelength smaller than the visible light hence has a greater frequency as compared to visible light. (frequency is inversely related to wavelength. hence smaller the wavelength , greater will be the frequency)

we also know that the radiant energy is directly proportional to the frequency. hence greater the frequency , greater will be the radiant energy.

Since the frequency is greater for ultraviolet light , it radiant energy is also greater


7 0
3 years ago
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On Earth, 1 kg = 9.8 N = 2.2 lbs. On the Moon, 1 kg = 1.6 N = 0.37 lbs. Use these relationships to answer the following question
romanna [79]

Answer:

(a) 490 N on earth

(b) 80 N on earth

(c) 45.4545 kg on earth

(d) 270.27 kg on moon

Explanation:

We have given 1 kg = 9.8 N = 2.2 lbs on earth

And 1 kg = 1.6 N = 0.37 lbs on moon

(a) We have given mass of the person m = 50 kg

As it is given that 1 kg = 9.8 N

So 50 kg = 50×9.8 =490 N

(b) Mass of the person on moon = 50 kg

As it is given that on moon 1 kg = 1.6 N

So 50 kg = 50×1.6 = 80 N

(c) We have given that weight of the person on the earth = 100 lbs

As it is given that 1 kg = 2.2 lbs on earth

So 100 lbs = 45.4545 kg

(d) We have given weight of the person on moon = 100 lbs

As it is given that 1 kg = 0.37 lbs

So 100 lbs \frac{100}{0.37}=270.27kg

8 0
3 years ago
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