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marissa [1.9K]
3 years ago
9

A ball is kicked at an angle of 35 degrees with the ground. What should be the initial velocity of the ball so that it hits a ta

rget that is 30 meters away at a height of 1.8 meters?
Physics
1 answer:
pashok25 [27]3 years ago
6 0

Answer:

18.5 M/S

Explanation:

trust me

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2 years ago
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How much heat is required to convert 500g of liquid water at 28°C into steam at 150°C? Take the specific heat capacity of water
JulijaS [17]

Answer:

1,327,063Joules

Explanation:

Heat energy is the energy needed to convert the state of a body from one phase to another.

According to the question, we want to calculate the total heat required to convert water into vapour (steam).

Note that before water can vapourize, it has to reach the boiling point first which is at 100°C. Heat energy needed to convert the water to 100°C is expressed as H1 = mcΔθ

m is the mass of the object in kg =0.5kg

c is the spcific heat capacity of water = 4183J/kg°C

Δθ is the change in temperature = 100-28 = 72°C

H1 = 0.5*4183*72

H1 = 150,588Joules

Energy required to convert the water to team H2 =mLsteam

Lsteam is the latent heat of vaporization = 2.26×10⁶J/kg

H2 = 0.5*2.26×10⁶

H2 = 1130000Joules

Heat energy needed to convert the water to 150°C is expressed as H3 = mcΔθ

m is the mass of the object in kg =0.5kg

c is the spcific heat capacity of steam= 1859J/kg°C

Δθ is the change in temperature = 150-100 = 50°C

H3 = 0.5*1859*50

H1 = 46,475Joules

Total Heat requires = H1+H2+H3 = 150,588Joules+1130000Joules+ 46,475Joules = 1,327,063Joules

8 0
3 years ago
The condition required to work to be done​
Dvinal [7]

Answer:

work=f(costheta)

Explanation:

work is done when a force acts on a body and displaces it on the direction of force

6 0
2 years ago
Help! (Look in the pic)
hjlf

Answer:

turtle

Explanation:

5 0
2 years ago
Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit
Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

  $= 822486 \times 10^{-8}$

  $=0.822 \times 10^{-2} \ km/s$

 = 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

  $= 673552 \times 10^{-8}$

  $=0.673 \times 10^{-2} \ km/s$

 = 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

  $= 3371 \times 153.86 \times 10^{-8}$

  = 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

  $= 153.84 \times 2371 \times 10^{-8}$

  $=0.364 \times 10^{-2} \ km/s$

 = 3.64 m/s

       

3 0
3 years ago
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