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cupoosta [38]
3 years ago
6

Please helppp

Physics
1 answer:
Y_Kistochka [10]3 years ago
6 0

Answer:

F = 17.3 kN

Explanation:

The normal force must support the weight of the car plus provide for the needed centripetal acceleration.

F = m(g + v²/R ) = 1000(9.8 + 15²/30) = 17,300

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A boat has a mass of 31,999 kg and a plane has a mass of 110,000 kg. What is the ratio of weight of the plane to the boat?
Gelneren [198K]

Answer:

3.44:1

Explanation:

We use ratios to find the ratio of weight.

 Mass of Plane= 110000 kg                          ∴ W= mg

 Mass of  Boat =31999 kg

\frac{weight of plane}{weight of boat}=\frac{110000*9.81}{31999*9.81}

 =\frac{3.44}{1}

SO 3.44:1

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Sea and land breezes over a large region that change direction with the seasons are called_____________.
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4.5 cm3 of water is boiled at atmospheric pressure to become 4048.3 cm3 of steam, also at atmospheric pressure. Calculate the wo
Mrrafil [7]

1. 408.4 J

The work done by a gas is given by:

W=p\Delta V

where

p is the gas pressure

\Delta V is the change in volume of the gas

In this problem,

p=1.01\cdot 10^5 Pa (atmospheric pressure)

\Delta V=4048.3 cm^3 - 4.5 cm^3 =4043.8 cm^3 = 4043.8 \cdot 10^{-6}m^3 is the change in volume

So, the work done is

W=(1.01\cdot 10^5 Pa)(4043.8 \cdot 10^{-6}m^3)=408.4 J

2. 10170 J

The amount of heat added to the water to completely boil it is equal to the latent heat of vaporization:

Q = m \lambda_v

where

m is the mass of the water

\lambda_v = 2.26\cdot 10^6 J/kg is the specific latent heat of vaporization

The initial volume of water is

V_i = 4.5 cm^3 = 4.5\cdot 10^{-6}m^3

and the water density is

\rho = 1000 kg/m^3

So the water mass is

m=\rho V_i = (1000 kg/m^3)(4.5\cdot 10^{-6}m^3)=4.5\cdot 10^{-3}kg

So, the amount of heat added to the water is

Q=(4.5\cdot 10^{-3}kg)(2.26\cdot 10^6 J/kg)=10,170 J

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