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lukranit [14]
3 years ago
14

A car travels 480 km in 12 hr. what is its velocity

Physics
2 answers:
pishuonlain [190]3 years ago
6 0

Answer:

The correct answer is 40

Explanation:

Divide 480 and 12 that  = 40v

Hoped I helped

Svet_ta [14]3 years ago
5 0

Answer:

Explanation:

velocity=s/t

given distance is 480 km we have to convert it into meter because the unit of distance is meter

1 km=1000 m

480 km=480000 m =4.8*10^5

1 hour= 3600 s

12 hour =43200 s  

now velocity=s/t

v=4.8*10^5/43200

v=11.11 m/s

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Explain the relationship between energy and the potential to do work
Alenkinab [10]
Potential energy is the energy possessed by a body by virtue of its position or configuration . And energy is the capacity of doing work since power is work divided by time therefore change in potential energy divided by time is equal to the power of the object.
4 0
2 years ago
A milk truck carries milk with density 64.6 lb/ft3 in a horizontal cylindrical tank with diameter 12 ft. (a) Find the force exer
7nadin3 [17]

Answer:

F = 351×10³lb

Explanation:

Given the density

ρg = 64.6lb/ft³

Diameter d = 12ft

The tank is horizontally cylindrical. The vertical distance from the top to the bottom of the tank is h = 12ft

The pressure in the tank is

P = ρgh = 64.6 × 12 = 775.2lb/ft²

The force exerted on one end of the tank is therefore F = PA = 775.2 × πd² = 775.2π×12²

F = 351×10³lb.

4 0
3 years ago
The device used to measure a masses of a body is kilogram . true or false​
olga nikolaevna [1]

Answer: false

Explanation:

While kilograms are the unit used to measure body mass, the device used is a scale.

Hope it helps :)

6 0
2 years ago
Light reflected from a glass surface is polarized along the plane perpendicular to the glass surface. Please select the best ans
leva [86]

I believe it would be true not 100% sure


7 0
3 years ago
A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate
exis [7]

Answer:

The velocity of water at the bottom, v_{b} = 28.63 m/s

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, P_{gauge} = 2.90 atm

Solution:

Now,

Atmospheric pressue, P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa

At the top, the absolute pressure, P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa

Now, the pressure at the bottom will be equal to the atmopheric pressure, P_{b} = 1 atm = 1.01\times 10^{5} Pa

The velocity at the top, v_{top} = 0 m/s, l;et the bottom velocity, be v_{b}.

Now, by Bernoulli's eqn:

P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}

where

h_{t} -  h_{b} = 12.8 m

Density of sea water, \rho = 1030 kg/m^{3}

\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} =  v_{b}

\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} =  v_{b}

v_{b} = 28.63 m/s

5 0
3 years ago
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