X = 1/2 a t square <br> Help me please

Determine the mass of fuel required for the expected energy consumption in the United States for the next 10 years:

alexandr1967 [171]

Answer:

(This will depend on the type of fuel, I will assume that the fuel is petrol)

First, let's find the expected energy consumption in the US for the next 10 years.

We know that in one year, a person consumes 3.5*10^11 joules.

There are 310,000,00 people on the US

Then the total consumption in one year is:

310,000,000*3.5*10^11 joules = 1.085*10^20 J

In 10 years the consumption is 10 times the consumption of a single year, then the expected energy consumption in the US for the next 10 years is:

10*1.085*10^20 J = 1.085*10^21 J

Now let's find the mass of fuel required.

We know that a liter of petrol has 31,536,000 joules of energy,

And a liter of petrol weights 0.75 kg

To find the number of liters of petrol that we need, we need to find the quotient between the expected energy consumption in the next 10 years and the energy of a single liter of petrol, this is:

N = (1.085*10^20 J)/(31,536,000 j) = 3.44*10^13

We will need 3.44*10^13 liters of petrol.

And the total mass of petrol will be:

M = 3.44*10^13*0.75 kg = 2.58*10^13 kg of fuel.

8 0

3 years ago

jim(mass=100kg) rollerblades on a smooth horizontal floor at a constant speed of 2.0 m/s for a distance of 5m in 5 seconds. What

azamat

Since Jim's speed is constant and he is moving in a straight line, he is not accelerating, and we know the net force on him is zero. There is no Force anywhere doing any work. So no power is being added to him or dissipated by him.

7 0

3 years ago

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A stationary police car emits a sound of frequen

bixtya [17]

Answer:

a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.

b) The frequency is 1404.08 Hz

Explanation:

If the police car is a stationary source, the frequency is:

$f_{a} =(\frac{v+v_{c} }{v} )f_{s}$ (eq. 1)

fs = frequency of police car = 1200 Hz

fa = frequency of moving car as listener

v = speed of sound of air

vc = speed of moving car

If the police car is a stationary observer, the frequency is:

$f_{L} =f_{a} (\frac{v}{v-v_{c} } )=(\frac{v+v_{c} }{v-v_{c} } )f_{s}$ (eq. 2)

Now,

fL = frequecy police car receives

fs = frequency police car as observer

a) The velocity of car is from eq. 2:

$1250=1200(\frac{v+v_{c} }{v-v_{c} } )\\1250(v-v_{c} )=1200(v+v_{c} )\\v_{c} =\frac{50*344}{2450} =7.02m/s$

b) Substitute eq. 1 in eq. 2:

$f_{L} =(\frac{v+v_{p} }{v-v_{c} } )(\frac{v+v_{c} }{v-v_{p} } )f_{s} =(\frac{344+20}{344-7.02} )(\frac{344+7.02}{344-20} )*1200=1404.08Hz$

7 0

3 years ago

if a person can jump maximum along distance of 3m ,on the earth how far could be jump on the moon where acceleration due to grav

allochka39001 [22]

Answer:

The person can jump 48 m on the Moon

Explanation:

The question parameters are;

The maximum long jump distance of a person on Earth, $R_{max}$ = 3 m

The acceleration due to gravity on the Moon = 1 ÷ 16 of that on Earth

The distance the person can jump on the Moon is given as follows;

A person performing a jump across an horizontal distance on Earth (under gravitational force) follows the path of the motion of a projectile

The horizontal range, $R_{max}$ , of a projectile motion is found by using the following formula

$R_{max} = \dfrac{u^2}{g}$

Where;

g = The acceleration due to gravity = 9.8 m/s²

Therefore, we have;

$R_{max} = 3 \, m = \dfrac{u^2}{9.8 \, m/s^2 }$

u² = 3 m × 9.8 m/s² = 29.4 m²/s²

Therefore, on the Moon, we have;

The acceleration due to gravity on the Moon, $g_{Moon}$ = 1/16 × g

∴ $g_{Moon}$ = 1/16 × g = 1/16 × 9.8 m/s² ≈ 0.6125 m/s²

$R_{max \ Moon} = \dfrac{u^2}{g_{Moon}} = \dfrac{29.4 \ m^2/s^2}{0.6125 \, m/s^2 } \approx 48 \, m$

The maximum distance the person can jump on the Moon with the same velocity which was used on Earth is $R_{max \ Moon}$ ≈ 48 m

8 0

3 years ago

PLEASE HELP ME A lens with a surface that curves outward like the exterior of a sphere is __________. (Points : 1) reflected ref

yanalaym [24]

Convex.

Concave curved inward (like how a cave foes in) and convex curves outward. Reflected and refracted do not apply to a lens.

3 0

3 years ago

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X = 1/2 a t square Help me please