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VikaD [51]
3 years ago
14

Consider a cloudless day on which the sun shines down across the United States. If 2073 kJ of energy reaches a square meter ( m

2 ) of the United States in one hour, how much total solar energy reaches the entire United States per hour
Physics
1 answer:
Mnenie [13.5K]3 years ago
5 0

The total amount of energy per hour is 2.039\cdot 10^{16} kJ

Explanation:

In this problem we are told that the amount of energy reaching a square meter in the United States per hour is

E_1 = 2073 kJ

The total surface area of the United States is

A=9.834\cdot 10^6 km^2

And converting into squared metres,

A=9.834\cdot 10^6 \cdot 10^6 = 9.834\cdot 10^{12} m^2

Therefore, the total energy reaching the entire United States per hour is given by:

E=AE_1 = (9.834\cdot 10^{12})(2073)=2.039\cdot 10^{16} kJ

Learn more about energy and power:

brainly.com/question/7956557

#LearnwithBrainly

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In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its sp
Slav-nsk [51]

Answer:

the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

Explanation:

Given that;

weight of vehicle = 4000 lbs

we know that 1 kg = 2.20462

so

m = 4000 / 2.20462 =  1814.37 kg

Initial velocity V_{i} = 60 mph = 26.8224 m/s

Final velocity V_{f} = 30 mph = 13.4112 m/s

now we determine change in kinetic energy

Δk = \frac{1}{2}m(  V_{i}² - V_{f}² )

we substitute

Δk = \frac{1}{2}×1814.37( (26.8224)² - (13.4112)² )

Δk = \frac{1}{2} × 1814.37 × 539.5808

Δk = 489500 Joules

we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule

so

Δk = 489500 / 3.6 × 10⁶

Δk = 0.13597 ≈ 0.136 kWh

Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

4 0
3 years ago
A bird flies at a speed of 2.3 m/s if it has 14 j of kinetic energy what is the mass
Sidana [21]
Kinetic Energy =  1/2 * mv²

Kinetic Energy = 14 J,  v = 2.3 m/s ,  m = ?

14  =      1/2 * m* 2.3²

14 = 0.5*m*2.3*2.3

m =  14 / (0.5*2.3*2.3)

m = 5.29 kg.

Mass = 5.29 kg.
7 0
3 years ago
Read 2 more answers
A lawn roller in the form of a thin-walled, hollow cylinder with mass M is pulled horizontally with a constant horizontal force
olga nikolaevna [1]

Answer:

a_{cm}=\frac{F}{2M}\\\\F_{fr}=\frac{F}{2}

Explanation:

Given the mass as M, the rotational inertia of the mower is;

I_{cm}=MR^2

-The roller doesn't slip while rolling;

v_{cm}=wR, a_{cm}=\alpha R

\sum F_x=Ma_x, -F+F_{fr}=-Ma_{cm}\\\\a_{cm}=\frac{F-Fr}{M}    \ \ \ \ \ \ \ \ eqtn1\\\\\sum \tau_{cm}=I_{cm}\alpha, F_{fr}(R)}=(MR^2)(\frac{a_{cm}}{R}), ->F_{fr}=Ma_c_m\ \ \  \ \ \ \ eqtn2\\\\a_{cm}=\frac{F-Ma_{cm}}{M}, ->F=2Ma_{cm}\\\\a_{cm}=\frac{F}{2M}\\\\\\\therefore F_{fr}=M(\frac{F}{2M})\\\\F_{fr}=\frac{F}{2}

6 0
3 years ago
A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck
Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing}) where M_{truck} is the mass of the truck, V_{truck} is velocity of the truck, V_{common} is the common velocity of moving and standing truck after collision and M_{standing} is the mass of the standing truck

Making V_{truck} the subject we obtain

V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}

Substituting M_{truck} as 25000 Kg, V_{common} as 22.3 m/s, M_{standing} as 2000 Kg we obtain

V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0
3 years ago
A child is playing on a swing. As long as he does not swing too high the time it takes him to complete one full oscillation will
Aleks [24]

Answer:

We know that for a pendulum of length L, the period  (time for a complete swing) is defined as:

T = 2*pi*√(L/g)

where:

pi = 3.14

L = length of the pendulum

g = gravitational acceleration = 9.8 m/s^2

Now, we can think on the swing as a pendulum, where the child is the mass of the pendulum.

Then the period is independent of:

The mass of the child

The initial angle

Where the restriction of not swing to high is because this model works for small angles, and when the swing is to high the problem becomes more complex.

7 0
3 years ago
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