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3 years ago

7

Leah is making a chart to compare and contrast renewable and non-renewable resources. Which of the following would best complete

the chart? (2 points)
Renewable Resources

Non-renewable Resources

__

Used faster than they can be replaced

Most sources do not emit greenhouse gases

Availability decreases over time

Includes solar power, wind power, and geothermal power

Includes fossil fuels, natural gas, and coal

a. Replaces all current forms of energy
b. New energy is created
c. Cannot be replaced
d. Can be replaced at the same rate or faster than they are used

2 answers:

Sedbober [7]3 years ago

6 0

Answer:

d. Can be replaced at the same rate or faster than they are used

Explanation:

I just took the test, and its the right answer. Your welcome ✌

BartSMP [9]3 years ago

4 0

Answer:

D

Explanation:

I took the test and got it right!

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Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug

Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

$F_D=\dfrac{1}{2}\rho C_DA v^2$

$C_D$ =Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

Now by putting the values

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

$\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}$

$\dfrac{F_D_1}{F_D_2}=0.444$

Percentage Change in the drag force

$\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100$

$\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100$

$\Delta F=\dfrac{1-0.444}{0.444}\times 100$

ΔF=125.22 %

Therefore force will increase by 125.22 %.

3 0

3 years ago

At a certain point in space, there is a potential of 800 V relative to zero. What is the potential energy of the system when a +

sammy [17]

To solve this problem we will apply the concepts related to electric potential and electric potential energy. By definition we know that the electric potential is determined under the function:

$V = \frac{k_e q}{r}$

$k_e$ = Coulomb's constant

q = Charge

r = Radius

At the same time

$U = \frac{k_e q_1q_2}{r}$

The values of variables are the same, then if we replace in a single equation we have this expression,

$U = Vq$

If we replace the values, we have finally that the charge is,

$V = 800V$

$q = 1\mu C$

$U = (800V)(1*10^{-6}C)$

$U = 8*10^{-4}J$

Therefore the potential energy of the system is $8*10^{-4} J$

7 0

3 years ago

What is the mass of a 100-N crate of delicious candy

Svetach [21]

Answer:

Explanation:

weight ( w ) = 100 N

mass (m) = ?

WE know

g = 10 m/s²

w = m * g

or

m = w / g

m = 100 / 10

m = 10 kg

hope it will help :)

6 0

3 years ago

Two teams of nine members each engage in tug-of-war. Each of the first team's members has an average mass of 68 kg and exerts an

diamong [38]

Answer:

(a) Acceleration = 0.1063 m/s^2 (Second team wins)

(b) Tension in rope = 65.106 N

Explanation:

Total mass of first team = 68 * 9 = 612 kg

Total force of first team = 1350 * 9 = 12150 N

Total mass of second team = 73 * 9 = 657 kg

Total force of seconds team = 1365 * 9 = 12285 N

Difference in force = 12285 - 12150 = 135 N (towards the second team as it has more force)

(a) For acceleration we get:

F = m * a

135 = (mass of both teams) * a

a = 135 / (612 + 657)

acceleration = 0.1063 m/s^2 (Second team wins)

(b) Since we know the acceleration of the first team (pulling being pulled towards the second team at an acceleration of 0.1063 m/s^2) , we can find out the force required to move them:

Force required for first team = mass of first team * acceleration

Force required = 612 * 0.1063

Force required = 65.106 N

This is the force exerted on the first team through the rope, so the tension in the rope will also be 65.106 N.

7 0

3 years ago

An accelerator produces a beam of protons with a circular cross section that is 2.0 mm in diameter and has a current of 1.0 mA.

rewona [7]

Answer:

the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

Explanation:

Given that;

diameter D = 2.0 mm

current I = 1.0 mA

K.E of each proton is 20 MeV

the number density of the protons in the beam = ?

Now, we make use of the relation between current and drift velocity

I = MeAv ⇒ 1 / eAv

The kinetic energy of protons is given by;

K = $\frac{1}{2}$ $m_{p}$ v²

v = √( 2K / $m_{p}$ )

lets relate the cross-sectional area A of the beam to its diameter D;

A = $\frac{1}{4}$ πD²

now, we substitute for v and A

n = I / $\frac{1}{4}$ πeD² ×√( 2K / $m_{p}$ )

n = 4I/π eD² × √( $m_{p}$ / 2K )

so we plug in our values;

n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )

n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵

n = 3.2 × 10¹³ m⁻³

Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

8 0

2 years ago