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2 years ago

14

Can someone please help?

1 answer:

OLga [1]2 years ago

3 0

Answer:

1: E

2: A

3: C

4: B

5: D

Explanation:

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Match each example to its type of energy.

PtichkaEL [24]

Solar energy - A

nuclear energy - B

fossil fuel energy - C

wind energy - D

geothermal energy - E

6 0

3 years ago

About three billion years ago, single-celled organisms called cyanobacteria lived in Earth’s oceans. They thrived on the ocean’s

pickupchik [31]

I think its Oxygen.
ancient cyanobacteria produced Earth's first oxygen-rich atmosphere, which allowed the eventual rise of eukaryotes. T<span>he chloroplasts of eukaryotic algae and plants are derived from cyanobacteria</span>

3 0

2 years ago

Read 2 more answers

A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision

Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is $a=2,800m/s^{2}$

Explanation:

<u>Known Data</u>

We will asume initial speed has a negative direction, $v_{i}=-30m/s$ , final speed has a positive direction, $v_{f}=26m/s$ , $\Delta t=20ms=0.020s$ and mass $m_{b}$ .

<u>Initial momentum</u>

$p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s$

<u>final momentum</u>

$p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s$

<u>Impulse</u>

$I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s$

<u>Average Force</u>

$F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}$

<u>Average acceleration</u>

$F=ma$ , so $a=\frac{F}{m_{b}}$ .

Therefore, $a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}$

8 0

3 years ago

A tank initially holds 100 gallons of salt solution in which 50 lbs of salt has been dissolved. A pipe fills the tank with brine

olga_2 [115]

Answer:

A. 171.24 Ibs

Explanation:

To find the amount of salt in the tank,

Let Q = Amount of salt in the mixture

And let 100 + (3-2)t = 100 + t be the volume of mixture at anytime t.

Rate of gain - Rate of loss = dQ / dt

Concentration of salt = Q / (100+t)

For the linear differential equation,

dQ / dt = 3(2) - 2 [Q/ (100 + t)]

dQ /dt + Q [2 / (100 + t)] = 6

The general solution of the linear differential equation is:

Q (i.f) = ∫ A(t) (i.f) dt + C

Therefore,

i.f = e ^ ∫ P(t) dt

And P(t) = 2 / (100 + t)

i.f = e ^ ∫ 2 / (100 + t)

= e ^ 2㏑ (100 + t)

= e ^ ㏑ (100 + t) ^2 = (100 + t) ^2

Q(100 + t) ^ 2 = ∫6 (100 + t) ^2 dt + C

Q(100 + t) ^2 = 2(100 + t) ^ 3 + C

When t = 0, Q = 50

Therefore,

50( 100) ^2 = 2(100) ^3 + C

C = -1.5 * 10 ^6

therefore, when t = 30,

Q (100 + 30) ^2 = 2(100 + 30) ^3 - 1.5 * 10 ^6

Q (400) ^2 = 2(130) ^3 - 1.5 * 10 ^6

Q = 171.24 Ibs

7 0

2 years ago

In a Young's double-slit experiment, 610-nm-wavelength light is sent through the slits. The intensity at an angle of 2.95° from

Sliva [168]

Answer:

spacing between the slits is 405.32043 × $10^{-9}$ m

Explanation:

Given data

wavelength = 610 nm

angle = 2.95°

central bright fringe = 85%

to find out

spacing between the slits

solution

we know that spacing between slit is

I = 4 $I_{0}$ × cos²∅/2

so

I/4 $I_{0}$ = cos²∅/2

here I/4 $I_{0}$ is 85 % = 0.85

so

0.85 = cos²∅/2

cos∅/2 = √0.85

∅ = 2 × $cos^{-1}$ 0.921954

∅ = 45.56°

∅ = 45.56° ×π/180 = 0.7949 rad

and we know that here

∅ = 2π d sinθ / wavelength

so

d = ∅× wavelength / ( 2π sinθ )

put all value

d = 0.795 × 610× $10^{-9}$ / ( 2π sin2.95 )

d = 405.32043 × $10^{-9}$ m

spacing between the slits is 405.32043 × $10^{-9}$ m

7 0

2 years ago