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3 years ago

The last multiple choice answer is

Physics

1 answer:

N76 [4]3 years ago

5 0

Answer:

Answer: C

Explanation:

The closer the city is by the ocean, the cooler it will get because of the winds it is created from the ocean. Meanwhile, Atlanta is slightly away from the ocean but wont be as cool as the Tybee Island. So the heat will hit more to the city away from the ocean.

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2. Which statement is correct about the Krebs cycle?

IRINA_888 [86]

The correct option is

a. Acetyl-CoA combines with a pyruvic acid to make glucose in the Krebs cycle.

Explanation:

The Krebs citric acid cycle happens within the mitochondrial matrix and generates a pool of energy (ATP, NADH, and FADH2) from the oxidization of pyruvate, the tip product of metabolism. Pyruvate is transported into the mitochondria and loses dioxide to make acetyl-CoA, a 2-carbon molecule.

5 0

3 years ago

Read 2 more answers

A river 1.00 mile wide flows with a constant speed of 1.00 mph. A man can row a boat at 2.00 mph. He crosses the river in a dire

gladu [14]

To solve this problem we will apply the geometric concepts of displacement according to the description given. Taking into account that there is an initial displacement towards the North and then towards the west, therefore the speed would be:

$V_T^2=v_N^2-v_W^2$

$V_T = \sqrt{v_N^2-v_W^2}$

Travel north 2mph and west to 1mph, then,

$V_T = \sqrt{2^2-1^2}$

$V_T = \sqrt{3}$

The route is done exactly the same to the south and east, so make this route twice, from the definition of speed we have to

$v= \frac{\Delta x}{t}$

$t = \frac{\Delta x}{v}$

$t = \frac{2*(1mile)}{\sqrt{3}mph}$

$t = 1.15hour$

Therefore the total travel time for the man is 1.15hour.

3 0

3 years ago

Two resistances, R1 and R2, are connected in series across a 9-V battery. The current increases by 0.450 A when R2 is removed, l

Rina8888 [55]

Answer:

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

Explanation:

Since the resistors R1 and R2 are connected in series, the current flowing through them when the 9 V battery is applied is 9/R1 + R2.

When the current increases by 0.450 A wen only R1 is in the circuit, the current is

9/R1 + R2 + 0.450 A = 9/R1 (1)

When the current increases by 0.225 A when only R2 is in the circuit, the current is

9/R1 + R2 + 0.225 A = 9/R2 (2)

equation (1) - (2) equals

9(1/R1 - 1/R2) = 0.450 A - 0.225

9(1/R1 - 1/R2) = 0.125

(1/R1 - 1/R2) = 0.125 A/9 = 0.0138

1/R1 = 0.0138 + 1/R2

R1 = R2/(1 + 0.0138R2) (3)

From (1)

9/R1 - 9/R1 + R2 = 0.450 A

9R2/[R1(R1 + R2)] = 0.450 A

R2/[R1(R1 + R2)] = 0.450 A/9 = 0.5

R2/[R1(R1 + R2)] = 0.5 (4)

From (3) R2/R1 = (1 + 0.0138R2) and from (4) R2/R1 = 0.5(R1 + R2). So,

(1 + 0.0138R2) = 0.5(R1 + R2)

0.5R1 + 0.5R2 = 1 + 0.0138R2

0.5R1 = 1 + 0.0138R2 - 0.5R2

0.5R1 = 1 - 0.4862R2 (5)

Substituting (3) into (5) we have

0.5R2/(1 + 0.0138R2) = 1 - 0.4862R2

R2 = (1 + 0.0138R2)(1 - 0.4862R2)

R2 = 1 - 0.4724R2 - 0.0067R2²

Collecting like terms, we have

0.0067R2² + 0.4724R2 + R2 - 1 = 0

0.0067R2² + 1.4724R2 - 1 = 0

Using the quadratic formula,

$R_{2} = \frac{-1.4724 +/-\sqrt{(1.4724)^{2} - 4 X 0.0067 X -1} }{2 X 0.0067} \\= \frac{-1.4724 +/-\sqrt{2.1680 + 0.0268} }{0.0268}\\= \frac{-1.4724 +/-\sqrt{2.1948} }{0.0268}\\= \frac{-1.4724 +/- 1.4815 }{0.0268}\\= \frac{-1.4724 + 1.4815 }{0.0268} or \frac{-1.4724 - 1.4815 }{0.0268}\\= \frac{0.0091 }{0.0268} or \frac{-2.9539}{0.0268}\\= 0.340 or -110.22$

We choose the positive answer.

So R2 = 0.340 Ω

From (5)

R1 = 0.5 - 0.9931R2

= 0.5 - 0.9931 × 0.340

= 0.5 - 0.338

= 0.162 Ω

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

5 0

3 years ago

player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. What is the hor

seropon [69]

Answer

22.5 m/s

Explanation

We shall use the trigonometric ratio cosine to find the horizontal component.

cos = adjacent/hypotenuse

Adjacent is the horizontal and hypotenuse is the fly speed.

cos 30° = horizontal / 26

horizontal velocity = 26 × cos 30°

= 26 × 0.866

= 22.5166

= 22.5 m/s

7 0

3 years ago

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A vector A⃗ has a magnitude of 40.0 m and points in a direction 20.0∘ below the positive x axis. A second vector, B⃗, has a magn

jolli1 [7]

The magnitude of the vector C is 96.32m

<h3>How to solve for the magnitude of vector c</h3>

Ax = AcosθA

= 40 cOS 20

= 37.59

Ay = AsinθA

-40sin20

= -13.68

Bx = B cos θ B

= 75Cos50

= 48.21

By = BsinθB

= 75sin50

= 57.45

Cx = AX + Bx

= 37.59 + 48.21

= 85.8

Cy = Ay + By

= -13.65 + 57.45

= 43.77

The magnitude is solved by

|c| = $\sqrt{Cx^{2}+Cy^{2} }$

= √85.8² + 43.77²

= 96.32m

The magnitude of the vector c is 96.32m

Read more on the magnitude of a vector here:

brainly.com/question/3184914

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6 0

2 years ago