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I think it’s B but I could be wrong. Please help!

What is the goal of a theory?

B. To analyze data from an experiment
D. To control the outcome of an experiment

I think I have already answer this question. Anyways please do let me know whether it is useful or not

Answer:

Explanation:

Area A = 4 pi R^2

lnA = ln(4 pi) + 2 ln R

dA/A = 2 dR/R = 2*(alpha)*dT

delta A = 2A*alpha*(delta T)

OR

A at T + deltaT

delta A = 4 pi (R + deltaR)^2 - A

= 4 pi R(1 + alpha*deltaT)^2 -A

= 4 pi R^2 *[1 + 2*alpha*deltaT +

(alpha*deltaT)^2] - A

= 2*A*alpha*deltaT + A*(alpha*deltaT)^2

(Ignore the second term in deltaT^2)

Answer:

[tex]thank \: you[/tex]

Answer:

joules/energy.

Explanation:

power is defined as the rate of work. Hence power=work/time, then you obtain Watts.

Work is the product of force and displacement.

Answer:

High pressure squeezes the hydrogen atoms together. They must be within 1x10-15 meters of each other to fuse. The sun uses its mass and the force of gravity to squeeze hydrogen atoms together in its core. We must squeeze hydrogen atoms together by using intense magnetic fields, powerful lasers or ion beams.

We have that the unit of the quantities are described

a.kg (m/s) (1/s)=Newton  

b. (kg/s) (m/s^2)=Newton

c. (kg/s) (m/s)^2=Newton  

d. (kg/s) (m/s)​=Newton

From the question we are give the units  

a.kg (m/s) (1/s)  

b. (kg/s) (m/s^2)  

c. (kg/s) (m/s)^2  

d. (kg/s) (m/s)​

a)  

Generally  

kg (m/s) (1/s) is Represented as Newton  

Where Newton is the unit Force and Weight

b)  

Generally  

(kg/s) (m/s) is the unit representation of Newton  

Therefore  

Newton

c)  

kg/s) (m/s) is the unit representation of Newton  

Therefore  

Newton

d)

Generally  

(kg/s) (m/s) is the unit representation of Newton  

Therefore  

Newton

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Question:-

Convert 4cm to m

Solution:-

Lets do

First note down the below formula

[tex]\boxed{\sf 1m=10^{2}cm}[/tex]

Then

[tex]\boxed{\sf 1cm=10^{-2}cm}[/tex]

Now we are given 4cm

So convert

[tex]\\ \sf\longmapsto 4\times 10^{-2}m[/tex]

If we simplify to decimal

[tex]\\ \sf\longmapsto 4cm=0.04m[/tex]

Hence done .

The  magnitude of their resultant vector is 4.6 meters/seconds

Since we are to add the  velocity vectors in order to  find the magnitude of their resultant vector.

Hence:

Resultant vector magnitude=5.8 meters/seconds + (1.2 meters/seconds)

Resultant vector magnitude=5.8 meters/seconds-1.2 meters/seconds

Resultant vector magnitude 4.6 meters/seconds

Inconclusion The  magnitude of their resultant vector is 4.6 meters/seconds

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Answer:

f = c / y      where y is the wavelength

f1 = 3 *10E8 m/s / .001 m = 3 * 10E11 / s

f2 = 3 * 10E8 m/s / 1 m = 3 * 10E8 / s

Answer:

it would be c.

Explanation:

starts at zero and isn't a straight shot up

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Answer:

Explanation:

The first question asks for the object's height. That is a y-dimension thing so we will use the one-dimensional formula:

[tex]v^2=v_0^2+2a[/tex]Δx and we are looking for delta x, which is the displacement of the object (here, the height since displacement in the y-dimension is height). Before we do that, though, we need to know the initial upwards velocity, which is found in:

[tex]v_{0y}=100sin(60)\\v_{0y}=90\frac{m}{s}[/tex]

And we also have to know from previous physics experience that the final velocity of an object at its max height is 0. Filling in:

[tex]0=(90)^2+2(-9.8)[/tex]Δx

[tex]0=8000-20[/tex]Δx and

-8000 = -20Δx so

Δx = 400 m

Now for the time it was in the air...we will use the simple equation

[tex]v=v_0+at\\0=90-9.8t\\-90=-9.8t\\t=9[/tex]

So the object is in the air for 9 seconds. Now we go to the other dimension, the x-dimension, to find out how far it goes horizontally. We will use the equation:

Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] but first we have to find the horizontal velocity, which is NOT the same as the upwards velocity.

[tex]v_{0x}=100cos(60)\\v_{0x}=50\frac{m}{s}[/tex]

And another thing we need to know prior to solving problems like this is that the acceleration in the x-dimension is ALWAYS 0. Filling in the equation:

Δx = [tex]50(9)+\frac{1}{2}(0)(9)^2[/tex]

Δx = 500 m

Answer:

infrared light

Explanation:

infrared light

Answer:

376 miles

Explanation:

speed = distance / time

therefore distance = speed * time

distance = 40 miles/hr * 9.4 hr

then

distance = 375 miles

Displacement = (14-20) Km = -6 Km

The displacement of a car is -6 Km South

Answer:

No because the heat that moved out of the water to make ice caused an even greater disorder in the environment around it.

Explanation:

Answer:

Y T and v

Explanation:

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In the human cannonball act makes use of of a special type of "cannon" that more safely ejects a person who acts as the "cannonball"

The world record for the longest flight by a human cannonball is approximately 194 feet and therefore the human cannon catapult that Bobo would need is one that would be able to propel him with an initial velocity of more than 120 kilometer per hour (more than the speed of the current record holder) with the cannon directed at an angle 45 degrees above the horizonal

The reason the values given above are correct is as follows:

The world record of the furthest range of the human cannonball is 193 ft. 8.8 inches. The maximum claim for the furthest range of a human cannonball is 200 ft. 4 in. Therefore, to safely set the world record would be to reach a range of 194 feet

The speed with which the world record was accomplished (120 km/jhr)  should be expected to be exceeded given, horizontal speed in free fall motion is constant (decreasing in the presence of drag), and the conditions are likely to be more or less similar

In order to reach the furthest range, R, at the possible speed, the angle of inclination of the "cannon" required is 45°, according to the range formula, which is given as follows;

[tex]Range, \ R = \dfrac{v^2 \times sin(2\cdot \theta)}{g}[/tex]

The maximum range is given where θ = 45, such that sin(2·θ) = sin(90°) = 1

A further range than the world record van be obtained using the above values

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Answer:

flowers

Explanation:

flowers cannot change

A. The S.I unit of mass is kg . therefore we have to convert 4000g into kg from gram

i.e 4000g = 4 kg

B. The S.I unit of length is meter. so we have to convert 2.4 km into meters,

i.e 2.4 km = 2400 m

#A

[tex]\boxed{\sf 1kg=1000g}[/tex]

[tex]\\ \sf\longmapsto 4000[/tex]

[tex]\\ \sf\longmapsto \dfrac{4000}{1000}[/tex]

[tex]\\ \sf\longmapsto 4kg[/tex]

#B

[tex]\boxed{\sf 1km=1000m}[/tex]

[tex]\\ \sf\longmapsto 2.4km=2.4(1000)[/tex]

[tex]\\ \sf\longmapsto 2400m[/tex]

Answer:

  2 seconds

Explanation:

Each second, it increases its speed by 6 m/s, so it will take 2 seconds for the speed to increase by 12 m/s.

Answer:

to sell and to buy

Explanation:

measurement is a comparison between unknown quantity with a known quantity

Answer:

v = 10 m/s

Explanation:

Let's assume the wheel does not slip as it accelerates.

Energy theory is more straightforward than kinematics in my opinion.

Work done on the wheel

W = Fd = 45(12) = 540 J

Some is converted to potential energy

PE = mgh = 4(9.8)12sin30 = 235.2 J

As there is no friction mentioned, the remainder is kinetic energy

KE = 540 - 235.2 = 304.8 J

KE = ½mv² + ½Iω²

ω = v/R

KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²

v = √(2KE / (m + I/R²))

v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6

v = 10.07968...

Answer:

With a force of 1 Newton, an object weighing 100 grams is dragged towards the planet's center. On Earth, an item with a mass of 100 grams has a gravitational pull of only 1%, or about kg.

Explanation:

i hope this helps

Answer:

kinetic energy = 1/2mv²

k.E =1/2 X 0.5 X 3²

k.e = 9/4

K.E = 2.25j

Answer:

1. Right hypochondriac region (upper, right).

2.  Epigastric region (upper, middle).

3. Left hypochondriac region (upper, left).

4. Right lumbar region (middle, right).

5. Umbilical region (centre).

6. Left lumbar region (middle, left).

7. Right iliac region (lower, right).

8. Hypogastric region (lower, middle).

9. Left iliac region (lower, left).

Explanation:

1. Right hypochondriac region (upper, right)- Liver, gallbladder, right kidney, small intestine.

2.  Epigastric region (upper, middle)- stomach, liver, pancreas, duodenum, spleen, adrenal glands

3. Left hypochondriac region (upper, left)- Spleen, colon, left kidney, pancreas

4. Right lumbar region (middle, right)- Gallbladder, liver, right colon

5. Umbilical region (centre)- Umbilicus, parts of the small intestine, duodenum

6. Left lumbar region (middle, left)- Descending colon and left kidney

7. Right iliac region (lower, right)- Appendix and cecum

8. Hypogastric region (lower, middle)- Bladder, reproductive organs, sigmoid colon

9. Left iliac region (lower, left)- Descending and sigmoid colon.

Learn more about terminology refer:

Average Velocity=Total Distance/Total Time

[tex]\\ \sf\longmapsto \dfrac{2.5}{9.1}[/tex]

[tex]\\ \sf\longmapsto 0.3m/s[/tex]