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Alisiya [41]
4 years ago
13

1. How is acceleration different from velocity?

Physics
1 answer:
Drupady [299]4 years ago
8 0

Answer: The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Whereas, acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities. The orientation of an object's acceleration is given by the orientation of the net force acting on that object.

-Here is a formula for acceleration.

-And due note there are multiple types of velocity, namely starting and final velocity there are multiple formulas. I urge you to learn them!

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What do the movements of stars and galaxies tell astronomers about how the universe formed?
natulia [17]
This could be Hubble's law, or something related to it. I think there's a possibly Doppler RED SHIFT in the optical spectra of stars etc as observed on the earth. It seems that they are accelerating away from the earth, and that the further away they are the faster they are moving.
It seems that this has been connected to the idea of "The Big Bang" theory of the origin of the universe which seems to have superceded Professor Sir Fred Hoye's Steady State theory of the universe.
There's some Special Relativity in this lot, too.
3 0
3 years ago
The length and width of a rectangular room are measured to be 3.92 ± 0.0035 m and 3.15 ± 0.0055 m. In this problem you can appro
Pavel [41]

Answer:

A)A=12.2480\ m^2

B)12.2480\pm 0.1029\ m^2

Explanation:

<u>Given:</u>

Length of the room l= 3.92 ± 0.0035

Width of the room w= 3.15 ± 0.0055

A) Let A be the area of the room

A=l\times w\\A=3.92\times3.15\\A=12.2480\ \rm m^2

B)We will calculate uncertainty in each dimension

%uncertainty in length=\dfrac{0.0035}{3.92}\times 100=0.0892\ %

%uncertainty in width =\dfrac{0.0055}{3.15}\times 100=0.0174%

The uncertainty in area will be sum of uncertainty in length and width

%uncertainty in Area=  %uncertainty in length + %uncertainty in width

%uncertainty in Area=0.0892\ % + 0.0174\ %

%uncertainty in Area=0.0106

Uncertainty in Area=0.0106\times 12.2480=0.1029\ \rm m^2

There Area is12.2480 ± 0.1029\ \rm m^2

7 0
3 years ago
An ideal gas is compressed at constant pressure p=1500 Pa from a volume Vi=0.4 m3 to Vf=0.25 m3. At the same time, heat is trans
Anon25 [30]

To solve this problem we will apply the first law of thermodynamics and we will make a balance between the heat transferred, its internal energy and the total work. Recall that for gases the definition of work can be expressed in terms of its pressure and volume. Let's start

dQ = dU +dW

Here,

dU = Internal Energy

dW = Work

But internal energy is unchanged, then

dQ = dW

dQ = PdV

Where

dV= Change in Volume

P = Pressure

Finally, the expression of the heat transferred can be expressed in terms of pressure and volume, so it would end up becoming

dQ = p(v_i-v_f)

Replacing,

dQ = (1500)(0.4-0.25)

dQ = 225J

Therefore the correct answer is B.

3 0
4 years ago
The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and
ira [324]

Complete question is;

a. Two equal sized and shaped spheres are dropped from a tall building. Sphere 1 is hollow and has a mass of 1.0 kg. Sphere 2 is filled with lead and has a mass of 9.0 kg. If the terminal speed of Sphere 1 is 6.0 m/s, the terminal speed of Sphere 2 will be?

b. The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and 9.0 kg, The terminal speed (in m/s) of Sphere 2 will now be

Answer:

A) V_t = 18 m/s

B) V_t = 10.39 m/s

Explanation:

Formula for terminal speed is given by;

V_t = √(2mg/(DρA))

Where;

m is mass

g is acceleration due to gravity

D is drag coefficient

ρ is density

A is Area of object

A) Now, for sphere 1,we have;

m = 1 kg

V_t = 6 m/s

g = 9.81 m/s²

Now, making D the subject, we have;

D = 2mg/((V_t)²ρA))

D = (2 × 1 × 9.81)/(6² × ρA)

D = 0.545/(ρA)

For sphere 2, we have mass = 9 kg

Thus;

V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρA))]

V_t = 18 m/s

B) We are told that The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1.

Thus;

Area of sphere 2 = 3A

Thus;

V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρ × 3A))]

V_t = 10.39 m/s

5 0
4 years ago
Most energy obtained from water is converted from _____.
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Potential energy behind dams
7 0
3 years ago
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