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kotykmax [81]
3 years ago
11

A car is sitting still. It accelerates to a constant speed then it decelerates again to zero speed. While the car is acceleratin

g how do the directions of the angular acceleration and angular velocity of one of the wheels compare
Physics
1 answer:
Kobotan [32]3 years ago
5 0

Answer:

in the acceleration process the quantity α and w must increase

the deceleration process the alpha quantity must constant  a direction opposite to the angular velocity

Explanation:

Acceleration and angular velocity are related to linear

           v = w xr

            a = αx r

The bold letters indicate vectors and the cross is a vector product, therefore if

we can see that the relationship between linear and angular variables is direct

therefore in the acceleration process the quantity α and w must increase as well as their linear counterparts

in the deceleration process the alpha quantity must constant as the linear acceleration and must have a direction opposite to the angular velocity

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Uan created a chart to help him study for a test.
yuradex [85]

Answer:

Y: transparent objects i think

6 0
3 years ago
Read 2 more answers
Find the currents flowing in the circuit in the figure below. (Assume the resistances are
Nitella [24]

By Kirchoff's law in left side loop

E_1 + E_2 = {r_1 + R_1 + R_4)I_1 + (R_2 + r_2) I_3

similarly kirchoff's law in right side loop

E_3 - E_4 - E_2 = (r_3 + r_4 + R_3)I_2 - (R_2 + r_2)I_3

also by junction law we know that

I_1 = I_2 + I_3

now by plug in all values we have

18 + 3 = (0.5 + 8 + 15)I_1 + (10 + 0.25)I_3

21 = 23.5I_1 + 10.25I_3

12 - 24 - 3 = (0.75 + 0.25 + 12)I_2 - (10+ 0.25)I_3

-15 = 13I_2 - 10.25I_3

So by solving above equations we have

I_1 = 0.492 A

I_2 = -0.428 A

I_3 = 0.920 A

6 0
4 years ago
a banana boat accelerates from 4.167 m/s ar 2.00m/s^2.How far has it traveled when it reaches 8.333 m/s
Nataly_w [17]

Answer: The distance that we'll be travelled is d=16.35m

Explanation: The main idea here is to use the equation of motion. We are given the acceleration of 2m/s^2,the initial and final velocies. The formula to use is v^2=u^2+2as. Now we have to substitute the values.

2^2= 8.33^2+2(2)d

d= 16.35m

8 0
4 years ago
Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles
Anna007 [38]

Question

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles  travelled in a straight line and some were deflected at different angles.

Which statement best describes what Rutherford concluded from the motion of the particles?

A) Some particles travelled through empty spaces between atoms and some particles were deflected by electrons.

B) Some particles travelled through empty parts of the atom and some particles were deflected by electrons.

C) Some particles travelled through empty spaces between atoms and some particles were deflected by small areas of high-density positive charge in atoms.

D) Some particles travelled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.    

Answer:

 

The right answer is C)    

Explanation:

In the experiment described above, a piece of gold foil was hit with alpha particles, which have a positive charge. Alpha particles <em>α</em> were used because, if the nucleus was positive, then it would deflect the positive particles. The principles of physics posit that electric charges of the same orientation repel.  

So most as expected some of the alpha particles went right through meaning that the gold atoms comprised mostly empty space except the areas that were with a dense population of positive charges. This area became known as the "nucleus".  

Due to the presence of the positive charges in the nucleus, some particles had their paths bent at large angles others were deflected backwards.

Cheers!

6 0
3 years ago
Read 2 more answers
A skier starts from rest at the top of a hill that is inclined 10.5° with respect to the horizontal. The hillside is 200 m long,
Svetlanka [38]

Answer:

d) 289.31 m

Explanation:

Energy provided by potential energy = mgh = m x 9.8x 200 sin10.5 = 357.18m

Energy used by friction = μmgcos 10.5 x 200 = .075 x m x 9.8 x cos 10.5 x200 = 144.54 m .

Energy used by friction on plain surface = μmg x d.( dis distance covered on plain ) =.075x m x 9.8 xd = .735 m d

To equate

357.18 m -144.54 m = .735 m d

d = 289.31 m .

4 0
3 years ago
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