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DiKsa [7]
3 years ago
12

a tire is rolling along a road, without slipping with a velocity v. a piece of tape is attached to the tire. When the tape is op

posite the road (at the top of the tire), its velocity with respect to the road is
Physics
1 answer:
Kryger [21]3 years ago
4 0

Answer:

The right solution will be the "2v".

Explanation:

For something like an object underneath pure rolling the speed at any point is calculated by:

⇒  v_{rolling}=v_{translational}+v_{rotational}

Although the angular velocity was indeed closely linked to either the transnational velocity throughout particular instance of pure rolling as:

⇒  \omega=\frac{v_{translational}}{r}

Significant meaning is obtained, as speeds are in the same direction. Therefore the speed of rotation becomes supplied by:

⇒  v_{rotational}=\omega \times r

On substituting the estimated values, we get

⇒                   =\frac{v_{translational}}{r} \times r

⇒                   =v_{translational}

So that the velocity will be:

⇒  v_{rolling}=v+v

⇒              =2v

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A 0.877 mol sample of N2(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the
MissTica

Answer : The value of q\text{ and }\Delta U is 286.2 J and 286.2 J respectively.

Explanation : Given,

Moles of sample = 0.877 mol

Change in temperature = 15.7 K

First we have to calculate the heat absorbed by the system.

Formula used :

q=n\times c_v\times \Delta T

where,

q = heat absorbed by the system = ?

n = moles of sample = 0.877 mol

\Delta T = Change in temperature = 15.7 K

c_v = heat capacity at constant volume of N_2 (diatomic molecule) = \frac{5}{2}R

R = gas constant = 8.314 J/mol.K

Now put all the given value in the above formula, we get:

q=0.877mol\times \frac{5}{2}\times 8.314J/mol.K\times 15.7K

q=286.2J

Now we have to calculate the change in internal energy of the system.

\Delta U=q+w

As we know that, work done is zero at constant volume. So,

\Delta U=q=286.2J

Therefore, the value of q\text{ and }\Delta U is 286.2 J and 286.2 J respectively.

8 0
3 years ago
Read 2 more answers
Ball A was carried to the top of a hill in a straight line, while ball B was carried in a longer, zigzag path. At the top of the
marissa [1.9K]

Answer: equal

Explanation:

Don’t know why honestly

5 0
3 years ago
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Lauren throws her first pitch of the season for her school's softball team. If the ball travels 80 feet and takes 1.58 seconds t
kow [346]

Answer:

<em><u>5</u></em><em><u>0</u></em><em><u>.</u></em><em><u>6</u></em><em><u>3</u></em><em><u> </u></em><em><u>f</u></em><em><u>t</u></em><em><u>/</u></em><em><u>s</u></em><em><u> </u></em><em><u>(</u></em><em><u>2</u></em><em><u>d</u></em><em><u>p</u></em><em><u>)</u></em>

Explanation:

Speed = Distance/Time

80/1.58 = 50.63291139

= <u>50.63</u><u> </u><u>f</u><u>t</u><u>/</u><u>s</u> (2dp)

6 0
3 years ago
Charge A and charge B are 3.00 m apart, and charge A is +1.44 C and charge B is +3.10 C. Charge C is located between them at a c
Sidana [21]

Answer:

the distance from charge A to C is r₁₃= 1.216 m

Explanation:

following Coulomb's law , the force exerted by 2 point charges between themselves is:

F= k*q₁*q₂/r₁₂² , where q is charge , r is distance and 1 and 2 represents the charge A and charge B respectively , k=constant

since C ( denoted as 3) is at equilibrium

F₁₃=F₂₃

k*q₁*q₃/r₁₃²=k*q₂*q₃/r₂₃²

q₁/r₁₃²=q₂/r₂₃²

r₁₃²/q₁=r₂₃²/q₂

r₂₃=r₁₃*√(q₂/q₁)

since C is at rest and is co linear with A and B ( otherwise it would receive a net force in either vertical or horizontal direction) , we have

r₁₃+r₂₃=d=r₁₂

r₁₃+r₁₃*√(q₂/q₁)=d

r₁₃*(1+√(q₂/q₁))=d

r₁₃=d/(1+√(q₂/q₁))

replacing values

r₁₃=d/(1+√(q₂/q₁)) = 3.00 m/(1+√(3.10 C/1.44 C)) = 1.216 m

thus the distance from charge A to C is r₁₃= 1.216 m

7 0
3 years ago
According to the definition of mechanical work, pushing on a rock accomplishes no work unless there is
LekaFEV [45]
There must be movement in the same direction as the force put on the object. Hope this helps!
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