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kkurt [141]
3 years ago
13

The magnitude of the Normal Force on a

Physics
1 answer:
lbvjy [14]3 years ago
7 0

Answer:

5. Is greater than mg, always

Explanation:

If the cone has an inclination of angle β, the sum of forces will be:

x-axis (centripetal axis):

N*sin β = m*ax  where ax is the centripetal acceleration

y-axis:

N*cos β - m*g = m*ay   where ay is the vertical acceleration. If the block starts falling down, ay will be negative. If the block starts sliding up, ay will be positive. If the block does not move up nor down, ay=0.

Solving for N:

N = \frac{m*g + m*ay}{cos \beta }

If ay is positive or zero, N will be greater than mg. If ay is negative, N will be less than mg.

If the block is sliding along a horizontal circular path (not up, nor down), ay = 0, so N will always be greater than mg.

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Change the following sentences into Passive Voice: (5M)
geniusboy [140]

Answer:

passive voice

many messengers all over the world was sent by emperor Ashoka to preach Buddhism.

Which pen was liked by you?

Has your passport size photo been taken for the application form?

A beautiful bicycle was given to me on my birthday by my father.

The plants is being watered by the gardener.

Paul said that he will never leave you and he will always be with you.

7 0
3 years ago
In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo
Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of 32^{\circ} (when light enters from the air.)

Explanation:

Let v_{1} denote the speed of light in the first medium. Let v_{\text{air}} denote the speed of light in the air. Assume that the light entered the boundary at an angle of \theta_{1} to the normal and exited with an angle of \theta_{\text{air}}. By Snell's Law, the sine of \theta_{1}\! and \theta_{\text{air}}\! would be proportional to the speed of light in the corresponding medium. In other words:

\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}.

When light enters a boundary at the critical angle \theta_{c}, total internal reflection would happen. It would appear as if the angle of refraction is now 90^{\circ}. (in this case, \theta_{\text{air}} = 90^{\circ}.)

Substitute this value into the Snell's Law equation:

\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}.

Rearrange to obtain an expression for the speed of light in the first medium:

v_{1} = v_{\text{air}} \cdot \sin(\theta_{1}).

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For 0 < \theta < 90^{\circ}, \sin(\theta) is monotonically increasing with respect to \theta. In other words, for \!\theta in that range, the value of \sin(\theta)\! increases as the value of \theta\! increases.

Therefore, compared to the medium in this question with \theta_{c} = 27^{\circ}, the medium with the larger critical angle \theta_{c} = 32^{\circ} would have a larger \sin(\theta_{c}). such that light would travel faster in that medium.

4 0
3 years ago
Helppppppppp plzzzzzz
seropon [69]
I think number 1 is incorrect I believe that answer is D. Number 6 I believe would be B. The rest seem to be correct.
4 0
3 years ago
Read 2 more answers
A rock is dropped straight down from roof. It hits the ground with a velocity of 28.2 m/sec. How high was the roof, and how long
notsponge [240]

Answer:

18 secs

Explanation:

6 0
3 years ago
A 14,700 N car is traveling at 25 m/s. The brakes are applied suddenly, and the car slides to a stop. The average braking force
Vlada [557]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

vo = 25 m/sec 
<span>vf = 0 m/sec </span>
<span>Fμ = 7100 N (Force due to friction) </span>
<span>Fg = 14700 N </span>
<span>With the force due to gravity, you can find the mass of the car: </span>
<span>F = ma </span>
<span>14700 N = m (9.8 m/sec²) </span>
<span>m = 1500 kg </span>
<span>Now, we can use the equation again to find the deacceleration due to friction: </span>
<span>F = ma </span>
<span>7100 N = (1500 kg) a </span>
<span>a = 4.73333333333 m/sec² </span>
<span>And now, we can use a velocity formula to find the distance traveled: </span>
<span>vf² = vo² + 2a∆d </span>
<span>0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d </span>
<span>0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d </span>
<span>-625 m²/sec² = (-9.466666666667 m/sec²) ∆d </span>
<span>∆d = 66.0211267605634 m </span>
<span>∆d = 66.02 m</span>
7 0
4 years ago
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