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Juli2301 [7.4K]
3 years ago
9

0. What is centripetal accleration?Derive relation for it​

Physics
2 answers:
NISA [10]3 years ago
4 0

Answer:

Centripetal acceleration is defined as the property of the motion of an object, traversing a circular path. Any object that is moving in a circle and has an acceleration vector pointed towards the center of that circle is known as Centripetal acceleration. ... Centripetal means towards the center.

FrozenT [24]3 years ago
4 0
Define as property of motion
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Which elements are alkali metals and which are alkaline earth metals
Irina18 [472]

Answer:

ix chemical elements that comprise Group 2 of the periodic table

Explanation:

Calcium, Magnesium, Barium, Beryllium, Strontium, and Radium are alkaline earth metals.

8 0
3 years ago
Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.9 km/h due east. Runner B is initia
Oksanka [162]

Answer:

B meet A 0.01 km east of flagpole

Explanation:

given data

distance A = 5.7 km  west

velocity V1 = 8.9 km/h

distance B = 4.5 km  east

velocity V2 = 7 km/h

to find out

How far runners from the flagpole, when paths cross

solution

we know A and B are 5.7 + 4.5 = 10.2 km apart

and we consider here B will run distance x km for meet

so time will be for B is

time B = distance / velocity

time B = x / 7     ...................1

and

for A distance for meet = ( 10.2 - x ) km

so time A = distance / velocity

time A = ( 10.2 - x )  / 8.9      .............2

now equating equation 1 and 2

time A = time B

x / 7  =   ( 10.2 - x )  / 8.9

x = 4.490

so distance of B run for meet is 4.490 km

so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km

so B meet A 0.01 km east of flagpole

4 0
3 years ago
A person stands at the base of a hill that is a straight incline making an angle φ with the horizontal. For a given initial spee
Brums [2.3K]

Answer:

θ = sin⁻¹\sqrt{2gd}

Explanation:

From one of the equations of motion, v² = u² + 2as.......... equation 1

Since the object thrown was moving against gravity, then the acceleration, a would change to -g and the initial velocity u would change to V₀ sin θ because the object is travelling at angle of θ to the horizontal. By inputting all these parameter into equation 1, we would arrive at:

v² = (u sin θ)² - 2gd

(u sin θ)² = 2gd

d = (u sin θ)²/2g

sin² θ = 2gd

sin θ = \sqrt{2gd}

θ = sin⁻¹ \sqrt{2gd}

4 0
3 years ago
Explain Thomsons model of an atom<br><br><br>please its aurgent fast​
Aleks04 [339]

Answer:

Thomson's model showed an atom that had a positively charged medium, or space, with negatively charged electrons inside the medium. After its proposal, the model was called a "plum pudding" model because the positive medium was like a pudding, with electrons, or plums, inside.

3 0
2 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
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