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GenaCL600 [577]
2 years ago
9

50 POINTS AND BRAINLEIST A Ferris wheel has a mechanical advantage of ___________________________ one and the trade off is that

it covers a greater distance.
A. greater than
B. less than
C. the same as
D. none of the above
Physics
2 answers:
kiruha [24]2 years ago
5 0

Answer:

C.) The same as one

Explanation:

Because the Ferris wheel has two bars at the side that keep it in place so it doesn't roll away. The Ferris wheel only moves in one place. If the ferris wheel did roll then you would have a greater than one. If it did roll it would be dangerous.

atroni [7]2 years ago
5 0

Answer:

C Same as one and covers a greater distance

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The velocity of a 1.3 kg block sliding down a frictionless inclined plane is found to be 1.26 m/s. 1.10 s later, it has a veloci
nasty-shy [4]

Answer:

\theta = 25.3^\circ

Explanation:

The acceleration of the block can be found by the kinematics equations:

v = v_0 + at\\5.88 = 1.26 + a(1.1)\\a = 4.2~m/s^2

Since the plane is frictionless, the only force acting on the block along the motion of the block is its weight.

F = mg\sin(\theta) = ma\\g\sin(\theta) = a\\(9.8)\sin(\theta) = 4.2\\\theta = 25.3^\circ

7 0
3 years ago
What is the final velocity of a body if it is moving with 13 m/s in 300 seconds and its acceleration is 30 m/s 2.
Nutka1998 [239]

Answer:

9013m/s

Explanation:

acceleration= v- u/ t

=> at = v-u

=> v = at + u

=> v =30*300+13

= 9013m/s

6 0
2 years ago
Why do we feel the force of gravity from the Earth but we don’t feel the force of gravity from the moon?
Zigmanuir [339]

Answer:

The gravitational force on the moon is less than on Earth because the strength of gravity is determined by an object's mass. The bigger the object, the bigger the gravitational force. Gravity is pretty much everywhere. We just feel it in different ways depending on our state of motion.

Explanation:

Hope this helped!!

8 0
2 years ago
Read 2 more answers
Temperature change time base problem. Suppose V = 24 V, I = 0.1 A, for water: mw = 51 gm, cw = 4.18 J/gm ∘K-1, for resistor: mr
nirvana33 [79]

Answer:

t = 444.125 sec

Explanation:

Given data:

V = 24 volt

I  = 0.1 ampere

mass of water mw = 51 gm

cr = 4.18 J/gm degree K^-1

mass of resistor = 8 gm

cr = 3.7 J/gm degree K^-1

we know that power is given as

Power P = VI

But P =E/t

so equating both side we have

\frac{E}{t} = VI

solving for t

t = \frac{E}{VI}

t = \frac{m_w C_w \Delta T}{VI}

t = \frac{51 \times 4.18 \times (5 -0)}{24\times 0.1}

t = 444.125 sec

5 0
3 years ago
Read 2 more answers
Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest
shutvik [7]

This is note the complete question, the complete question is:

One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?

Answer:  3.04 D

Explanation:

when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.

An image of 0.47m from the eye will be ( 47 - 2 )

i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.

Also, the image distance will become negative

gap between lense and eye = 2cm

Therefore;

image distance d₁ = - 45cm = - 0.45m

object distance  d₀ = 21 - 2 = 19cm = 0.19m

P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)

P = 1/f =  5.26315789 - 2.22222222

P = 1/f = 3.04093567 ≈ 3.04 D

5 0
3 years ago
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