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arlik [135]
3 years ago
15

A student of weight 652 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude

of the normal force on the student from the seat is 585 N. (a) What is the magnitude of at the lowest point
Physics
1 answer:
miskamm [114]3 years ago
8 0

Answer:

Explanation:

Given

Weight of person W=652\ N

At highest point Magnitude of the normal force F=585\ N

net force at highest point

F_{net}=W+F_c

where F_c= centripetal force

F_c=\dfrac{mv^2}{r}

F_{net}=F_{n}= Normal Force

585=652+F_c

F_c=-67\ N

Negative sign shows force is in upward direction

At bottom point centripetal force is towards the bottom

F_n=F_c+W

F_n=652+67

F_n=719\ N  

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The Huka Falls on the Waikato River is one of New Zealand's most visited natural tourist attractions. On average, the river has
OverLord2011 [107]

Answer:

(a) V = 0.75 m/s

(b) V = 0.125 m/s

Explanation:

The speed of the flow of the river can be given by following formula:

V = Q/A

V = Q/w d

where,

V = Speed of Flow of River

Q = Volume Flow Rate of River

w = width of river

d = depth of river

A = Area of Cross-Section of River = w d

(a)

Here,

Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s

w = 20 m

d = 20 m

Therefore,

V = (300 m³/s)/(20 m)(20 m)

<u>V = 0.75 m/s</u>

<u></u>

(b)

Here,

Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s

w = 60 m

d = 40 m

Therefore,

V = (300 m³/s)/(60 m)(40 m)

<u>V = 0.125 m/s</u>

8 0
3 years ago
The Olympic record for a marathon is 2.00 hour 9.00 minute 21.0 seconds if the average speed of the runner achieving this record
iren2701 [21]
(2.00 hours) x (3,600 seconds/hour)  =  7,200 seconds

(9.00 minutes) x (60 seconds/minute)  =  540 seconds

The record time = (7,200 + 540 + 21) = 7,761 seconds

Distance  =  (speed) x (time)

                 = (5.436 m/s) x (7,761 sec)  =<span>  42,188.8 meters
________________________________________________
</span>
The official length of the marathon run is  42,195 meters.
If we divide that by the record time in the question, we get

                       5.4368... m/s .

Rounded to the nearest thousandth, that's  5.437 m/s.

If the question had given the speed as  5.437  instead of  5.436 ,
then we would have calculated the distance to be 

            (5.437 m/s) x (7,761 sec)  =<span>  42,196.6 meters,

4.6 meters closer to the official distance than the answer we did get.
</span>
3 0
2 years ago
Your shadow follows you everywhere. How is it formed? Is it always the same size or shape? Do you always have one shadow?
Yanka [14]
Formed from an object blocking the light to a certain area.

Is not always the same size or shape, it varies depending on the position of the light.

You can have more than one shadow of you have more than one light source.
8 0
3 years ago
A 0.4 kg soccer ball approaches a player with a velocity of 18 m/s to the east. The player strikes
Zigmanuir [339]

Answer:

16Newton sec

Explanation:

Ft = m( v + u)

= 0.4( 22 + 18)

= 0.4 x 40

= 16newton sec

7 0
3 years ago
What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?
IgorLugansk [536]

Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

We are asked to calculate the acceleration of the block.

Let the masses of two bodies are denoted as m_{1} \ and\ m_{2}\ respectively

Let\ m_{1} =1 kg\ and\ m_{2} =2 kg

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

                             m_{1} g-T=m_{1} a        [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction [\mu]=0.13

Hence\ the\ frictional\ force\ F=\mu N

                                                     F=\mu m_{2} g

Hence the net force acting on the body having mass 2 kg-

                                  T-\mu m_{2} g=m_{2} a  [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

                           m_{1} g-T=m_{1}a             [1]

                           T-\mu m_{2}g= m_{2} a   [2]

                           ---------------------------------------------------

                           [m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a

                           a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g

                           a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2

                           a=\frac{0.74}{3} *9.8\ m/s^2

                           a=2.417 m/s         [ans]

6 0
3 years ago
Read 2 more answers
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