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1 year ago

A figure skater glides along a circular path of radius 6.70 m. If she coasts around one half of the circle, find the following.

Physics

1 answer:

adoni [48]1 year ago

8 0

(a) Her distance from the starting location is 21.05 m.

(b) The length of the path she skated is 21.05 m.

<h3>Distance of the skater from the starting position</h3>

The distance around a complete circular path is calculated as 2πr.

The distance for a half circle is calculated as ¹/₂ x 2πr = πr

Distance from the starting location = π x 6.7 m = 21.05 m

The length of the path she skated is the same as her distance from the starting location = 21.05 m.

Learn more about distance round a circle here: brainly.com/question/3100527

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A butcher grinds 5 and 3/4 lb of meat then sells it for 2 and 2/3 pounds to the customer what is the maximum amount me that the

KiRa [710]

Answer:

The maximum amount of meat that the butcher can sell is $3\frac{1}{12}\:lb$

Explanation:

The maximum amount can be found by taking the difference of mixed numbers.

$5\frac{3}{4}-2\frac{2}{3}\\\\\mathrm{Subtract\:the\:numbers:}\:5-2=3\\\\\mathrm{Combine\:fractions:\:}\frac{3}{4}-\frac{2}{3}=\frac{1}{12}\\\\=3\frac{1}{12}\\$

Best Regards!

8 0

2 years ago

Packages having a mass of 6 kgkg slide down a smooth chute and land horizontally with a speed of 3 m/sm/s on the surface of a co

nalin [4]

Answer:

t = 1.02 s

Explanation:

The computation of the time required is shown below:

The package speed for belt is

= 3 - 1

= 2 m/s

Moreover, the decelerative force would be acted on the block i.e u.m.g

So, the decelerative produced

= 0.2 × 9.81

= 1.962 m/s^2

And, final velocity = 0

v = u - at

here

V = 0 = final velocity

u = 2 m/s

so,

0 = 2 - 1.962 × t

t = 1.02 s

5 0

2 years ago

suppose that you look into a photometer's eyepiece and the fluorescent disks appear to be equal in intensity. If the distance be

d1i1m1o1n [39]

Use the Inverse square law, Intensity (I) of a light is inversely proportional to the square of the distance(d).

I=1/(d*d)

Let Intensity for lamp 1 is L1 distance be D1 so on, L2 D2 for Intensity for lamp 2 and its distance.

L1/L2=(D2*D2)/(D1*D1)

L1/15=(200*200)/(400*400)
L1=15*0.25
L1=3.75 <span>candela</span>

7 0

2 years ago

A caterpillar crawls 20 feet in 5 minutes. What is the caterpillars crawling speed?

Pavel [41]

The equation is
s= d/t

In this case you would have to write it out as:
s= 20/5
Speed = 4

6 0

3 years ago

A 5.00 kg crate is on a 21.0 degree hill. Using X-Y axes tilted down the plane, what is the y-component of the normal force?

Rama09 [41]

Answer:

The y-component of the normal force is 45.74 N.

Explanation:

Given that,

Mass of the crate, m = 5 kg

Angle with hill, $\theta=21^{\circ}$

We need to find the y component of the normal force. We know that the y component of the normal force is given by :

$F_y=F\ \cos\theta\\\\F_y=mg\ \cos\theta\\\\F_y=5\times 9.8\ \cos(21)\\\\F_y=45.74\ N$

So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.

7 0

2 years ago