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Ierofanga [76]
1 year ago
12

a large sphere is on a horizontal field on a sunny day. at a certain time the shadow reaches out a distance of 10 m from the poi

nt where the sphere touches the ground. at the same instant a meter stick (held vertically with one end on the ground) casts a shadow of 2 m. what is the radius of the sphere in meters? (assume the sun's rays are parallel and the meter stick is a line segment.)
Physics
1 answer:
Mars2501 [29]1 year ago
5 0

The radius of the sphere in meters is ,r = 10\sqrt{5} -20

Think about the angle the ground and the shadow make. Since the sun's beams are parallel, the angle created by the stick's shadow is also equal. Since the stick is 1 m high and its shadow is 2 m long, we know that the stick's angle is arctan 1/2. Therefore, by thinking of a right-angled triangle,

r/10 = tan [arctan(1/2)] = tan (1/2)

Since, tan (θ/2) = 1-cos(θ) / sin(θ)

we find that,

r/10 = \sqrt{5} -2

Hence, r = 10\sqrt{5} -20

So, the radius of the sphere in meters is ,r = 10\sqrt{5} -20

Learn more about radius (r) of the sphere here;

brainly.com/question/14100787

#SPJ4

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A rectangular block floats in pure water with 0.400 in. above the surface and 1.60 in. below the surface. When placed in an aque
labwork [276]

Answer:

specific gravity = 0.8

specific gravity of  solution  = 2

Explanation:

given data

rectangular block above water  = 0.400 in

rectangular block below water = 1.60 in

material floats below water = 0.800 in

solution

first we get here specific gravity of block  that is

specific gravity = block vol below ÷ total block vol × specific gravity  water   ..............1

put here value we get

specific gravity =  \frac{1.60}{1.60+0.400}  × 1

specific gravity = 0.8

and now we get here specific gravity of  solution  that is express as

specific gravity of  solution  = total block vol ÷ block vol below × specific gravity  block   ........................2

put here value we get

specific gravity of  solution  = \frac{1.60+0.400}{0.800} × 0.8

specific gravity of  solution  = 2

6 0
3 years ago
An electrical motor spins at a constant 1975.0 rpm. If the armature radius is 7.112 cm, what is the acceleration of the edge of
stealth61 [152]

Answer:

Option D is the correct answer.

Explanation:

Since value of angular acceleration is constant, the body has only centripetal acceleration.

Centripetal acceleration

               a=\frac{v^2}{r}=\frac{(r\omega )^2}{r}=r\omega ^2

We have radius = 7.112 cm = 0.07112 m

Frequency, f = 1975 rpm = 32.92 rps

Angular frequency, ω = 2πf = 2 x π x 32.92 = 206.82 rad/s

Substituting in centripetal acceleration equation,

              a=r\omega ^2=0.07112\times 206.82^2=3042.17m/s^2

Option D is the correct answer.                

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Answer:

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Explanation:

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Answer:

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Explanation:

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