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Sauron [17]
3 years ago
12

What is the volume of a hollow ball whose other radius is 200m and inner radius of 100m? What is the mass if it is iron?

Mathematics
1 answer:
kati45 [8]3 years ago
5 0

Solution :

Volume of ball = Volume of outer ball - Volume of inner ball

V = \dfrac{4\pi R^3}{3} - \dfrac{4 \pi r^3}{3}\\\\V = \dfrac{4\pi}{3}( R^3 - r^3 )\\\\V = \dfrac{4\pi}{3}( 200^3 - 100^3 )\\\\V = \dfrac{4\pi}{3}(2^3 - 1^3) \times 10^6\\\\V = 2.93 \times 10^7 \ m^3

Now, density of iron is, d = 7300 kg/m³.

So, mass of iron ball is :

m = d\times V\\\\m = 7300\times 2.93 \times 10^7 \ kg\\\\m =  2.14 \times 10^{11}\ kg

Hence, this is the required solution.

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A man hits a baseball when it is 4 ft above the ground with an initial velocity of 120 ft/sec. The ball leaves the bat at a 30 d
Sauron [17]
<span>et us assume that the origin is the floor right below the 30 ft. fence

To work this one out, we'll start with acceleration and integrate our way up to position.

At the time that the player hits the ball, the only force in action is gravity where: a = g (vector)
ax = 0
ay = -g (let's assume that g = 32.8 ft/s^2. If you use a different value for gravity, change the numbers.

To get the velocity of the ball, we integrate the acceleration
vx = v0x = v0cos30 = 103.92
vy = -gt + v0y = -32.8t + v0sin40 = -32.8t + 60

To get the positioning, we integrate the speed.
x = v0cos30t + x0 = 103.92t - 350
y = 1/2*(-32.8)t² + v0sin30t + y0 = -16.4t² + 60t + 4

If the ball clears the fence, it means x = 0, y > 30

x = 0 -> 103.92 t - 350 = 0 -> t = 3.36 seconds

for t = 3.36s,
y = -16.4(3.36)^2 + 60*(3.36) + 4
= 20.45 ft

which is less than 30ft, so it means that the ball will NOT clear the fence.


Just for fun, let's check what the speed should have been :)

x = v0cos30t + x0 = v0cos30t - 350
y = 1/2*(-32.8)t² + v0sin30t + y0 = -16.4t² + v0sin30t + 4

x = 0 -> v0t = 350/cos30
y = 30 ->
-16.4t^2 + v0t(sin30) + 4 = 30
-16.4t^2 + 350sin30/cos30 = 26
t^2 = (26 - 350tan30)/-16.4
t = 3.2s

v0t = 350/cos30 -> v0 = 350/tcos30 = 123.34 ft/s

So he needed to hit the ball at at least 123.34 ft/s to clear the fence.

You're welcome, Thanks please :)
</span>
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