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Ymorist [56]
3 years ago
9

A 6 cm diameter drill bit is used to drill a cylindrical hole through the middle of a sphere of radius 5 cm. What is the volume

of the resulting object?
Physics
1 answer:
ira [324]3 years ago
3 0

Answer:

2.4086 * 10^(-4)  m^3

Explanation:

Volume of sphere = 4/3 * pi * r1^3

Volume of cylinder = pi*r2^2*h

r1 = 0.05 m

r2 = 0.03 m

Hence,

Resulting Volume = Volume of sphere - Volume of cylinder

= \frac{4}{3}*pi*(0.05)^3 - pi*(0.03)^2*(0.1)\\= 5.236 * 10^(-4)  - 2.8274*10^(-4) \\= 2.4086 * 10^(-4) m^3

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A mass m at the end of a spring vibrates with a frequency
Wittaler [7]

Answer:

m = 0.59 kg.

Explanation:

First, we need to find the relation between the frequency and mass on a spring.

The Hooke's law states that

F = -kx

And Newton's Second Law also states that

F = ma = m\frac{d^2x}{dt^2}

Combining two equations yields

a = -\frac{k}{m}x

The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.

\omega = \sqrt{\frac{k}{m}}

And given that ω = 2πf

the relation between frequency and mass becomes

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}.

Let's apply this to the variables in the question.

0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg

6 0
3 years ago
A man of weight Wman is standing on the second floor and is pulling on a rope to lift a box of weight Wbox from the floor below.
slavikrds [6]

Answer:

Explanation:

See the attached figure . See the forces acting on man pulling up the box .

Man is stationary so net force acting on man is zero .

T + R = Wman

R is the reaction force of the ground of second floor  .

R = Wman - T

3 0
3 years ago
I NEED HELP RIGHT NOW PLEASE I will  30pts and A brainly to anyone who helps me with questions 2-13!!!!!!!
neonofarm [45]
True True False True False False True I hope I helped on the first few
3 0
3 years ago
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Find the acceleration of a car with the mass of 1,200 kg and a force of
Mashutka [201]

Answer:9.17 m/s^2

Explanation:

mass=1200kg

Force=11 x 10^3 N

Acceleration=force ➗ mass

Acceleration=11 x 10^3 ➗ 1200

Acceleration=9.17

Acceleration=9.17 m/s^2

3 0
3 years ago
The drawing shows a large cube (mass = 28.6 kg) being accelerated across a horizontal frictionless surface by a horizontal force
MrRissso [65]

Answer:

P= 454.11 N

Explanation:

Since P is the only horizontal force acting on the system, it can be defined as the product of the acceleration by the total mass of the system (both cubes).

P= (M+m)*a\\a = \frac{P}{28.6 +4.3}\\a = \frac{P}{32.9}

The friction force between both cubes (F) is defined as the normal force acting on the smaller cube multiplied by the coefficient of static friction. Since both cubes are subject to the same acceleration:

F = m * a*\mu \\F= 4.3*0.710*\frac{P}{32.9}\\F=3.053*\frac{P}{32.9}

In order for the small cube to not slide down, the friction force must equal the weight of the small cube:

3.053*\frac{P}{32.9} = 4.3 * g\\\\P = \frac{4.3*9.8*32.9}{3.053} \\P= 454.11 N

The smallest magnitude that P can have in order to keep the small cube from sliding downward is 454.11 N

8 0
3 years ago
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