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3 years ago

Ahh ahhh ahhh ahhh ahhh ahhh ahhh ahhh ahhh ahha aggg agg

Physics

2 answers:

krek1111 [17]3 years ago

6 0

Agreeable. you have good points there.

timama [110]3 years ago

4 0

shhhh hahahaha ahhhhhhhh ahhhh

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Which statements best describe science? Check all that apply. A)Science uses beliefs and opinions to construct explanations. B)S

Maru [420]

Answer:

Science is supported by facts and processes.

Science involves observation and experimentation.

Science continually changes and is constantly updated.

6 0

3 years ago

Read 2 more answers

You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius

Gennadij [26K]

Answer:

t = 0.0735 m

Explanation:

Angular acceleration of the flywheel is given as

$\alpha = 3 rad/s^2$

now after t = 8 s the speed of the flywheel is given as

$\omega = \alpha t$

$\omega = 3 \times 8$

$\omega = 24 rad/s$

now rotational kinetic energy of the wheel is given as

$K = \frac{1}{2}I\omega^2$

$K = \frac{1}{2}(\frac{1}{2}mR^2)(24^2)$

$800 = \frac{1}{4}m(0.23)^2(24^2)$

$m = 105 kg$

now we have

$m = \rho (\pi R^2) t$

$105 = 8600(\pi \times 0.23^2) t$

$t = 0.0735 m$

4 0

3 years ago

One of the BEST measures of success for businesses and their entrepreneurs is

Sidana [21]

Are there choices or no??

6 0

3 years ago

Read 2 more answers

Two fans are watching a baseball game from different positions. One fan is located directly behind home plate, 18.3 m from the b

EleoNora [17]

Answer:

0.317 s

Explanation:

distance of fan A = 18.3 m

distance of fan B = 127 m

speed of sound (s) = 343 m/s

What is the time difference between hearing the sound at the two locations?

time (T) = distance / speed

time for sound to reach fan A = 18.3 / 343 = 0.053 s

time it takes for sound to reach fan B = 127 / 343 = 0.370 s
time difference = 0.370 - 0.053 = 0.317 s

5 0

3 years ago

A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout

nika2105 [10]

Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
So, we can write the following equation:

$E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)$

As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
So, on the outer surface of the shell there must be a charge that be the difference between them:

$Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C$

Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

$E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C$

As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.

6 0

3 years ago