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Westkost [7]
3 years ago
9

can y'all please help me I really need you guys or just give me an idea or wrote the first paragraph please​

Physics
1 answer:
elixir [45]3 years ago
7 0
Write about some struggles you’ve faced and what they can do to avoid those struggles. Maybe also what you personally learned from the struggle too.
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A regulation basketball has a 15 cm diameter and may be appropriated as a thin spherical shell.
REY [17]

Solution :

From the given data,

For the spherical shell is $\frac{k^2}{R^2}= \frac{2}{3}$

where x is the radius of gyration and the acceleration of a rolling body on an inclined plane = a

Therefore,

$a =\frac{g \sin \theta}{1+ \frac{k^2}{R^2}} $

  $= \frac{g \sin \theta}{1 +\frac{2}{3}}$

  $= \frac{3}{5} g \sin \theta$

  = 0.6 x 9.81 x sin ( 29.7)

  $= 2.913 \ m/s^2$

$h = \frac{1}{2} a(\Delta t)^2$

$\Delta t = \sqrt{\frac{2h}{a}}$

$\Delta t = \sqrt{\frac{2 \times 2.9}{2.913}}$

Δt = 1.411 s

6 0
3 years ago
Is this answer correct?
alisha [4.7K]
Yes the answer is correct
4 0
3 years ago
A string is wrapped around a pulley with a radius of 2.0 cm. The pulley is initially at rest. A constant force of 50 N is applie
Ray Of Light [21]

Answer:

0.20kg-m^2

Explanation:

Let the linear velocity of the rope(=of pulley) is v m/s

Using kinematic equation

=> v = u + at

=>v = 0 + 4.9a

=>v = 4.9a ------------ eq1

By v^2 = u^2 + 2as

=>v^2 = 0 + 2 x v/4.9 x 1.2

=>4.9v^2 - 2.4v = 0

=>v(4.9v - 2.4) = 0

=>v = 2.4/4.9 = 0.49 m/s

Thus by v = r x omega

=>omega = v/r = 0.49/0.02 = 24.49 rad/sec

BY W = F x s = 50 x 1.2 = 60 J

=>KE(rotational) = W = 1/2 x I x omega^2

=>60 = 1/2 x I x (24.49)^2

=>I = 0.20 kg-m^2

5 0
3 years ago
What drives plate tectonics in the earths interior
xxTIMURxx [149]
The convection going on with the magma in the asthenosphere
4 0
3 years ago
Read 2 more answers
What is the most likely outcome of increasing the number of slits per unit distance on a diffraction grating?1) lines become nar
Nina [5.8K]

It's important to know that diffraction gratings can be identified by the number of lines they have per centimeter. Often, more lines per centimeter is more useful because the images separation is greater when this happens. That is, the distance between lines increases.

<h2>Therefore, the answer is 2.</h2>
8 0
2 years ago
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