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zlopas [31]
3 years ago
9

What is the frequency of an event?

Physics
2 answers:
Delvig [45]3 years ago
5 0

Answer:

Expected Occurrence.

GRADPOINT

konstantin123 [22]3 years ago
4 0

"Frequency" just means "often-ness" ... how often something happens.
It's always expressed as

        <em>(number of happenings) / (some period of time) .</em>


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A ball thrown vertically upward is caught by the thrower after 2.93 s. Find the initial velocity of the ball. The acceleration o
almond37 [142]

Answer:

The initial velocity of the ball is 28.714 m/s

Explanation:

Given;

time of flight of the ball, t = 2.93 s

acceleration due to gravity, g = 9.8 m/s²

initial velocity of the ball, u = ?

The initial velocity of the ball is given by;

v = u + (-g)t

where;

v is the final speed of the ball at the given time, = 0

g is negative because of upward motion

0 = u -gt

u = gt

u = (9.8 x 2.93)

u = 28.714 m/s

Therefore, the initial velocity of the ball is 28.714 m/s

7 0
3 years ago
which of the following laboratory tool would be most appropiate for measuring the approximate volume of a liquid?
SIZIF [17.4K]
Beaker would be most appropriate for measuring the approximate volume of a liquid.
7 0
2 years ago
A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio
natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}

For the maximum velocity of the ball at the top of the vertical circular motion,

v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}

where g is the acceleration due to gravity,

Substituting the known values,

\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

8 0
1 year ago
Convert 200in/10s into m/s (1m = 39.37in)
stiv31 [10]
5.08




hope you got it right !
7 0
3 years ago
2 kilogram bowling ball sits on top of a building that is 40 m the object has k e or g p e or both be sure to include the correc
Licemer1 [7]
GPE=mass*GFS*height
2kg*9.8N/Kg*40m -I've used the Grabitational field strength of the earth's
surface

=784J
7 0
3 years ago
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