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3 years ago

Write a use of reverberation

Physics

1 answer:

MakcuM [25]3 years ago

3 0

Hello there!

definition of reverberation: "Prolongation of a sound; resonance."

reverberation in a sentence: The school's special effects team added reverberation to the actress's monologue to make her speech more dramatic and the school play a little more interesting.

I hope this helps, have a great day! ^-^

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A motorcycle has a speed of 30 m / s. After braking, it decelerates with constant deceleration A = -3.0 m / s ^ 2

marin [14]

$a = \displaystyle\frac{v-u}{t}$

$-3 = \displaystyle\frac{v - 30}{3}$

$v -30 = -9$

$v = 21 m/s$

4 0

3 years ago

Read 2 more answers

A 2.964 kilogram truck strikes a fence at 7.00 meters/second and comes to

Makovka662 [10]

10.92N

Explanation:

Given parameters:

Mass of truck = 2.964kg

Velocity of truck = 7m/s

Time taken = 1.9s

Unknown:

Average force on the car = ?

Solution:

According to newton's third law of motion "action and reaction are equal and opposite".

The force with which the truck struck the fence is the same as the force the fence acted on the truck with but in another direction.

From newton's second law:

Force = mass x acceleration

We know that acceleration is the change in velocity with time;

acceleration = $\frac{change in velocity }{time }$

Force = mass x $\frac{change in velocity }{time }$

Force = $\frac{2.964 x 7}{1.9}$ = 10.92N

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

3 0

3 years ago

To what potential should you charge a 3.0 μf capacitor to store 1.0 j of energy?

Nimfa-mama [501]

The energy stored in a capacitor is given by:
$U= \frac{1}{2}CV^2$
where
U is the energy
C is the capacitance
V is the potential difference

The capacitor in this problem has capacitance
$C=3.0 \mu F = 3.0 \cdot 10^{-6} F$
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
$V= \sqrt{ \frac{2U}{C} }= \sqrt{ \frac{2 \cdot 1.0 J}{3.0 \cdot 10^{-6}F} }=816 V$

8 0

3 years ago

Two point charges each have a value of 3.0 c and are separated by a distance of 4.0 m. what is the electric field at a point mid

swat32

Place a test charge in the middle. It is 2cm away from each charge.
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point.
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out.
THIS IS A TRICK QUESTION.
THe electric field exactly midway between them = 0/Q = 0.
But if the point moves even slightly you need the following formula
F= (1/4Piε)(Q1Q2/D^2)
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)

4 0

3 years ago

You push a table 8 meters for 16 minuutes and do 6720 j of work how much power do ypu use

Yuri [45]

P=W/t

P=Power
W=Work
t=Time

Convert 16 minutes in seconds:
16 mins = 960 secs

P=6720/960=7.23 W [Watt]

3 0

3 years ago