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Anastasy [175]
3 years ago
12

In a sound wave, the maximum displacement of the air molecule from it's undisturbed position is called _______________.

Physics
1 answer:
professor190 [17]3 years ago
6 0
It's called amplitude...
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find the mass of a ball on a roof 30 meters high, if the ball's gravitational potential energy is 58.8 joules
Sidana [21]
We are given the gravitational potential energy and the height of the ball and is asked in the problem to determine the mass of the ball. the formula to be followed is PE = mgh where g is the gravitational acceleration equal to 9.81 m/s^2. substituting, 58.8 J = m*9.8 m/s^2 * 30 m; m = 0.2 kg.
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2 years ago
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which statement best describes how changes behave a.opposites charges repel against each other and like charges attract to each
Anon25 [30]

I think it is B hope this helps

God Bless

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2 years ago
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An electron of mass 9.11×10−31 kgkg leaves one end of a TV picture tube with zero initial speed and travels in a straight line t
ki77a [65]

Explanation:

We will use the equations of constant acceleration to find out a_{x} and time t.

As we know that the initial speed is zero. So

(a)  

v_{0x} = 0

x - x_{o} = 1.25×10^{-2}m

v_{x} = 3.3×10^{6}m/s

v^{2} _{x} =  v^{2} _{x_{o} } + 2a_{x} (x - x_{o} )

a_{x} = \frac{v^{2} _{x} - v^{2} _{ox} }{2(x - x_{o}) }

   = \frac{(3.3 * 10^{6})^{2}  - 0 }{2(1.25 * 10^{-2}) }

   = 4.356×10^{14} m/s²

(b)

v_{x} = v_{ox} + a_{x}t

t = v_{x} - vo_{x}/a_{x}

t = \frac{3.00 * 10^{6} }{4.356*10^{14} } = 6.8870×10^{-9}s

(c)

ΣF_{x} = ma_{x}

       = (9.11×10^{-31})(4.356×10^{14}m/s²)

       = 3.968×10^{-16} N

6 0
2 years ago
A reference point allows you to determine the _____________ of an object.
marin [14]

Answer:

a reference point allows you to determine the motion of an object :)

example: when your in a car and u look out the window and your moving fast than the car next to you. It seems that way because the car next to you is actually moving much slower.

8 0
2 years ago
A 0.0750kg teddy bear is dropped from a deck that is 3.00m above the ground. What will be the velocity of the teddy bear as it s
FrozenT [24]

The velocity of the teddy bear as it strikes the ground is 7.67 m/s.

<h3>Velocity of the teddy when it strikes the ground</h3>

The velocity of the teddy when it strikes the ground is calculated from principle of conservation of energy as shown below.

K.E(bottom) = P.E(top)

¹/₂mv² = mgh

v² = 2gh

v = √2gh

where;

  • h is height of fall of the teddy
  • g is acceleration due to gravity

v = √(2 x 9.8 x 3)

v = 7.67 m/s

Thus, the velocity of the teddy bear as it strikes the ground is 7.67 m/s.

Learn more about conservation of energy here: brainly.com/question/166559

#SPJ1

4 0
1 year ago
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