When atoms bond with each other, they exchange A.) Protons B) Neutrons C)Electrons

A person began running due east and covered 15 km in 2.0 hr. What is the average velocity of the person?

g100num [7]

Velocity = distance / time

v = (15*1000)m / (2*60*60)s

v=2.08 m / s

6 0

3 years ago

compared to an object that does not roll, but instead slides without friction, should a rolling object be released from the same

Zinaida [17]

Answer:

from a lesser hight

Explanation:

because you need less force

hope its right if it is mark brainlyest ;)

6 0

2 years ago

The following are examples of physical properties except

Ugo [173]

<h2>Answer:</h2><h3>D. ability to react with oxygen</h3><h2>Explanation:</h2>

Im not sure this in your choices but if it is, this is the answer.

I hope I've helped.

6 0

3 years ago

Read 2 more answers

Can someone awnser this

Svetllana [295]

Answer:

The force of gravity exerts a downward force. The floor exerts an upward force. Since these two forces are of equal magnitude and in opposite directions, they balance each other.

6 0

3 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0

3 years ago