Ask question

Ask question

All categories

julia-pushkina [17]

3 years ago

14

Chloe is playing the flute some distance away from a crowd. If the atmospheric temperature is 15°C, what is the speed of the sou

nd traveling from the flute?
A.) 340 meters/second

B.) 330 meters/second

C.) 320 meters/second

D.) 310 meters/second

E.) 300 meters/second

1 answer:

borishaifa [10]3 years ago

7 0

I believe the answer to your question is A. 340 meters/second hope i helped

You might be interested in

Samuel tripped while playing basketball and skinned his knee on the concrete. Many of the skin cells on that knee were killed or

kvv77 [185]

The answer is B. Old cells need to create new cells to replenish the ones that were lost.

3 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

Zina [86]

Answer:

charge C = greatest net force

charge B = the smallest net force

ratio = 9 : 1

Explanation:

we know that in Electrostatic Forces, when 2 charges are at same sign then they repel each other and if they are different signed charges then they attract each other

so as per Coulomb's formula of Electrostatic Forces

F = $\frac{k\ q_1\ q_2}{r^2}$ .....................1

and here k is 9 × $10^9$ N.m²/c² and we consider each charge at distance d

so two charge force at A to B is

F1 = $\frac{k\ q^2}{d^2}$

and force between charges at A to C, at 2d distance

F1 = $\frac{k\ q^2}{(2d)^2}$ = $\frac{k\ q^2}{4d^2}$

force between charges at A to D, 3d distance

F1 = $\frac{k\ q^2}{(3d)^2}$ = $\frac{k\ q^2}{9d^2}$

so

Charge a It receives force to the left from b and c and to the right from d

so at a will be

F(a) = -F1 - F2 + F3 ....................2

put here value

F(a) = $-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

solve it

F(a) = $\frac{k\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

F(a) = $-\frac{41}{36}\ F1$ = 1.13 F1

and

Charge b It receives force to the right from a and d and to the left from c

F(b) = F1 - F1 + F2 ....................3

F(b) = $\frac{k\ q^2}{d^2}-\frac{k\ q^2}{d^2}+\frac{k\ q^2}{4d^2}$

F(b) = $\frac{1}{4} \ F1$ = 0.25 F1

and

Charge c It receives forces to the right from all charges.

F(c) = F2 + F 1 + F 1 ....................4

F(c) = $\frac{k\ q^2}{4d^2}+\frac{k\ q^2}{d^2}+\frac{k\ q^2}{d^2}$

F(c) = $\frac{9}{4} \ F1$ = 2.25 F1

and

Charge d It receives forces to the left from all charges

F(d) = - F3 - F2 -F 1 ....................5

F(d) = $-\frac{k\ q^2}{9d^2}-\frac{k\ q^2}{4d^2}-\frac{k\ q^2}{d^2}$

so

F(d) = $-\frac{49}{36} \ F1$ = 1.36 F1

and

now we get here ratio of the greatest to the smallest net force that is

ratio = $\frac{2.25}{0.25}$

ratio = 9 : 1

5 0

3 years ago

Which best describes the second law of thermodynamics? energy is not created nor destroyed, but it can change into matter. energ

otez555 [7]

<span>the answer is energy is not created nor destroyed, but it can change from one energy form to another.</span>

3 0

3 years ago

Read 2 more answers

The two forces acting on a skydiver are __________ and air _____________.

Stolb23 [73]

Answer:

Gravity and

Air resistance

Explanation:

The two forces acting on a skydiver are gravitational force and air resistance.

Gravitational force is a force that tends to pull all massive bodies towards the center of the earth. It works on all bodies that has mass. The larger or bigger the mass, the more the pull of gravity on the body.

Air resistance is the drag of air on a body as it passes to it. It is resisting force.

When a sky diver jumps out of a plane, he/she encounters both gravity and air resistance.
It soon balances both force and attain terminal velocity.
Air resistance is a frictional force that opposes motion.
This frictional force pushes in the opposite direction of motion
Motion direction is downward due to the celerity caused by gravity.

3 0

3 years ago

Read 2 more answers

A 0.144 kg baseball is moving towards home plate with a speed of 43.0 m/s when it is bunted. the bat exerts an average force of

Ganezh [65]

As per impulse momentum theorem we know that

$F\Delta t = m(v_f - v_i)$

now here we will have

$F = 6.50 \times 10^3 N$

t = 1.30 ms

m = 0.144 kg

$v_i = -43 m/s$

now we need to find final speed using above formula

$(6.50 \times 10^3)(1.30 \times 10^{-3}) = 0.144 ( v_f - (-43))$

$v_f = 15.7 m/s$

so final speed is given as above

6 0

3 years ago