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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Physics

2 answers:

Natasha_Volkova [10]3 years ago

7 0

Ok srry bout last time but the answer is A) kelvin: time the reason is because everything else is a SI. Hope this helps and TURTLE

Diano4ka-milaya [45]3 years ago

4 0

Im not sure but i think its a or b im pretty sure

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When water waves meet, they can combine to form new waves. In constructive waves, a ________ amplitude wave is formed. In destru

Sever21 [200]

Answer:

In constructive waves, a greater amplitude wave is formed. In destructive waves, a wave with a smaller amplitude is formed. (option A)

Explanation:

Interference is called the superposition or sum of two or more waves. Depending mainly on the wavelengths, amplitudes and the relative distance between them, there are two types of interference: constructive or destructive.

Constructive interference occurs when there are two waves of identical or similar frequency (both have motions equal to an even number of similar wavelengths) and overlap the peak of one with the peak of the other. These effects add together and make a wave of greater amplitude. All of this is possible because the waves were in the same phase in the beginning (in the same position).

Destructive interference occurs in the opposite case to constructive. When the crest of one wave overlaps the valley of the other, they cancel out since they are in different phases when they overlap (they were in different positions). That is, as in the case of constructive waves they were added, in the case of destructive waves they cancel out (subtract).

So, In constructive waves, a greater amplitude wave is formed. In destructive waves, a wave with a smaller amplitude is formed.

3 0

3 years ago

Select the correct answer.

wolverine [178]

The answer is C in this question.

5 0

3 years ago

Read 2 more answers

A piece having a length of 4.0 cm was cut from a much longer, uniform rod. The piece has a volume of 3.0 cm3 and a mass of 24 g.

Gre4nikov [31]

Answer: 96

Explanation:

6 0

3 years ago

the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during

Ivanshal [37]

Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

Mass of the baseball after struck by the bat, M = 900 g

Velocity of the baseball after struck by the bat, v = 47 m/s

According to the conservation of momentum,

$Mv+mu=Mv_1+mv_2$

(900 x 47) + (200 x -30) = (900 x $v_1$ ) + (200 x $v_2$ )

36300 = (900 x $v_1$ ) + (200 x $v_2$ )

$9v_1 + 2v_2 = 363$ ..............(i)

$9v_1 = 363 - 2v_2$

$v_1=\frac{363 - 2v_2}{9}$

The mathematical expression for the conservation of kinetic energy is

$\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2$

$\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2$ ................(ii)

$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$

$21681 = 9v_1^2+2v_2^2$

Substituting (i) in (ii)

$21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2$

$(363-2v_2)^2+18v_2^2=195129$

$(363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0$

$22v_2^2-145v_2-63360=0$

Solving the equation, we get

$v_2=96 \ m/s, -30 \ m/s$

The negative velocity is neglected.

Therefore, substituting 96 m/s for $v_2$ in (i), we get

$v_1=\frac{363-(2 \times 96)}{9}$

= 19

Thus, only impulse of importance is used to find final velocity.

8 0

3 years ago

How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C?

Svet_ta [14]

Answer:

229,098.96 J

Explanation:

mass of water (m) = 456 g = 0.456 kg

initial temperature (T) = 25 degrees

final temperature (t) = - 10 degrees

specific heat of ice = 2090 J/kg

latent heat of fusion =33.5 x 10^(4) J/kg

specific heat of water = 4186 J/kg

for the water to be converted to ice it must undergo three stages:

the water must cool from 25 degrees to 0 degrees, and the heat removed would be Q = m x specific heat of water x change in temp

Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J

the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp

Q = 0.456 x 33.5 x 10^(4) = 152760 J

the water must cool further to -10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp

Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J

The quantity of heat removed from all three stages would be added to get the total heat removed.

Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J

6 0

3 years ago