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masya89 [10]
3 years ago
10

Which statement can be made about waves A and B? Wave A has a lower amplitude than Wave B. Wave A has a higher amplitude than Wa

ve B. Wave A has a lower frequency than Wave B. Wave A has a higher frequency than Wave B.
Physics
2 answers:
gayaneshka [121]3 years ago
8 0

Answer:

b

Explanation:

got it right on edge

givi [52]3 years ago
5 0

Answer:

Wave A has a higher amplitude than Wave B.

Explanation:

please check the attached image for the waves

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pls help asap Create a Punnett square to model asexual reproduction of bacteria with the dominate trait T, recessive trat t. Cre
JulsSmile [24]
Ok you need a punnett square? its gonna be easy, create the square first, Then put the letters in their, When done message me

5 0
4 years ago
At what rate is 444kg object accelerating if a force of 7,700N is applied to it?
vlabodo [156]

Answer:

646,800

Explanation:

If you multiply the 7,700N and the 444kg your answer would be 646,800.

7 0
4 years ago
Which runner finished the 100 m race in the least amount of time?
Nana76 [90]

Answer:

Use the drop-down menus to answer each question.

Which runner finished the 100 m race in the least amount of time?

✔ Ming

Which runner stopped running for a few seconds during the race?

✔ Chloe

At what distance did Anastasia overtake Chloe in the race?

✔ 40 m

8 0
3 years ago
A cylinder with rotational inertia I1=2.0kg·m2 rotates clockwise about a vertical axis through its center with angular speed ω1=
Mrac [35]

Answer:

<em>a) 0.67 rad/sec in the clockwise direction.</em>

<em>b) 98.8% of the kinetic energy is lost.</em>

Explanation:

Let us take clockwise angular speed as +ve

For first cylinder

rotational inertia I = 2.0 kg-m^2

angular speed ω = +5.0 rad/s

For second cylinder

rotational inertia I = 1.0 kg-m^2

angular speed = -8.0 rad/s

The rotational momentum of a rotating body is given as = Iω

where I is the rotational inertia

ω is the angular speed

The rotational momenta of the cylinders are:

for first cylinder = Iω = 2.0 x 5.0 = 10 kg-m^2 rad/s

for second cylinder = Iω = 1.0 x (-8.0) = -8 kg-m^2 rad/s

The total initial angular momentum of this system cylinders before they were coupled together = 10 + (-8) = <em>2 kg-m^2 rad/s</em>

When they are coupled coupled together, their total rotational inertia I_{t} = 1.0 + 2.0 = 3 kg-m^2

Their final angular rotational momentum after coupling = I_{t}w_{f}

where I_{t} is their total rotational inertia

w_{f} = their final angular speed together

Final angular momentum = 3 x w_{f} = 3w_{f}

According to the conservation of angular momentum, the initial rotational momentum must be equal to the final rotational momentum

this means that

2 =  3w_{f}

w_{f} = final total angular speed of the coupled cylinders = 2/3 = <em>0.67 rad/s</em>

From the first statement, <em>the direction is clockwise</em>

b) Rotational kinetic energy = \frac{1}{2} Iw^{2}

where I is the rotational inertia

w is the angular speed

The kinetic energy of the cylinders are:

for first cylinder = \frac{1}{2} Iw^{2} = \frac{1}{2}*2*5^{2} = 25 J

for second cylinder = \frac{1}{2}*1*8^{2} = 32 J

Total initial energy of the system = 25 + 32 = 57 J

The final kinetic energy of the cylinders after coupling = \frac{1}{2}I_{t}w^{2} _{f}

where

where I_{t} is the total rotational inertia of the cylinders

w_{f} is final total angular speed of the coupled cylinders

Final kinetic energy =  \frac{1}{2}*3*0.67^{2} = 0.67 J

kinetic energy lost = 57 - 0.67 = 56.33 J

percentage = 56.33/57 x 100% = <em>98.8%</em>

6 0
3 years ago
A spring-loaded toy gun is used to shoot a ball of mass 1.5 kg straight up in the air, as shown. The spring has spring constant
asambeis [7]

Answer:

Part a)

U = 15.5 J

Part b)

h = 1.05 m

Part c)

v = 4.55 m/s

Explanation:

Part a)

Initial potential energy stored in the spring is given by

U = \frac{1}{2}kx^2

now we have

k = 777 N/m

x = 20.0 cm

now from above formula we will have

U = \frac{1}{2}(777)(0.20)^2

U = 15.5 J

Part b)

By mechanical energy conservation law we can say that initial spring potential energy stored = final gravitational potential energy at the top position

\frac{1}{2}kx^2 = mgh

15.5 = 1.5(9.8)h

h = 1.05 m

So maximum height from gun is 1.05 m

Part c)

For muzzle velocity of the ball we can use the energy conservation again

According to which the final kinetic energy of the ball = initial spring energy stored in it

\frac{1}{2}mv^2 = \frac{1}{2}kx^2

(\frac{1}{2})(1.5 v^2) = 15.5

v = v = 4.55 m/s

4 0
3 years ago
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