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masya89 [10]
3 years ago
12

Ike is at the beach watching the waves in the ocean. Ike notices that some of the waves are short. Other waves are very tall and

come up high above the water. Which of these is always true about two waves that are different heights? A. They are made of different materials. B. They are traveling at different speeds. C. They have different wavelengths. D. They have different amplitudes.
Physics
1 answer:
aniked [119]3 years ago
3 0

Ike is at the beach watching the waves in the ocean. Ike notices that some of the waves are short. Other waves are very tall and come up high above the water. Two waves that are different heights because They have different amplitudes.

Answer: Option (D) is correct

Explanation:

The different heights of the waves are due to their different amplitudes. The Amplitude of a particular wave depends upon the amount of energy being carried by waves. It the waves carry more energy than their amplitude will be higher.

But if energy carried by a wave is less than the wave will have a low amplitude. The Amplitude shows the distance covered from the rest position to peak position.

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Which statement is true?
miv72 [106K]
B might be the correct answer
8 0
3 years ago
A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s
Naddika [18.5K]

Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

A)

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is:   w_{c} =m_{c} g   and maximum tension is T=2.50 m_{c} g=1.41*10^4N

total mass and weight is :

M =m_{c}+ m_{b} =740kg+550kg=1290 kg

w_{M} =1.2650*10^4N

∑F_{y} =ma_{y}

T-M_{g} =Ma_{y}

a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg

=0.645m/s^2

B)

maximum acceleration

a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0

using y-y_{0} =v_{oy} t+1/2(a_{y} )t^2

to solve for t

t=\sqrt{2(y-y_{0} )/a_{y} }

t=\sqrt{2(119m)/0.645m/s^2} =19.20s

6 0
3 years ago
Calculate the magnitude of the total impulse applied to the car to bring it to rest
n200080 [17]

Answer:

Total impulse = mv = Initial momentum of the car

Explanation:

Let the mass of the car be 'm' kg moving with a velocity 'v' m/s.

The final velocity of the car is 0 m/s as it is brought to rest.

Impulse is equal to the product of constant force applied to an object for a very small interval. Impulse is also calculated as the total change in the linear momentum of an object during the given time interval.

The magnitude of impulse is the absolute value of the change in momentum.

|J|=|p_f-p_i|

Momentum of an object is equal to the product of its mass and velocity.

So, the initial momentum of the car is given as:

p_i=mv

The final momentum of the car is given as:

p_f=m(0)=0

Therefore, the impulse is given as:

|J|=|p_f-p_i|=|0-mv|=|-mv|=mv

Hence, the magnitude of the impulse applied to the car to bring it to rest is equal to the initial momentum of the car.

5 0
3 years ago
The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1
Vaselesa [24]

Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2}) (1)

Where:

F - Explosive force, measured in newtons.

\Delta s - Barrel length, measured in meters.

m - Mass of the shell, measured in kilograms.

v_{o}, v_{f} - Initial and final speeds of the shell, measured in meters per second.

If we know that m = 1250\,kg, v_{o} = 0\,\frac{m}{s}, v_{f} = 750\,\frac{m}{s} and \Delta s = 15\,m, then the explosive force experienced by the shell inside the barrel is:

F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}

F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}

F = 23437500\,N

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

6 0
2 years ago
What is the concentration of the 100 g of solution having 25 g of solute?
Alexus [3.1K]

Answer:

70%

Explanation:

it's 70% hope it helps

8 0
2 years ago
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