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Digiron [165]
3 years ago
6

If the wave speed is 20 m/s and the wave length is 2 meters find the frequency ?

Physics
1 answer:
timofeeve [1]3 years ago
3 0

Answer:

10

Explanation:

v=f×lander

20=f2

f=10

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A 225-kg object and a 525-kg object are separated by 3.80 m. (a) Find the magnitude of the net gravitational force exerted by th
pav-90 [236]

Answer:F_{net}=3.383\times 10^{-7}\ N

Explanation:

Given

Mass of first object m_1=225\ kg

Mass of second object m_2=525\ kg

Distance between them d=3.8\ m

m_3=61\ kg object is placed between them

So force exerted by m_1 on m_3

F_{13}=\frac{Gm_1m_3}{1.9^2}

F_{13}=\frac{6.674\times 10^{-11}(225\times 61)}{1.9^2}

F_{13}=2.5374141274×10^{−7}\ N

Force exerted by m_2\ on\ m_3

F_{23}=\frac{Gm_2m_3}{1.9^2}

F_{23}=\frac{6.674\times 10^{-11}(525\times 61)}{1.9^2}

F_{23}=5.920632964\times 10^{-7}\ N

So net force on m_3 is

F_{net}=F_{23}-F_{13}

F_{net}=5.920632964\times 10^{-7}-2.5374141274\times 10^{-7}

F_{net}=3.383\times 10^{-7}\ N

i.e. net force is towards m_2

(b)For net force to be zero on m_3, suppose

So force exerted by m_1 and m_2 must be equal

F_{13}=F_{23}

\Rightarrow \frac{Gm_1m_3}{x^2}=\frac{Gm_2m_3}{(3.8-x)^2}

\Rightarrow \frac{m_1}{x^2}=\frac{m_2}{(3.8-x)^2}

\Rightarrow (\frac{3.8-x}{x})^2=\frac{m_2}{m_1}

\Rightarrow \frac{3.8-x}{x}=\sqrt{\frac{525}{225}}

\Rightarrow 3.8-x=1.52752x

\Rightarrow 3.8=2.52x

\Rightarrow x=1.507\ m

4 0
2 years ago
Which of the following BEST explains why stability is an important skill for athletes?
Irina-Kira [14]

Answer;

B. Without good stability, athletes are too clumsy to perform well.

Explanation;

Stability refers to the resistance to both linear and angular acceleration, or resistance to disruption of equilibrium.

Developing and enhancing core strength and stability helps athletes to maximize their power output and enhance game day performance. Additionally, Improving core strength and stability can also help athletes reduce their risk of injury.

8 0
3 years ago
Read 2 more answers
Velocity and acceleration are both vectors; they have a direction. What is thedirection of the velocity and acceleration vectors
just olya [345]

Answer:

d. Both A & C

Explanation:

- The velocity of the ball is a vector, whose magnitude indicates its rate of change of position, and it has a direction. In this case, the ball is moving upward, therefore the direction of motion is upward, so the direction of the velocity is upward as well.

- The acceleration of the ball is a vector, whose magnitude indicates the rate of change of the velocity. The direction of the acceleration is:

-- positive if the the magnitude of the velocity is increasing

-- negative if the magnitude of the velocity is decreasing

For a ball thrown upward, the acceleration is given by the acceleration of gravity, g=9.8 m/s^2. This acceleration points downward, and it is constant during the entire motion. In particular, it does not change direction, as it is always directed downward. Therefore, the acceleration of the ball is downward.

So, the correct answer is

d.Both A & C

Since A and C are both true:

a.The velocity vector is directed upward.

c.The acceleration is directed downward.

3 0
3 years ago
A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially relea
grandymaker [24]

A mass weighing 32 pounds stretches a spring 2 feet.

(a) Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s.

(b) How many complete cycles will the mass have completed at the end of 4 seconds?

Answer:

A = 1.803 ft

Period = \frac{\pi}{2} seconds

8 cycles

Explanation:

A mass weighing 32 pounds stretches a spring 2 feet;

it implies that the mass (m) = \frac{w}{g}

m= \frac{32}{32}

= 1 slug

Also from Hooke's Law

2 k = 32

k = \frac{32}{2}

k = 16 lb/ft

Using the function:

\frac{d^2x}{dt} = - 16x\\\frac{d^2x}{dt} + 16x =0

x(0) = -1        (because of the initial position being above the equilibrium position)

x(0) = -6          ( as a result of upward velocity)

NOW, we have:

x(t)=c_1cos4t+c_2sin4t\\x^{'}(t) = 4(-c_1sin4t+c_2cos4t)

However;

x(0) = -1 means

-1 =c_1\\c_1 = -1

x(0) =-6 also implies that:

-6 =4(c_2)\\c_2 = - \frac{6}{4}

c_2 = -\frac{3}{2}

Hence, x(t) =-cos4t-\frac{3}{2} sin 4t

A = \sqrt{C_1^2+C_2^2}

A = \sqrt{(-1)^2+(\frac{3}{2})^2 }

A=\sqrt{\frac{13}{4} }

A= \frac{1}{2}\sqrt{13}

A = 1.803 ft

Period can be calculated as follows:

= \frac{2 \pi}{4}

= \frac{\pi}{2} seconds

How many complete cycles will the mass have completed at the end of 4 seconds?

At the end of 4 seconds, we have:

x* \frac{\pi}{2} = 4 \pi

x \pi = 8 \pi

x=8 cycles

5 0
2 years ago
A skier of mass 103 kg comes down a slopeof constant angle 32◦with the horizontal.What is the force on the skier parallel tothe
LuckyWell [14K]

Answer:

534.9 N

Explanation:

The skier weight is his mass times gravitational acceleration g

W = mg = 103 * 9.8 = 1009.4 N

This weight can be divided into 2 components, one perpendicular and the other parallel to the 32-degree slope. The parallel component would equal to

Wsin(32^0) = 1009.4sin(32^0) = 534.9 N

3 0
3 years ago
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