Currents that occur in deep water are:

A surface receiving sound is moved from its original position to a position three times farther away from the source of the soun

vazorg [7]

Answer:

c. nine times as low.

Explanation:

Sound intensity is defined as the acoustic power transferred by a sound wave per unit of normal area to the direction of propagation:

$I\propto \frac{1}{A}$

Since the sound wave has a spherical wavefront of radius r, then the area is given by:

$A=4\pi r^2$

Here r is the distance from the source of the sound. Thus sound intensity decreases as:

$I\propto \frac{1}{4\pi r^2}\\\\I'\propto \frac{1}{4\pi r'^2}\\\\I'\propto \frac{1}{4\pi (3r)^2}\\\\I'\propto \frac{1}{9} \frac{1}{4\pi r^2}\\\\I'\propto \frac{1}{9} I$

6 0

4 years ago

How many of each component are shown in the diagram? Check all that apply.

g100num [7]

Can you show the diagram please

3 0

3 years ago

The cheetah is one of the fastest-accelerating animals, because it can go from rest to 16.2 m/s (about 36 mi/h) in 2.4 s. If its

Scilla [17]

<h2>Power of cheetah is 5576.85 W = 7.48 hp</h2>

Explanation:

Power is the ratio of energy to time.

Here we need to consider kinetic energy,

Mass, m = 102 kg

Initial velocity = 0 m/s

Final velocity = 16.2 m/s

Time, t = 2.4 s

Initial kinetic energy = 0.5 x Mass x Initial velocity² = 0.5 x 102 x 0² = 0 J

Final kinetic energy = 0.5 x Mass x Final velocity² = 0.5 x 102 x 16.2² = 13384.44 J

Change in energy = Final kinetic energy - Initial kinetic energy

Change in energy = 13384.44 - 0

Change in energy = 13384.44 J

Power = 13384.44 ÷ 2.4 = 5576.85 W = 7.48 hp

Power of cheetah is 5576.85 W = 7.48 hp

7 0

3 years ago

Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f

elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

$W_T+W_F=K_f-K_i$ (1)

Where:

$W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}$

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to $\frac{1}{2}mv^{2}$ , so if the initial velocity $v_i$ is equal to zero, the initial kinetic energy $K_i$ is equal to zero.

Then, replacing the values on the equation and solving for $v_f$ , we get:

$W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}$

$\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f$

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0

3 years ago

Newton’s 3rd law of motion states that “every action has an equal and opposite reaction.” A golf ball was hit with a force of 20

DedPeter [7]

Answer:

<u>Example of Newton's III law</u>

In the, golf the ball was hit by a club with certain force. As the club hits the ball it's the action. When the ball flies away its the reaction.
When a person swings a golf club at the ball, when it hits the ball, it causes the ball to roll up the face of the club and into the air towards the target.

8 0

4 years ago