NEED HELP!!!!!

Fiesta28 [93]

l will help and if you get it correctly I will say you are welcome

6 0

3 years ago

A high fountain of water is located at the center of a circular pool. Not wishing to get his feet wet, a student walks around th

Alisiya [41]

Answer:

The fountain is 3.43 m high.

Explanation:

Circumference of the pool = 15 m.

C = 2 $\pi$ r

where C is the circumference and r its radius.

r = $\frac{C}{2\pi }$

= $\frac{15}{2(\frac{22}{7}) }$

r = 2.3864

radius of the pool = 2.40 m

So that the height of the fountain, h, can be determined by applying trigonometric function.

Tan θ = $\frac{opposite}{adjacent}$

Tan 55 = $\frac{h}{2.4}$

h = Tan 55 x 2.4

= 1.4282 x 2.4

= 3.4277

h = 3.43 m

The height of the fountain is 3.43 m.

8 0

3 years ago

What is the highest temperature ever recorded on earth

Llana [10]

Therefore the world's record high temperature of 134.0°F (56.7°C) is held by Furnace Creek Ranch in Death Valley, California. That global high temperature was attained on July 10, 1913.

4 0

4 years ago

Consider a basketball player spinning a ball on the tip of a finger. If a player performs 1.99 J of work to set the ball spinnin

scoundrel [369]

To solve this problem it is necessary to apply the concepts related to rotational kinetic energy, the definition of the moment of inertia for a sphere and the obtaining of the radius through the circumference. Mathematically kinetic energy can be given as:

$KE= I\omega^2$

Where,

I = Moment of inertia

$\omega =$ Angular velocity

According to the information given we have that the radius is

$\Phi= 2\pi r$

$0.749m = 2\pi r$

$r = 0.1192m$

With the radius obtained we can calculate the moment of inertia which is

$I = \frac{2}{3}mr^2$

$I = \frac{2}{3}(0.624)(0.1192)^2$

$I = 5.91*10^{-3} kg \cdot m^2$

Finally, from the energy equation and rearranging the expression to obtain the angular velocity we have to

$\omega = \sqrt{\frac{2KE}{I}}$

$\omega = \sqrt{\frac{2(1.99)}{5.91*10^{-3}}}$

$\omega = 25.95rad/s$

Therefore the angular speed will the ball rotate is 25.95rad/s

8 0

3 years ago

A simple pendulum is made from a 0.54-m-long string and a small ball attached to its free end. The ball is pulled to one side th

Serga [27]

Answer:

0.37sec

Explanation:

Period of oscillation of a simple pendulum of length L is:

T = 2 π × √ (L /g)

L=length of string 0.54m

g=acceleration due to gravity

T-period

T = 2 x 3.14 x √[0.54/9.8]

T = 1.47sec

An oscillating pendulum, or anything else in nature that involves "simple harmonic" (sinusoidal) motion, spends 1/4 of its period going from zero speed to maximum speed, and another 1/4 going from maximum speed to zero speed again, etc. After four quarter-periods it is back where it started.

The ball will first have V(max) at T/4,

=>V(max) = 1.47/4 = 0.37 sec

3 0

4 years ago