harold threw a baseball 3 meters. harold is 2 meters tall. how fast did he throw the ball and how long was it in the air

What two factors will most likely change the speed of a mechanical wave?

iVinArrow [24]

Period and wavelength

8 0

3 years ago

A 0.8 m length of wire is formed into a single turn, square loop in which there is a current of 12 A. The loop is placed in magn

rewona [7]

Answer:

0.073 N-m

Explanation:

i = 12 A, l = 0.8 m, B = 0.12 T

The circumference of the loop is 0.8 m.

Let r be the radius of the loop.

2 x 3.14 x r = 0.8

r = 0.127 m

Maximum Torque = i x A x B

Maximum Torque = 12 x 3.14 x 0.127 x 0.127 x 0.12 = 0.073 N-m

8 0

3 years ago

Select all that apply. An acorn falls from a tree. Which of the following statements is true? The force of gravity is acting on

DedPeter [7]

The net force on the acorn is less than the force of gravity.

6 0

3 years ago

Read 2 more answers

A beam of light in air is incident at an angle of 30º to the surface of a rectangular block of clear plastic (n = 1.46). The lig

Aneli [31]

Answer:

θ = 30°

Explanation:

Firts, the angle when the beam of light passes through the block cam be calculated using Snell Law:

$n_{1}sin(\theta_{1}) = n_{2}sin(\theeta_{2})$

<u>Where</u>:

n₁: is the index of refraction of the incident medium (air) = 1

θ₁: is the incident angle = 30°

n₂: is the medium 2 (plastic) = 1.46

θ₂: is the transmission angle

Hence, θ₂ is:

$sin(\theta_{2}) = \frac{n_{1}*sin(\theta_{1})}{n_{2}} = \frac{1*sin(30)}{1.46} = 0.34 \rightarrow \theta_{2} = 20.03 ^{\circ}$

Now, when the beam of light re-emerges from the opposite side, we have:

n₁: is the index of refraction of the incident medium (plastic) = 1.46

θ₁: is the incident angle = 20.03°

n₂: is the medium 2 (air) = 1

θ₂: is the transmission angle

Hence, the angle to the normal to that surface (θ₂) is:

$sin(\theta_{2}) = \frac{n_{1}*sin(\theta_{1})}{n_{2}} = \frac{1.46*sin(20.03)}{1} = 0.50 \rightarrow \theta_{2} = 30 ^{\circ}$

Therefore, we have that the beam of light will come out at the same angle of when it went in, since, it goes from air and enters to a plastic medium and then enters again in this medium to go out to air again. This was proved using the Snell Law.

I hope it helps you!

5 0

3 years ago

A radar station detects an airplane approaching directly from the east. At first observation, the range to the plane is 375.0 m

Gekata [30.6K]

Answer:

819.78 m

Explanation:

<u>Given:</u>

OA = range of initial position of the airplane from the point of observation = 375 m
OB = range of the final position of the airplane from the point of observation = 797 m
$\theta$ = angle of the initial position vector from the observation point = $43^\circ$
$\alpha$ = angle of the final position vector from the observation point = $123^\circ$
$\vec{AB}$ = displacement vector from initial position to the final position

A diagram has been attached with the solution in order to clearly show the position of the plane.

$\vec{OA} = OA\cos \theta \hat{i}+OA \sin \theta \hat{j}\\\Rightarrow \vec{OA} = 375\ m\cos 43^\circ \hat{i}+375\ m\sin 43^\circ \hat{j}\\\Rightarrow \vec{OA} = (274.26\ \hat{i}+255.75\ \hat{j})\ m\\\vec{OB} = OB\cos \alpha \hat{i}+OB \sin \alpha \hat{j}\\\Rightarrow \vec{OB} = 797\ m\cos 123^\circ \hat{i}+797\ m\sin 123^\circ \hat{j}\\\Rightarrow \vec{OB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m$

Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:

$\vec{AB}=\vec{OB}-\vec{OA}\\\Rightarrow \vec{AB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m-(274.26\ \hat{i}+255.75\ \hat{j})\ m\\\Rightarrow \vec{AB} = (-708.34\ \hat{i}+412.67\ \hat{j})\ m$

The magnitude of the displacement vector = $\sqrt{(-708.34)^2+(412.67)^2}\ m = 819.78\ m$

Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.

8 0

3 years ago