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3 years ago

If an object falls from rest and lands 3.2 seconds later, how many meters did it fall

Physics

1 answer:

DIA [1.3K]3 years ago

6 0

Answer:

50.2 meters

Explanation:

When an object falls, it goes a distance d in time t according to the formula:

$d=\frac{1}{2}*g*t^{2}$

d is the distance in meter, g is the acceleration due to gravity with the value of 9.8 $m/s^{2}$ , t is the time in seconds

Therefore, $d = \frac{1}{2}*9.8*(3.2)^{2}$

d= 50.176 ≈ 50.2m

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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:

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

Where:

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 racecar accelerates uniformly from 18.5 mil to 46.1 m/s in 2.47 seconds.

laiz [17]

Answer:

The acceleration of the racecar is $\mathbf{11.17~m/s^2}$

Explanation:

Uniformly Accelerated Motion

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Following the definition above, the acceleration is defined as:

$\displaystyle a=\frac{v_f-v_o}{t}$

Where a is the constant acceleration, vo the initial speed, vf the final speed, and t the time.

The racecar goes from vo=18.5 m/s to vf=46.1 m/s in t=2.47 seconds, thus the acceleration is:

$\displaystyle a=\frac{46.1-18.5}{2.47}$

$\displaystyle a=\frac{27.6}{2.47}$

$a = 11.17~m/s^2$

The acceleration of the racecar is $\mathbf{11.17~m/s^2}$

5 0

3 years ago

A 68.0-kg person jumps from rest off a 2.20-m-high tower straight down into the water. Neglect air resistance during the descent

Margarita [4]

Answer:

$F= 1333.767\,N$

Explanation:

The velocity of the swimmer just before touching the water is:

$v = -\sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.20\,m)}$

$v \approx 6.569\,\frac{m}{s}$

The average force exerted on the diver by the water is determined by the use of the Principle of Energy Conservation and the Work-Energy Theorem:

$(68\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (2.20\,m) +\frac{1}{2}\cdot (68\,kg)\cdot (6.569\,\frac{m}{s} )^{2}-F\cdot(2.20\,m) = 0\,J$

$F= 1333.767\,N$

6 0

3 years ago

Read 2 more answers

One Affective way to develop the feeling function in one’s personality is to

Mademuasel [1]

Answer:

Explanation:

I think it’s a. Because, I normally think about how someone else feels if I want to say something.

5 0

2 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:

Given:

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