You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. Unfortunately the archer stands on an elevated platform of unknown height. However, you find the arrow stuck in the ground 67.0 m away, making a 3.00 ∘ angle with the ground.How fast was the arrow shot?

Answers

Answer 1

Answer:

To develop this problem it is necessary to apply the kinematic equations that describe displacement, velocity and clarification.

By definition we know that velocity is defined as the change of position due to time, therefore

$V = (d)/(t)$

Where,

d = Distance

t = Time

Speed can also be expressed in vector form through its components $V_x$ and $V_y$

In the case of the horizontal component X, we have to

$V_x = (d)/(t)$

Here d means the horizontal displacement, then

$t = (d)/(V_x)$

$t = (67)/(V_x)$

At the same time we have that the vertical component of the velocity is

$V_y = gt$

Here,

g = Gravity

Therefore using the relation previously found we have that

$V_y = g (67)/(V_x)$

The relationship between the two velocities and the angle can be expressed through the Tangent, therefore

$tan\theta = (V_y)/(V_x)$

$tan \theta = (g (67)/(V_x) )/(V_x)$

$tan 3 = (9.8(67)/(V_x) )/(V_x)$

$tan 3 = (9.8*67)/(V_x^2)$

$V_x^2 = (9.8*67)/(tan 3)$

$V_x= \sqrt{ (9.8*67)/(tan 3)}$

$V_x = 111.93m/s \hat{i}$

This is the horizontal component, we could also find the vertical speed and the value of the total speed with the information given,

Then $V_y,$

$V_y = g (67)/(V_x)$

$V_y = 9.8*(67)/(111.93)$

$V_y = 5.866m/s\hat{j}$

$|\vec{V}| = √(111.93^2+5.866^2)$

$|\vec{V}| = 112.084m/s$

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A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 27.0 m/s by a 7850 N braking force acting opposite the car’s motion. What is the car's velocity after 2.52s?
How far does the car move during the 2.52 s?
How long does it take the car to come to a complete stop?

Answers

Answer:

19.1 m/s

58.1 m

8.60 s

Explanation:

Take north to be positive and south to be negative.

Use Newton's second law to find the acceleration.

∑F = ma

-7850 N = (2500 kg) a

a = -3.14 m/s²

Given:

v₀ = 27.0 m/s

a = -3.14 m/s²

Find: v given t = 2.52 s

v = at + v₀

v = (-3.14 m/s²) (2.52 s) + 27.0 m/s

v = 19.1 m/s

Find: Δx given t = 2.52 s

Δx = v₀ t + ½ at²

Δx = (27.0 m/s) (2.52 s) + ½ (-3.14 m/s²) (2.52 s)²

Δx = 58.1 m

Find: t given v = 0 m/s

v = at + v₀

0 m/s = (-3.14 m/s²) t + 27.0 m/s

t = 8.60 s

Which statement about thin lenses is correct? In each case, we are considering only a single lens. A. A diverging lens always produces a virtual inverted image. B. A converging lens always produces a real inverted image. C. A converging lens sometimes produces a real erect image. D. A diverging lens produces a virtual erect image only if the object is located within the focal point of the lens. E. A diverging lens always produces a virtual erect image.

Answers

A diverging lens always produces a virtual erect image.

The general lens formula is given as;

$(1)/(F) = (1)/(U) + (1)/(V)$

Where;

U = object distance
V = image distance
F = focal length of the lens

A lens can be converging or diverging.

A converging lens produces a virtual image when the object is placed in front of the focal point. The image can also be real when the object is placed beyond focal point.

The image produced by a diverging lens is always virtual and upright.

Thus, we can conclude that a diverging lens always produces a virtual erect image.

Learn more here:brainly.com/question/11788630

Answer:

E) true. The image is always virtual and erect

Explanation:

In this exercise we are asked to find the correct statements,

for this we can use the constructor equation

1 / f = 1 / p + 1 / q

where f is the focal length, p the distance to the object and q the distance to the image

In diverging lenses, the focal length is negative and the image is virtual and erect

In convergent lenses, the positive focal length, if the object is farther than the focal length, the image is real and inverted, and if the object is at a shorter distance than the focal length, the image is virtual and straight.

With this analysis let's review each statement

A) False. The image is right

B) False. The type of image depends on where the object is with respect to the focal length

C) False. The real image is always inverted

D) False. The image is always virtual

E) true. The image is always virtual and erect

The distance in the x direction between two control points on a vertical aerial photograph is 4.5". If the distance between these same two points is 3.6" on another photograph having a scale of 1:24,000, determine the scale of the first vertical aerial photograph. Of the focal length of the camera is 6"and the average elevation at these points is 100 ft, determine the flying height from which each photograph was taken

Answers

Answer:

Use proportions to find the scale of the first photo, then use that scale and other given information to fill in the equation

S=f/(H-h)

Where:

S = scale of the photo

f = focal length of the camera (in feet)

H = flying height

h = average elevation

A sound wave travels with a velocity of 330 m/s and has a frequency of 500 Hz. What is itswavelength?

Answers

Answer:

Wavelength = 0.66 meters

Explanation:

Given the following data;

Speed = 330 m/s

Frequency = 500 Hz

To find the wavelength;

Mathematically, wavelength is calculated using this formula;

$Wavelength = \frac {speed}{frequency}$

Substituting into the equation, we have;

$Wavelength = \frac {330}{500}$

Wavelength = 0.66 meters

Define the term energy density of a body under strain

Answers

Answer:

Please mark as Brainliest!!

Explanation:

Strain energy is defined as the energy stored in a body due to deformation. The strain energy per unit volume is known as strain energy density and the area under the stress-strain curve towards the point of deformation. When the applied force is released, the whole system returns to its original shape.

What is an inexpensive, portable, and common way to assess body fat in the fitness industry?DEXA
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Skinfold testing
Hydrostatic weighing

Answers

Answer: Skinfold testing

Explanation:

Skinfold testing, is also referred to as calliper testing and it's used to know the body fat percentage. Skinfold testing is an inexpensive, portable, and common way to assess body fat in the fitness industry.

It is typically done with the use of caliper tapes, marker pens which makes it cheap. Skinfold testing isn't usually accurate which is as a result of human errors.

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