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Stolb23 [73]
3 years ago
13

a 320-foot tree casts a 130-foot shadow on horizontal ground. a girl lying on the grass at the tip of the shadow is looking at a

bird nesting at the top of the tree. at what angle is the girl looking up? assume that the tree grows at a right angle to the ground g

Physics
1 answer:
mixer [17]3 years ago
7 0

Answer:\theta =67.88^{\circ}

Explanation:

Given

Height of tree is H=320 ft

length of shadow L=130 ft

let \theta be the angle of elevation

using trigonometric relation

\tan \theta =\frac{H}{L}

\tan \theta =\frac{320}{130}

\theta =tan^{-1}(2.461)

\theta =67.88^{\circ}

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Steel is very stiff, and the Young's modulus for steel is unusually large, 2.0×1011 N/m2. A cube of steel 25 cm on a side suppor
TEA [102]

Answer:

Force (normal) = 833.85 N Compression = 1.67 x 10⁻⁸ m

Explanation:

Given data Young's Modulus (Y) = 2 x 10¹¹ N/m², Length of one side of cube = 25 cm = 0.25 m, mass of load = 85 kg

Normal force is the force exerted upon an object that is in contact with another stable object. This force would be applied by the surface onto the object in the same vector and is used to keep the object stable while it rests on a surface.

We know from Newton's Second Law that

F = ma where m is the mass and a is the acceleration (<em>in this case due to gravity</em>) hence, the normal opposing force to the load applied by the surface would be equal to the force applied on the surface by the weight of the load on the surface, So

F (normal) = M (load) x a = 85 x 9.81 = 833.85 N

Compression is the change in length of an object by the exertion of force upon it. Using the Young's Modulus formula we can find this change in the cube of steel. The Young's Modulus is given by

Y = (F/A)/(ΔL/L), where Y is the Young's Modulus, F is the Force being applied on the object, A is the cross sectional area on which the said force is applied, ΔL is the change in length due to said force being applied and L is the original Length of the side of the cross sectional area.

Solving this for ΔL, we can re- arrange the equation

ΔL = (F x L)/(Y x A) since area of square is L x L we can simplify the equation to get

ΔL = (F)/(Y x L), substitute the values

ΔL = (833.85)/(2 x 10¹¹ x 0.25) = 1.67 x 10⁻⁸m

6 0
4 years ago
A box sits at the edge of a spinning disc. The radius of the disc is 0.5 m, and it is initially spinning at 5 revolutions per se
Furkat [3]

Answer:

a) α = 0.375 rad/s²

b) at = 0.1875 m/s²

c) ac =79 m/s²  

d) θ = 52 rad

Explanation:

The uniformly accelerated circular movemeis a circular path movement in which the angular acceleration is constant.

Tangential acceleration is calculated as follows:

at = α*R     Formula (1)

Centripetal acceleration is calculated as follows:

ac =ω² *R   Formula (2)

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t  Formula (3)

θ=  ω₀*t + (1/2)*α*t² Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed  ( rad/s)

R : radius of the circular path (m)

at:  tangential acceleration, (m/s²)

ac: centripetal acceleration, (m/s²)

Data:

R= 0.5 m  : radius of the disk

t₀=0 , ω₀ = 5 rev/s  

1 revolution = 2π rad

ω₀ = 5*(2π)rad/s  =10π rad/s  = 31.42 rad/s

ωf = 2*(2π)rad/s  =4π rad/s  = 12.57 rad/s

t = 8 s

(a) angular acceleration of the box

We replace data in the formula (3)

ωf= ω₀ + α*t

2 = 5 + α*(8)

2 -5 = α*(8)

-3 = (8)α

α=3 /8

α = 0.375 rad/s²

(b) Tangential acceleration of the box

We replace data in the formula (1)z

at =(α)*R

at = (0.375)*(0.5)

at = 0.1875 m/s²

c) Centripetal acceleration of the box at  t = 8 s

We replace data in the formula (2)

ac =ω² *R

ac =(12.57)² *(0.5)

ac = 79 m/s²  

d) Radians that the box has rotated over after t = 8 s

We replace data in the formula (4)

θ = ω₀*t + (1/2)*α*t²

θ = (5)*(8)+ (1/2)*( 0.375)*(8)²

θ = 52 rad

9 0
3 years ago
A skydiver free falls from rest. How long does it take for the skydiver to reach 7 m/s?
aev [14]

Answer:

<em>The skydiver needs 0.71 seconds to reach 7 m/s</em>

Explanation:

<u>Free Fall Motion </u>

When an object is dropped in free air (no friction) from a certain height h, it follows a free-fall motion, whose acceleration is due exclusively to gravity. The speed at a moment t when the object is dropped (from rest) is:

V_f=gt

We need to find How long does the skydiver needs to reach 7 m/s. We solve for t

\displaystyle t=\frac{V_f}{g}

\displaystyle t=\frac{7}{9.8}

t=0.71\ sec

The skydiver needs 0.71 seconds to reach 7 m/s

5 0
3 years ago
Two large, parallel, nonconducting sheets of positive charge face each other. What is at points (a) to the left of the sheets, (
alexira [117]

Answer:

a)The electric Field will be zero at the point between the sheets

b)E_1=\dfrac{\sigma}{\epsilon_0}

c)E_2=\dfrac{\sigma}{\epsilon_0}

Explanation:

Let \sigma be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field  so the Electric Flux due to it is zero.

Now using Gauss law we have, E be the electric Field at the distance r from the sheet then

E\times 2A=\dfrac{\sigma A}{\epsilon_0}\\E=\dfrac{\sigma}{2\epsilon_0}

The Field will be away from the sheet and perpendicular to it.

a) The Electric Field between them

E_1=\dfrac{\sigma}{2\epsilon_0}-\dfrac{\sigma}{2\epsilon_0}\\=0

b)The Electric Field to the right of the sheets

E_1=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}

c)The Electric Field to the left of the sheets

E_2=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}

3 0
3 years ago
Before collecting a specimen of urine or feces, the nursing assistant asks the nurse or consults the lab for?
bulgar [2K]

Before collecting a specimen of urine or feces, the nursing assistant asks the nurse or consults the lab for which storage and delivery method to use.

There are different tests that need to be performed on a urine or feces specimen. It depends on the patients that which test would be asked for him to be done.

The storage and delivery methods for urine and feces in the case of different tests has to be consulted by the nurse or consultant.

Incase of an emergency test, whose results are immediately required the method of delivery is a fast mode one. For patients that have a mild disease or are not at risk, other reliable methods can be used. Each specimen should be properly labeled for proper checking.

To learn more about urine specimen, click here:

brainly.com/question/28329488

#SPJ4

8 0
2 years ago
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