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tigry1 [53]
3 years ago
15

A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed of 10 m/s. If air

resistance is negligible, the distance from the edge of the base of the cliff to where the rock hits the level ground below the cliff is most nearly
a. 20 m

b. 5 m

c. 10 m

d. 40 m
Physics
1 answer:
Gwar [14]3 years ago
8 0

Answer:

option A

Explanation:

given,

height of the cliff = 20 m

initial speed = 10 m/s

distance from the edge of the base of the cliff  = ?

now,

time taken to cover vertical distance

initial vertical speed = 0 m/s

using equation of motion

s = u t + \dfrac{1}{2}gt^2

s =\dfrac{1}{2}gt^2

t = \sqrt{\dfrac{2s}{g}}

t = \sqrt{\dfrac{2\times 20}{9.8}}

  t = 2 s

in same time the ball will travel horizontal distance

horizontal velocity of ball = 10 m/s

distance = s x t

d = 10 x 2

d = 20 m

Hence, the correct answer is option A

You might be interested in
Is everyone in your class able to hear a quiet sound equally well?
Sergio039 [100]

Answer:

No

Explanation:

Loudness describes how people perceive sound (see loudness). ... If people could hear equally well at all frequencies, the contour lines would be flat because the same measured sound intensity would be perceived to be equally loud regardless of the sound frequency. In fact, people do not hear as well at low frequencies.

8 0
3 years ago
Under the assumption that the beam is a rectangular cantilever beam that is free to vibrate, the theoretical first natural frequ
BartSMP [9]

Answer:

a) Δf = 0.7 n , e)   f = (15.1 ± 0.7) 10³ Hz

Explanation:

This is an error about the uncertainty or error in the calculated quantities.

Let's work all the magnitudes is the SI system

The frequency of oscillation is

        f = n / 2π L² √( E /ρ)

where n is an integer

Let's calculate the magnitude of the oscillation

       f = n / 2π (0.2335)² √ (210 10⁹/7800)

       f = n /0.34257 √ (26.923 10⁶)

       f = n /0.34257    5.1887 10³

       f = 15.1464 10³ n

a) We are asked for the uncertainty of the frequency (Df)

       Δf = | df / dL | ΔL + df /dE ΔE + df /dρ Δρ

in this case no  error is indicated in Young's modulus and density, so we will consider them exact

       ΔE = Δρ = 0

       Δf = df /dL  ΔL

       df = n / 2π   √E /ρ   | -2 / L³ | ΔL

       df = n / 2π 5.1887 10³ | 2 / 0.2335³) 0.005 10⁻³

       df = n 0.649

Absolute deviations must be given with a single significant figure

        Δf = 0.7 n

b, c) The uncertainty with the width and thickness of the canteliver is associated with the density

 

In your expression there is no specific dependency so the uncertainty should be zero

The exact equation for the natural nodes is

          f = n / 2π L² √ (E e /ρA)

where A is the area of ​​the cantilever and its thickness,

In this case, they must perform the derivatives, calculate and approximate a significant figure

        Δf = | df / dL | ΔL + df /de  Δe + df /dA  ΔA

        Δf = 0.7 n + n 2π L² √(E/ρ A) | ½  1/√e | Δe

               + n / 2π L² √(Ee /ρ) | 3/2 1√A23  |

the area is

        A = b h

        A = 24.9  3.3  10⁻⁶

        A = 82.17 10⁻⁶ m²

        DA = dA /db ΔB + dA /dh Δh

        dA = h Δb + b Δh

        dA = 3.3 10⁻³ 0.005 10⁻³ + 24.9 10⁻³ 0.005 10⁻³

        dA = (3.3 + 24.9) 0.005 10⁻⁶

        dA = 1.4 10⁻⁷ m²

let's calculate each term

         A ’= n / 2π L² √a (E/ρ A) | ½ 1 /√ e | Δe

         A ’= n/ 2π L² √ (E /ρ)      | ½ 1 / (√e/√ A) |Δe

        A ’= 15.1464 10³ n ½ 1 / [√ (24.9 10⁻³)/ √ (81.17 10⁻⁶)] 0.005 10⁻³

        A '= 0.0266  n

        A ’= 2.66 10⁻² n

       A ’’ = n / 2π L² √ (E e /ρ) | 3/2  1 /√A³ |

       A ’’ = n / 2π L² √(E /ρ) √ e | 3/2  1 /√ A³ | ΔA

       A ’’ = n 15.1464 10³ 3/2 √ (24.9 10⁻³) /√ (82.17 10⁻⁶) 3 1.4 10⁻⁷

       A ’’ = n 15.1464 1.5 1.5779 / 744.85 1.4 10⁴

       A ’’ = 6,738 10²

we write the equation of uncertainty

     Δf = n (0.649 + 2.66 10⁻² + 6.738 10²)

The uncertainty due to thickness is

    Δf = 3 10⁻² n

The uncertainty regarding the area, note that this magnitude should be measured with much greater precision, specifically the height since the errors of the width are very small

     Δf = 7 10² n

 d)    Δf = 7 10² n

e) the natural frequency n = 1

       f = (15.1 ± 0.7) 10³ Hz

7 0
3 years ago
16)
VMariaS [17]

Answer:

a.proton, proton

hope this helped :) have a goodday

6 0
3 years ago
Suppose we wrap a string around the surface of a uniform cylinder of radius 38.0 cm that is supported by an axle passing through
liraira [26]

Answer:

Angular acceleration = 23.68 rad / s²

Explanation:

Given that,

acceleration = 9m/s²

Therefore acceleration of string is 9m/s²

since string is constant in length

cylinder of radius 38.0 cm = 0.38m

Angular acceleration = a / r

Angular acceleration = 9 / 0.38

                                   = 23.68 rad / s²

Angular acceleration = 23.68 rad / s²

4 0
3 years ago
Which force attracts all matter to each other
likoan [24]

Gravity is the force that attracts all matter to each other.

Explanation:

Sir Isaac Newton discovered Gravity when he saw a falling apple while thinking about the forces of nature.

Gravity is a fundamental force that causes objects to have weight. Gravity acts on all matter and is a function of both mass and distance. Each object attracts every other object with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force of attraction is, however, negligible between most objects because of their small size.

Gravitational force is given as:

F = \frac{Gm1m2}{r^{2} }

Where G is gravitational constant and is equal to 6.674×10−11 m³⋅kg⁻¹⋅s⁻²

m₁ and m₂ are the masses of the two objects.

r is the distance between the two objects.

The gravity is what makes an apple fall on the ground and gravity is the force that keeps us on the ground.

Keywords: gravity, Newton, Force, weight

Learn more about gravitational force from brainly.com/question/14321566

#learnwithBrainly

8 0
3 years ago
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