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3 years ago

A trip is taken that passes through the following points in order

Physics

1 answer:

Inessa05 [86]3 years ago

5 0

Answer:

B) - 5.0 m

Explanation:

B is located on a positive location, 15m from the starting point A. Hence, since E is located a positive distance 10m from A, the difference becomes 10 - 15 = - 5.0 m

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PART ONE

Korvikt [17]

Answer:

1) 460.5 N

2) 431.7 N

Explanation:

Draw a free body diagram. There are four forces on the hammer:

Applied force 62.5 N in the +x direction, 30 cm from the ground

Reaction force Rᵧ in the +y direction, at the point of contact

Reaction force Rₓ in the +x direction, at the point of contact

Reaction force F at 31° from the vertical, 4.75 cm to the left of the point of contact

Part One

To find F, sum the moments about the point of contact:

∑τ = Iα

(62.5 N) (30 cm) − (F cos 31°) (4.75 cm) = 0

F = 460.5 N

Part Two

To find Rₓ and Rᵧ, sum the forces in the x and y directions.

∑Fₓ = ma

62.5 N − F sin 31° + Rₓ = 0

Rₓ = 174.7 N

∑Fᵧ = ma

-F cos 31° + Rᵧ = 0

Rᵧ = 394.7 N

The net reaction force at the point of contact is:

R = √(Rₓ² + Rᵧ²)

R = 431.7 N

7 0

3 years ago

Please helppppppppppp

Leni [432]

Answer: 4

Explanation:

when energy is greater then frequency should increase and wavelength should decrease.

Therefore answer is 4

4 0

3 years ago

A wheel rotates clockwise 6 times per second. What will be its angular displacement after 7 seconds? Answer should be rounded to

solong [7]

Answer:

The frequency of the wheel is the number of revolutions per second:

f= \frac{N_{rev}}{t}= \frac{10}{1 s}=10 Hz

And now we can calculate the angular speed, which is given by:

\omega = 2 \pi f=2 \pi (10 Hz)=62.8 rad/s in the clockwise direction.

Explanation:

5 0

3 years ago

Read 2 more answers

A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.00 rad/s^2. At the instant th

xxMikexx [17]

Answer:

ac= 15.07 m/s²

Explanation:

The Wheel rotates with a constant angular acceleration.:

Centripetal acceleration is calculated as follows:

ac =ω² *R Formula (1)

ac =v² / R Formula (2)

Kinematics of the wheel

We apply the equations of circular motion uniformly accelerated :

ωf²= ω₀² + 2αθ Formula (3)

v = ω* R Formula (4)

Where:

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

α : angular acceleration (rad/s²)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

v: tangential velocity of a point on the rim ( m/s)

R : radius of wheel (m)

ac: centripetal acceleration, (m/s²)

Data:

D = 40.0 cm : diameter of the wheel

R = D/2= 40.0 cm/ 2 = 20 cm = 0.2m

α = 3.00 rad/s^2

ω₀ = 0

n = 2 revolutions : number of revolutions

θ =2πn (rad) = 2π*2 (rad) = 4π rad

Calculate of the ωf

We replace data in the formula (3)

ωf²= ω₀² + 2αθ

ωf²= 0 + 2(3)(4π)

ωf²= 24π

$w_{f} = \sqrt{24\pi }$

ωf = 8.68 rad/s

Calculate of the v

We replace data in the formula (4)

v = ω*R

v = (8.68)*(0.2)

v = 1.736 m/s

Calculate of the ac

We replace data in the formula (1)

ac = ( ω)²*(R)

ac = (8.68)²*(0.2)

ac = 15.07 m/s²

We replace data in the formula (2)

ac = v²/ R

ac = ( 1.736 )²/(0.2)

ac = 15.07 m/s²

7 0

4 years ago

The range of audible frequencies is from about 20.0 hz to 2.00×104 hz . what is range of the wavelengths of audible sound in air

TiliK225 [7]

The wavelength is related to the frequency by the relationship:
$\lambda= \frac{v}{f}$
where v is the wave speed and f is its frequency.

The speed of sound in air is v=344 m/s. The lowest frequency is f=20.0 Hz, so the corresponding wavelength is
$\lambda_1 = \frac{v}{f_1}= \frac{344 m/s}{20.0 Hz}=17.2 m$
The highest frequency is $f_2 = 2.00 \cdot 10^4 Hz$ , so the corresponding wavelength is
$\lambda_2 = \frac{v}{f_2}= \frac{344 m/s}{2.00 \cdot 10^4 Hz}=0.017 m$

Therefore, the range of wavelengths of audible sound in air is
[0.017 m - 17.2 m]

4 0

3 years ago