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3 years ago

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And a hydraulic system piston one has the surface area of 100 cm² and piston to has a surface area of 900 cm² piston one exerts

a pressure of 10 PA on the fluid in the hydraulic lift what is the fluid pressure on piston to

1 answer:

Georgia [21]3 years ago

8 0

Hope this helps you.

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Calculate the speed of a gamma ray with a frequency of 3.0 x 10^19 Hz and a wavelength of 1.0 x 10^-11 m.

Ulleksa [173]

Answer:

Speed of gamma rays = 3 x 10⁸ m/s

Explanation:

Given:

Frequency of gamma ray = 3 x 10¹⁹ Hz

Wavelength of gamma rays = 1 x 10⁻¹¹ meter

Find:

Speed of gamma rays

Computation:

Velocity = Frequency x wavelength

Speed of gamma rays = Frequency of gamma ray x Wavelength of gamma rays

Speed of gamma rays = [3 x 10¹⁹][1 x 10⁻¹¹]

Speed of gamma rays = 3 x [10¹⁹⁻¹¹]

Speed of gamma rays = 3 x [10⁸]

Speed of gamma rays = 3 x 10⁸ m/s

6 0

3 years ago

One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

Alja [10]

Answer:

$\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}$

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

$f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}$

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

$\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}$

$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

$\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}$

4 0

3 years ago

What is the momentum of a 0.1-kg mass moving with a speed of 5 m/s

Andrej [43]

This question is wrong because in momontum we will write acceleration instead of speed. suppose acceleration is 5m/s2 then
P= ma
then put values

4 0

3 years ago

Define :density٬archimedes principle

olga2289 [7]

Answer:

density is defined as the amount of mass contained in unit volume of a body .its si unit is kg/m*3

5 0

3 years ago

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

5 0

3 years ago

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