what is the output distance of a machine with an input distance of 3.0 centimeters and an ideal mechanical advantage of 12

1. (30 pts) Let x(t) = cos(πt/2) be a continuous-time signal,

posledela

Given that the function of the wave is f(x) = cos(π•t/2), we have;

a. The graph of the function is attached

b. 4 units of time

c. Even

d. 4.935 J/kg

e. 1.234 W/kg

<h3>How can the factors of the wave be found?</h3>

a. Please find attached the graph of the signal created with GeoGebra

b. The period of the signal, T = 2•π/(π/2) = 4

c. The signal is even, given that it is symmetrical about the y-axis

d. The energy of the signal is given by the formula;

$\frac{1}{2} \cdot \mu^{2} \cdot \omega ^{2} \cdot \: {a}^{2} \times \lambda$

Which gives;

E = 0.5 × 1.571² × 1² × 4 = 4.935 J/kg

e. The power of the wave is given by the formula;

E = 0.5 × 1.571² × 1² × 4 × 0.25 = 1.234 W/kg

Learn more about waves here:

brainly.com/question/14015797

8 0

3 years ago

A rectangular block of steel measures 4cm*2cm*1.5cm*. its mass is 93.6g

nata0808 [166]

Answer:

b

Explanation:

5 0

3 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

4 years ago

Brainly the unit of current in which 6,240,000,000,000,000,000 electrons flow past a given point in a circuit in one second is t

BigorU [14]

In electronics, the SI unit for current is Ampere. It is the amount of charge in Coulombs per unit time. It is named after the father of electrodynamics, Andre-Marie Ampere. Also, the current can be easily determined through the Ohm's Law, which states that current is equal to volts divided by the resistance. The answer is letter D.

5 0

3 years ago

Read 2 more answers

A spring with a rest length of 0.7 m has a spring constant of 70 N/m. It is stretched and now has a length of 2.5 m. What is the

pentagon [3]

Answer:

113.4 J

Explanation:

Elastic Potential Energy

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

$\displaystyle PE = \frac{1}{2}k(\Delta x)^2$

The spring has a natural length of 0.7 m and a spring constant of k=70 N/m. When the spring is stretched to a length of 2.5 m, the change of length is

Δx = 2.5 m - 0.7 m = 1.8 m

The energy stored in the spring is:

$\displaystyle PE = \frac{1}{2}70(1.8)^2$

PE = 113.4 J

7 0

3 years ago