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2 years ago

A speed-time graph is shown below:

Physics

1 answer:

VladimirAG [237]2 years ago

5 0

Answer: $0.5 m/s^{2}$

Explanation:

Average acceleration $a_{ave}$ is the variation of velocity $\Delta V$ over a specified period of time $\Delta t$ :

$a_{ave}=\frac{\Delta V}{\Delta t}}$

Where:

$\Delta V=V_{f}-V_{o}$ being $V_{o}=0 cm/s$ the initial velocity and $V_{f}=4 cm/s$ the final velocity (according to the information given from the described graph)

$\Delta t=8 s$

Then:

$a_{ave}=\frac{4 cm/s -0 cm/s}{8 s}}$

$a_{ave}=0.5 m/s^{2}$

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A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next

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Complete Question

A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?

Answer:

$V=1.4*10^5m/s$

Explanation:

From the question we are told that:

Electric field $B=1.5*10N/C$

Distance $d=2 x 10^{-3}$

At negative plate

Generally the equation for Velocity is mathematically given by

$V^2=2as$

Therefore

$V^2=\frac{2*e_0E*d}{m}$

$V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}$

$V=\sqrt{19.2*10^9}$

$V=1.4*10^5m/s$

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3 years ago

Which of the following statements is true? A. X-rays have longer wavelengths than microwaves. B. Radio waves have shorter wavele

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D is correct, Gamma rays have the shortest wavelength.

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3 years ago

A force of 9 pounds stretches a spring 1 foot. A mass weighing 6.4 pounds is attached to the spring, and the system is then imme

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Answer:

$\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0$

Explanation:

let $m$ be the mass attached, let $k$ be the spring constant and let $\beta$ be the positive damping constant.

-By Newton's second law:

$m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}$

where $x(t)$ is the displacement from equilibrium position. The equation can be transformed into:

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3 years ago

A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr

rjkz [21]

Answer:

Part A

I = 1.4 mW/m²

Part B

β = 91.46 dB

Explanation:

Part A

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^{2}}$

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

$I = \frac{34}{4\pi (44)^{2}}$

I = 0.0013975 W/m²

≈ I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

Part B

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value. It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;

β $= 10log_{10}\frac{I}{I_{0}}$ dB