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dangina [55]
3 years ago
5

A certain car can accelerate from 0 to 60 mph in 7.9 s. What is the car's average acceleration in mph/s?

Physics
1 answer:
Anna007 [38]3 years ago
7 0

Answer:

<em>a = 7.6\ mph/s</em>

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the velocity of an object changes uniformly in time.

The equation that describes the change of velocities is:

v_f=v_o+at

Where:

a   = acceleration

vo = initial speed

vf  = final speed

t    = time

Solving the equation [for a:

\displaystyle a=\frac{v_f-v_o}{t}

The car accelerates from vo=0 to vf=60 mph in t=7.9 s, thus the acceleration is:

\displaystyle a=\frac{60 \ mph-0}{7.9}

a = 7.6\ mph/s

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The answer is C. that liquids and gases both take the shape of their container.  

Think of it this way, if you take an ice cube and put it in your glass, it will stay in its shape and stay that way until it melts.  But if you put liquid or a gas into a glass, it will take the shape of the glass that it is put into.  

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In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that
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Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m

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3\lambda/2 = L

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)

a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0

For this equation to be equal to zero, sin(59.94t) = 0. So,

59.94t = \pi\\t = \pi/59.94 = 0.0524~s

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s

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3 years ago
What gas law applies to the situation you have described above. Why?
Allushta [10]

If you give us the situation described then I'll be able to help.

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3 years ago
A cabbie is trying to stop when he notices a fare is whistling them over. The
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  • K.E=18750J
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\boxed{\sf K.E=\dfrac{1}{2}mv^2}

\\ \sf\longmapsto 18750=\dfrac{1}{2}2100v^2

\\ \sf\longmapsto 18750=1050v^2

\\ \sf\longmapsto v^2=\dfrac{18750}{1050}

\\ \sf\longmapsto v^2=17.85m^2

\\ \sf\longmapsto v=\sqrt{17.85}

\\ \sf\longmapsto v=4.1m/s

7 0
3 years ago
An assault rifle fires an eight-shot burst in 0.40 s. Each bullet has a mass of 7.5 g and a speed of 300 m/s as it leaves the gu
myrzilka [38]

Answer:

The average recoil force on the gun during that 0.40 s burst is 45 N.

Explanation:

Mass of each bullet, m = 7.5 g = 0.0075 kg

Speed of the bullet, v = 300 m/s

Time, t = 0.4 s

The change in momentum of an object is equal to impulse delivered. So,

F\times t=mv\\\\F=\dfrac{mv}{t}

For 8 shot burst, average recoil force on the gun is :

F=\dfrac{8mv}{t}\\\\F=\dfrac{7.5}{1000}\cdot\dfrac{300}{0.4}\cdot8\\\\F=45\ N

So, the average recoil force on the gun during that 0.40 s burst is 45 N.

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