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3 years ago

Jenise is buying a car for $13,085. She wants $9,400 in upgrades to the car.

Physics

2 answers:

VARVARA [1.3K]3 years ago

5 0

I think it will be 405

inysia [295]3 years ago

3 0

Answer:

$55.6

Explanation:

sorry sorry sorry sorry sorry

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Total mechanical energy (The sum of kinetic and potential energy

victus00 [196]

Answer:

Emechanical=mgh+ $\frac{1}{2}$ mν²

Explanation:

The equation for the total mechanical energy is:

Emechanical=Epotential+Ekinetic

In which,

Epotential=mgh; m: mass of the body, g: gravity; h: height

Ekinetic= $\frac{1}{2}$ mν²; m: mass of the body, ν: velocity of the body

So,

Emechanical=mgh+ $\frac{1}{2}$ mν²

5 0

3 years ago

Nerve cells transmit electric signals through their long tubular axons. These signals propagate due to a sudden rush of Na+ ions

iren [92.7K]

Answer:

$I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A$

Explanation:

For this case we have the following info given:

Number of Na+ ions $5.7 x10^{11} ions$

Each ion have a charge of +e and the crage of the electron is $1.6 x10^{-19}C$

The time is given $t = 7 ms$ if we convert this into seconds we got:

$t = 7ms * \frac{1s}{1000 ms}= 0.007s$

Now we can use the following formula given from the current passing thourhg a meter of nerve axon given by:

$Q = Ne$

Where N represent the number of ions, e the charge of the electron and Q the total charge

If we replace on this case we have this:

$Q= 5.7x10^{11} * (1.6 x10^{-19}C) = 9.12x10^{-8} C$

And from the general definition of current we know that:

$I =\frac{Q}{t}$

And since we know the total charge Q and the time we can replace:

$I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A$

The current during the inflow charge in the meter axon for this case is $13.02 \mu A$

3 0

3 years ago

The magnitude of the electric current is directly proportional to the _____ of the electric field.

Sedbober [7]

Is proportional to the CHANGE.

hope this helps

3 0

3 years ago

Read 2 more answers

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

erik [133]

Answer:

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

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

This means that the relation between the wavelength and the length of the string is

$3\lambda/2 = L$

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:

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

4 0

3 years ago

A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250

Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30$

coefficient of static friction $\mu =0.35$

coefficient of kinetic friction $\mu _k=0.25$

distance traveled $d=77.3 cm$

spring constant $k=35 N/m$

work done by gravity+work done by friction=Energy stored in Spring

$mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}$

$mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}$

$0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}$

$x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}$

$x=0.234 m$

$x=23.4 cm$

6 0

3 years ago