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1 year ago

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Cori uses 475 J of energy from her muscles to push a bar 1 m on a weight machine at the gym. Between the bar’s motion , the heat

generated by the machine, and all other types of work done by machine, ( blank) J of output work will be done.

2 answers:

Softa [21]1 year ago

8 0

Answer:

Cori uses 475 J of energy from her muscles to push a bar 1 m on a weight machine at the gym. Between the bar’s motion , the heat generated by the machine, and all other types of work done by machine, ( 475 ) J of output work will be done.

Explanation:

on edge

dimulka [17.4K]1 year ago

5 0

Answer:

475

Explanation:

Cori does not exert any more force than 475 J, so 475 is the answer.

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Consider the waveform expression. y (x, t) = ym sin (0.333x + 5.36 + 585t) The transverse displacement (y) of a wave is given as

Sonja [21]

Explanation:

The waveform expression is given by :

$y(x,t)=y_m\ sin(0.333x+5.36+585t)$ ...........(1)

Where

y is the position

t is the time in seconds

The general waveform equation is given by :

$y(x,t)=y_m\ sin(kx+\phi+\omega t)$ ..........(2)

Where

$k=\dfrac{2\pi}{\lambda}$

$\omega=2\pi f$

On comparing equation (1) and (2) we get :

$0.333=\dfrac{2\pi}{\lambda}$

$\lambda=18.86\ m$

$585=2\pi f$

f = 93.10 Hz

Time period, $T=\dfrac{1}{f}$

$T=\dfrac{1}{0.010}$

T = 0.010 s

Phase constant, $\phi=5.36\ radian$

Hence, this is the required solution.

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2 years ago

Question 2 (1 point)

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Answer:

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Explanation:

3p - 3e = 0 and that leaves 2 neutrons so it will be neutral

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2 years ago

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An activity that is relatively short in time (< 10 seconds) and has few repetitions predominately uses the _____________ ener

tensa zangetsu [6.8K]

An activity that is relatively short in time <10 seconds and has few repetitions predominantly uses the ATP/PC energy system. The cellular respiration procedure that changes food energy into ATP which is a form of energy is largely reliant on oxygen obtainability. During exercise the source and request of oxygen obtainable to muscle is unnatural by period and strength and by the individual’s cardiorespiratory suitability level.
Steps of the ATP-PC system:

1. Primarily, ATP kept in the myosin cross-bridges which is microscopic contractile parts of muscle is broken down to issue energy for muscle shrinkage. This action consents the by-products of ATP breakdown which are the adenosine diphosphate and one single phosphate all on its own.

2. Phosphocreatine is then broken down by the enzyme creatine kinase into creatine and phosphate.

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2 years ago

A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through

Svet_ta [14]

Answer:

The speed is $\sqrt{2}v_{0}$ .

(a) is correct option.

Explanation:

Given that,

Potential difference $V= V_{0}$

Speed $v = v_{o}$

If it were accelerated instead

Potential difference $V'=2V_{0}$

We need to calculate the speed

Using formula of initial work done on proton

$W = q V$

We know that,

$\Delta W=\Delta K.E$

$q V=\dfrac{1}{2}mv^2$

Put the value into the formula

$q V_{0}=\dfrac{1}{2}mv_{0}^2$

$v_{0}^2=\dfrac{2qV_{0}}{m}$ ....(I)

If it were accelerated instead through a potential difference of $2 V_{0}$ , then it would gain a speed will be given as :

Using an above formula,

$v_{0}'^2=\dfrac{2qV_{0}}{m}$

Put the value of $V_{0}$

$v_{0}'^2=\dfrac{2q\times2V_{0}}{m}$

$v_{0}'=\sqrt{\dfrac{4qV_{0}}{m}}$

$v_{0}'=\sqrt{2}v_{0}$

Hence, The speed is $\sqrt{2}v_{0}$ .

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1 year ago

An electron in a vacuum chamber is fired with a speed of 9800 km/s toward a large, uniformly charged plate 75 cm away. The elect

melisa1 [442]

Answer:

The plate's surface charge density is $-8.056\times10^{-9}\ C/m^2$

Explanation:

Given that,

Speed = 9800 km/s

Distance d= 75 cm

Distance d' =15 cm

Suppose we determine the plate's surface charge density?

We need to calculate the surface charge density

Using work energy theorem

$W=\Delta K.E$

$W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{i}^2$

Here, final velocity is zero

$W=0-\dfrac{1}{2}mv_{i}^2$ ...(I)

We know that,

$W=-Fd$

$W=-E\times e\times d$

$W=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d$ ...(II)

From equation (I) and (II)

$-\dfrac{1}{2}mv_{i}^2=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d$

Charge is negative for electron

$\lambda=\dfrac{mv^2\epsilon_{0}}{(-e)d}$

Put the value into the formula

$\lambda=-\dfrac{9.1\times10^{-31}\times(9800\times10^{3})^2\times8.85\times10^{-12}}{1.6\times10^{-19}\times(75-15)\times10^{-2}}$

$\lambda=-8.056\times10^{-9}\ C/m^2$

Hence, The plate's surface charge density is $-8.056\times10^{-9}\ C/m^2$

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2 years ago