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Elanso [62]
3 years ago
13

A student moves a box across the floor by exerting 56.7 N of force and doing 195 J of

Physics
1 answer:
Paha777 [63]3 years ago
8 0

Answer:

A. 3.4 m

Explanation:

Given the following data;

Force = 56.7N

Workdone = 195J

To find the distance

Workdone is given by the formula;

Workdone = force * distance

Making "distance" the subject of formula, we have;

Distance = \frac {workdone}{force}

Substituting into the equation, we have;

Distance = \frac {195}{56.7}

Distance = 3.4 meters.

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What is the S.I. unit of drift velocity and electron mobility?
DanielleElmas [232]
Drift velocity is equal to displacement of the moving object per unit time. The SI unit for displacement is meters while that of time is second. Hence the derived SI unit of velocity is meter per second. This also applies to electron mobility which relates to the displacement per unit time of a moving electron 
7 0
3 years ago
Draw velocity-ime graph for uniform moion of an object , when initially body is at rest.
Klio2033 [76]
When the body is at rest, its speed is zero, and the graph lies on the x-axis.

When the body is in uniform motion, the speed is constant, and the graph is a horizontal line, parallel to the x-axis and some distance above it.

It's impossible to tell, based on the given information, how these two parts of the
graph are connected.  There must be some sloping (accelerated) portion of the graph
that joins the two sections, but it cannot be accounted for in either the statement
that the body is at rest or that it is in uniform motion, since acceleration ... that is,
any change of speed or direction ... is not 'uniform' motion'.
8 0
3 years ago
A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60
Bumek [7]

Answer:

1.2cm

Explanation:

V=(2ev/m)^1/2

=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2

=6.2x10^5m/s

Radius of resulting path= MV/qB

= 1.67*10^-27x6.92*10^6/1.6*10^-16 x0.6

=0.012m

=1.2cm

5 0
3 years ago
Read 2 more answers
A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid
Sergeeva-Olga [200]

Answer:

8.37*10^5 rpm

Explanation:

Given that rotational kinetic energy = 4.66*10^9J

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = \frac{1}{2} mr ^2

Rotational K.E is illustrated as (K.E)_{rt} = \frac{1}{2} I \omega^2

\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }

\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }

\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }

\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }

\omega = 87636.04

\omega = 8.76*10^4 rad/s

Since 1 rpm = \frac{2 \pi}{60}  rad/s

\omega = 8.76*10^4(\frac{60}{2 \pi})

\omega = 836518.38

\omega = 8.37 *10^5 rpm

3 0
3 years ago
For a ∆x of 0.1mm, what is ∆px, the uncertainty in the transverse momentum of a photon passing through a slit (where uncertainty
Colt1911 [192]

Answer:

0.53\times 10^{-30}kgms^{-1}

Explanation:

Uncertainty principle say that the position and momentum can not be measured simultaneously except one relation which is described below,

\Delta x\Delta p=\frac{h}{4\pi }

Given that the uncertainty in x is 0.1 mm.

Therefore,

\Delta p=\frac{6.626\times 10^{-34} }{4\times 3.14\times 1\times 10^{-4}m }\\\Delta p=0.53\times 10^{-30}kgms^{-1}

Therefore, uncertainty in the transverse momentum of photon is 0.53\times 10^{-30}kgms^{-1}

6 0
3 years ago
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