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KatRina [158]
3 years ago
9

A climber pulls herself 8 meters upwards with a force of 150 Newtons. If it takes her 16 seconds to cover the 8 meters, how much

power is she using?
Physics
1 answer:
mr Goodwill [35]3 years ago
3 0

Answer:

 P = 75 W

Explanation:

given,

Distance, L = 8 m

Force,F = 150 N

Time, t = 16 s

Work by the climber

Work done = Force x displacement

W = F. L

W = 150 x 8

W = 1200 J

We know,

Power =\dfrac{Work\ done}{time}

P =\dfrac{1200}{16}

 P = 75 W

Hence, Power climber is using to climb is equal to 75 W.

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a student moves a box across the floor by exerting 23.3 N of force and doing 47.2 J of work on the box. How far does the student
kozerog [31]
Work = Force x Distance
47.2J = 23.3N x d
d = 47.2/23.3
d = 2.0258 m

hope this helps :P
8 0
3 years ago
The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 ​5 ​​ m/s then ho
Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10​⁵​​ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

F = \frac{MV}{t}

solving this two equations together;

\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9}  \ s \ = 1.639 \ ns

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0
3 years ago
The moon Phobos orbits Mars
dlinn [17]

Answer:

The moon Phobos orbits Mars

(mass = 6.42 x 1023 kg) at a distance

of 9.38 x 106 m. What is its period of

orbit?

Explanation:

Answer: 27.9816 x 10^3 is the period of orbit

8 0
3 years ago
How fast, in rpm, would a 5.6 kg, 25-cm-diameter bowling ball have to spin to have an angular momentum of 0.26 kgm2/s
solong [7]

Answer:

71 rpm

Explanation:

Given that:

Angular momentum (L) = 0.26

Diameter = 25cm = 0.25 cm

Radius, r = (d/2) = 0.125m

Mass = 5.6 kg

Moment of inertia (I) = 2mr² / 5

I = (2 * 5.6 * 0.125^2) / 5

= 0.175

= 0.175 / 5

= 0.035 kgm²

Angular speed (w) ;

w = L / I

w = 0.26 / 0.035

= 7.4285714

= 7.429 rad/s

w = (7.429 * 60/2π)

w = 445.74 / 2π rpm

w = 70.941724

Angular speed = 70.94 rpm

= 71 rpm

5 0
3 years ago
Which of the following wavelengths will produce standing waves on a string that is 3.5 m long?
denpristay [2]

In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:

\lambda=\frac{2}{n} L


The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:

\lambda=\frac{2}{1} \cdot 3.5 m=7.0 m


The wavelength of the 2nd harmonic is:

\lambda=\frac{2}{2} \cdot 3.5 m=3.5 m


The wavelength of the 4th harmonic is:

\lambda=\frac{2}{4} \cdot 3.5 m=1.75 m


It is not possible to find any integer n such that \lambda=5 m, therefore the correct options are A, B and D.

3 0
3 years ago
Read 2 more answers
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