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3 years ago

In order to increase the amount of work done, we need to:

1 answer:

Nadusha1986 [10]3 years ago

8 0

Option (D) is the correct one.

In order to increase the amount of work done, we need to increase the force applied to the object.

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A woman is standing in the ocean, and she notices that after a wave crest passes, five more crests pass in a time of 54.0 s. The

timama [110]

Answer:

a) f=0.1 Hz ; b) T=10s

c)λ= 36m

d)v=3.6m/s

e)amplitude, cannot be determined

Explanation:

Complete question is:

Determine, if possible, the wave's (a) frequency, (b) period, (c) wavelength, (d) speed, and (e) amplitude.

Given:

number of wave crests 'n'= 5

pass in a time't' 54.0s

distance between two successive crests 'd'= 36m

a) Frequency of the waves 'f' can be determined by dividing number of wave crests with time, so we have

f=n/t

f= 5/ 54 => 0.1Hz

b)The time period of wave 'T' is the reciprocal of the frequency

therefore,

T=1/f

T=1/0.1

T=10 sec.

c)wavelength'λ' is the distance between two successive crests i.e 36m

Therefore, λ= 36m

d) speed of the wave 'v' can be determined by the product of frequency and wavelength

v= fλ => 0.1 x 36

v=3.6m/s

e) For amplitude, no data is given in this question. So, it cannot be determined.

5 0

3 years ago

A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul

Allisa [31]

Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

$KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2$

also we know that

$I = \frac{2}{5}mR^2$

$w= \frac{v}{R}$

Now kinetic energy is given by

$KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2$

$KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2$

$KE = \frac{7}{10}mR^2w^2$

$KE = \frac{7}{10}*150*(0.200)^2(50)^2$

$KE = 10500 J$

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

So in the above case work done on sphere is 10500 J

7 0

3 years ago

A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A.

Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

$\Delta K = \Delta E_{p} = q\Delta V$

<u>Where</u>:

K: is the kinetic energy

$E_{p}$ : is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV

$\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J$

Now, the kinetic energy of the particle as it moves through point B is:

$\Delta K = K_{f} - K_{i}$

$K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J$

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!

8 0

3 years ago

What are the characteristics of an ideal transformer

Vsevolod [243]

It is an imaginary transformer which has no core loss, no ohmic resistance and no leakage flux. The ideal transformer has the following important characteristic. The resistance of their primary and secondary winding becomes zero. The core of the ideal transformer has infinite permeability.

4 0

3 years ago

which of the following professionals would most likely be called upon to treat a person suffering from a dangerous level of schi

Likurg_2 [28]

<span>The primary physician in cases of schizophrenia is the psychiatrist.

hope this helps, Terrakmiller80! :-)</span>

7 0

3 years ago