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Vitek1552 [10]
3 years ago
10

A 30kg bucket of water is lifted vertically 3.0m. How much work was done on the bucket during this movement

Physics
2 answers:
Alex3 years ago
3 0

Answer:

Explanation:HELPPPPPPP

nikdorinn [45]3 years ago
3 0

Answer:900joules

Explanation:

mass=30kg

Height=3m

g=10m/s^2

Work=mass x g x height

Work=30 x 10 x 3

Work=900joules

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FIGURE 1 shows part of a mass spectrometer. The whole arrangement is in a vacuum. Negative ions of mass 2.84 x 10-20 kg and char
yuradex [85]

Yes, the ions can exit slit P without being deflected, if the electric field strength is 170.6 N/C

Explanation:

When the ions are inside the container, they are subjected to two forces, with directions opposite to each other:

  • The force due to the electric field, whose magnitude is F_E=qE, where q is the charge of the ion and E is the strength of the electric field
  • The force due to the magnetic field, whose magnitude is F_B=qvB, where v is the speed of the ions and B is the strength of the magnetic field

The ions will move straight and undeflected if the two forces are equal and opposite. By using Fleming Left Hand rule, we notice that the magnetic force on the (negative) ions point upward: this means that the electric field must be also upward (so that the electric force on the ions is downward). Then, the two forces are balanced if

F_E = F_B

which translates into

qE=qvB\\\rightarrow v = \frac{E}{B}

Therefore, if the speed of the ions is equal to this ratio, the ions will go undeflected.

We can even calculate the value of E at which this occurs. In fact, we know that the ions are earlier accelerated by a potential difference V=-3000 V, so we have that their kinetic energy is given by the change in electric potential energy:

qV=\frac{1}{2}mv^2

where

q=-2.0\cdot 10^{-19}C\\m=2.84\cdot 10^{-20}kg

Solving for v, the speed,

v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(-2.0\cdot 10^{-19})(-3000)}{2.84\cdot 10^{-20}}}=205.6 m/s

And since the magnetic field strength is

B = 0.83 T

The strength of the electric field must be

E=vB=(205.6 m/s)(0.83 T)=170.6 N/C

Learn more about electric and magnetic fields:

brainly.com/question/8960054

brainly.com/question/4273177

brainly.com/question/3874443

brainly.com/question/4240735

#LearnwithBrainly

7 0
3 years ago
A cyclist rides 6.2 km east, then 9.28 km in a direction 27.27 degrees west of north, then 7.99 km west. A. What is the magnitud
Pani-rosa [81]

Answer:

Explanation:

given,

cyclist ride  6.2 km east and then 9.28 km in the direction of 27.27° west of north and then 7.99 km west.

vertical component = 9.28 cos∅

                                = 9.28 cos 27.27°

                                = 8.24 km

horizontal axis component = 9.28 sin ∅

                                             = 9.28 sin 27.27°

                                             = 4.5 km

distance of the final point from the origin

                            = 7.99 -(6.2-4.5)

                            = 6.29 km

displacement

d = \sqrt{6.29^2+8.24^2}

d = 10.37 km

b) tan \theta = \dfrac{6.29}{8.24}

θ = 37.36°

4 0
3 years ago
A car goes from 5 m/s to 25 m/s in 6 s. What is the acceleration of the car?
damaskus [11]

Answer:

5m/s

Explanation:

3 0
3 years ago
What is meant by the statement that a laser beam is coherent, mono-chromatic and parallel?
Artyom0805 [142]

Answer:

Laser light however contains only one wavelength. This property makes lasers monochromatic, meaning of one color. Another property of lasers is that all the wavelengths are in phase, meaning they wave together. This property is called coherency. Laser light travels in the same direction, parallel to one another.

8 0
3 years ago
Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

q_1 = 21.0\mu C

q_1 = 47.0\mu C

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

E = \frac{kq}{x}^2

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2

solving for x we get

\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}

x = 0.200 m

b)electric field at half way mean x =0.25

E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}

E = 3.84*10^{-4} N/C

6 0
3 years ago
Read 2 more answers
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