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Tatiana [17]
3 years ago
10

By how much should the pressure of a litre of water be changed to compress it by 0.10% ?​

Physics
1 answer:
wel3 years ago
8 0

Answer:

it is given that the water is to be compressed by 0.10%. Therefore, the pressure on the water should be

2.2 × 10^6 Nm ^ -2 .

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A pair of closely spaced parallel conducting plates, charged with equal and opposite electric charges, produces a uniform electr
Sedbober [7]

Answer:

Plate B.

Explanation:

If the direction of the electric field is from plate A to plate B, then this means that plate A is positively charged and plate B is negatively charged. If we are to move an electron between the plates, then we should place the electron on plate B, so the negatively charged electron can be attracted by the positive charges on plate A.

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3 years ago
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A student presses a book between his hands, as the drawing indicates. The forces that he exerts on the front and back covers of
Aleksandr-060686 [28]

Answer: 39.5 N

Explanation:

2 x force x static friction coefficient = weight  (weight=2μsF)

force = ?

static friction coefficient = 0.416

weight = 32.9N

substituting the values above into the equation we get

2 x F x 0.416 = 32.9

F = 32.9 / (2 x 0.416)

F = 39.5N

4 0
3 years ago
the circumference of Earth's orbit is 938 million kilometres how fast does earth travel in km/h in one year​
allsm [11]

Answer:

about 940 million km each year

Explanation:

3 0
3 years ago
Points A and B lie within a region of space where there is a uniform electric field that has no x- or z-component; only the y-co
liraira [26]

Answer:

(a) Ey is negative

(b) The magnitude of the electric field is E = 171.429 V/m

(c) The potential difference between points B and C is 17.1429 V

Explanation:

(a) Here, we have the potentials given by;

V_A - V_B = +12.0V with point A at y = 8.00 cm and point B at point y = 15.0 cm

where point B is at a higher potential than point A, that is the electric potential is from;

B with y = 15.0 cm to A with y = 8.0 cm which means

E_y decreases as y increases or E_y  is negative.

(b) The magnitude of the electric field is given by

The work done to move a charge from B to A is

W_{BA} = - \Delta U where

\Delta U = U_a -U_b = q_0E(y_b-y_a)

V_{BA} = \frac{\Delta U}{q_0} = \frac{q_0E(y_b-y_a)}{q_0}  = E(y_b-y_a)

∴ E = \frac{V_{BA}}{(y_b-y_a)}

E = \frac{12 \hspace{0.09cm}V}{(0.015\hspace{0.09cm} m -0.008\hspace{0.09cm} m)}

E = 171.429 V/m

(c) Here we have point C x = 5.00 cm and y = 5.00 cm

Therefore we have the distance from B to C given by

y_b-y_c = 15.00 \hspace{0.09cm}cm - 5.00  \hspace{0.09cm}cm = 10.00 \hspace{0.09cm} cm

Where 10.00 cm = 0.01 m

E = V/Δy

Therefore, V = Δy·E

For V_{BC}, Δy = y_b-y_c  = 0.01 \hspace{0.09cm} m and we have,

V_{BC} = E\times (y_b-y_c)

V_{BC} = 171.429\times (0.015-0.005) = 17.1429\hspace{0.09cm}V

7 0
3 years ago
Gravitational Potential Energy = mass x gravity x height
Blizzard [7]
M= gpe / gh
G= gpe / mh
H=gpe / mg

7 0
4 years ago
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