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2 years ago

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An object with charge q=−6.00×10−9C is placed in a region of uniform electric field and is released from rest at point A. After

the charge has moved to point B, 0.500 m to the right, it has kinetic energy 3.00×10−7J.
(a) If the electric potential at point A is +30.0 V, what is the electric potential at point B?
(b) What are the magnitude and direction of the electric field?

1 answer:

shtirl [24]2 years ago

8 0

Answer:

its a

Explanation:

used google

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Sharks are generally negatively buoyant; the upward buoyant force is less than the weight force. This is one reason sharks tend

Tresset [83]

Answer:

8.67807 N

34.7123 N

Explanation:

m = Mass of shark = 92 kg

$\rho_{se}$ = Density of seawater = 1030 kg/m³

$\rho_{f}$ = Density of freshwater = 1000 kg/m³

$\rho_{sh}$ = Density of shark = 1040 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Net force on the fin is (seawater)

$F_n=mg-V_s\rho_{se}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{se}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1030\times 9.81\\\Rightarrow F_n=8.67807\ N$

The lift force required in seawater is 8.67807 N

Net force on the fin is (freshwater)

$F_n=mg-V_s\rho_{f}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{f}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1000\times 9.81\\\Rightarrow F_n=34.7123\ N$

The lift force required in a river is 34.7123 N

6 0

2 years ago

A bucket of water experiencing a gravitational force of 525 N is pulled up from a water well. The net force in the y-direction i

lukranit [14]

Answer:

6n!!!!!!!!!!!!!!!!!!

Explanation:

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2 years ago

En la etiqueta de un bote de fabada de 350 g, leemos que su aporte energético es de 1630 kj por cada 100 g de producto a) La can

fredd [130]

Answer:

(a) 153.37 g

(b) 5705 kJ

Explanation:

(a) To find the amount of bean needed by a man you first calculate the equivalence in beans to 2500kJ

$2500kJ*\frac{100g}{1630kJ}=153.37\ g$

Thus, 153.37 g has the energy needed by a man that needs 200kJ per day.

(b) The amount of energy per pot of bean is given by:

$E=350g*\frac{1630kJ}{100g}\\\\E=5705\ kJ$

Thus, the energy is 5705kJ

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2 years ago

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i

Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

$V = 4\frac{kQ}{r}$

here

r = distance of corner of the square from it center

$r = \frac{a}{\sqrt2}$

$r = \frac{10nm}{\sqrt2} = 7.07 nm$

$Q = e = -1.6 * 10^{-19} C$

now the net potential is given as

$V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}$

$V = 0.815 V$

now potential energy of alpha particle at this position

$U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J$

Now at the mid point of one of the side

Electrostatic potential is given as

$V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}$

here we know that

$r_1 = \frac{a}{2} = 5 nm$

$r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}$

$r_2 = 11.2 nm$

now potential is given as

$V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}$

$V = -0.576 - 0.257 = -0.833 V$

now final potential energy is given as

$U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J$

Now work done in this process is given as

$W = U_f - U_i$

$W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}$

$W = -7 * 10^{-22} J$

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2 years ago

A bucket of water experiencing a gravitational force of 525 N is pulled from a water well. Net force in the Y direction is 45 N

vivado [14]

Answer:

T = 570 N

Explanation:

Given that,

The gravitational force acting on a bucket of water = 525 N

Net force in the Y direction is 45 N

We need to find the magnitude of the force of tension. It can be calculated as :

45 = T - 525

T = 525 + 45

T = 570 N

Hence, the force of tension is 570 N.

7 0

2 years ago