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3 years ago

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Two positively charged particles are separated on an axis by 0.1 meters, and q1 has a 6 µC charge and q2 has a 2 µC charge. Reca

ll that k = 8.99 × 109 N • meters squared over Coulombs squared.. What is the force applied between q1 and q2 in N? In which direction does particle q2 want to go?

2 answers:

ZanzabumX [31]3 years ago

8 0

Answer:

10.8 N & away from particle q1, in a straight line to the right

Explanation:

•°•°•°

Vlada [557]3 years ago

5 0

Answer:

Force between two charges is 10.8 N

both charges are positive nature charge so they both repel each other and hence q2 will move away from charge q1

Explanation:

As we know that the force between two charge particles is given by Coulomb's law of electrostatics

So we will say it as

$F = \frac{kq_1q_2}{r^2}$

here we know that

$q_1 = 6\mu C$

$q_2 = 2\mu C$

$d = 0.1 m$

$F = \frac{8.99 \times 10^9(6 \mu C)(2\mu C)}{0.1^2}$

$F = 10.8 N$

Since both charges are positive nature charge so they both repel each other and hence q2 will move away from charge q1

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Pleaseeeee help me with b, c, and d. There are no angles.

taurus [48]

Answer:

a. 150 J

b. 150 J

c. 0 J

d. 0 J

Explanation:

The given parameters are;

The horizontal force with which the man pulls the canister, F = 50 N

The distance he moves the vacuum cleaner, d = 3.0 m

a. Work done, W = Force applied, F × Distance moved by the force, d

Therefore, for the work done by the 50 N force on the canister, we have;

W = 50 N × 3.0 m = 150 N·m = 150 J

b. Given that he pulls the canister at a constant speed, we have;

The acceleration of the canister, a = 0 m/s²

Therefore, the net force on the canister, $F_{NET}$ = F - $F_{Friction}$ = m × a

Where;

m = The mass of the canister

a = The acceleration of the canister

F = The applied force = 50 N

$F_{Friction}$ = The force of friction

∴ $F_{NET}$ = m × a = m × 0 m/s² = 0 N

Therefore;

$F_{NET}$ = F - $F_{Friction}$ = 0 N

From which we have;

F = $F_{Friction}$ = 50 N (The applied force, F is equal to the force of friction,

The work done by friction = The force of friction × The distance in which the force of friction acts

∴ The work done by friction = $F_{Friction}$ × d - 50 N × 3.0 m = 150 J

The work done by friction = 150 J

c. The normal force, N acts perpendicular to the force of friction

The distance the canister moves in the perpendicular direction, $d_p$ = 0 m

∴ The work done by the normal direction = N × $d_p$ = N × 0 m = 0 J

The work done by the normal direction = 0 J

d. The vacuum weight, W, acts on the same line as the normal force but in the opposite direction to the normal force, N

Therefore, the weight, W, acts perpendicular to the line of motion of canister

The distance the canister moves in the direction of the weight, $d_{wieght}$ = 0 m

Therefore, the work done by the weight = W × $d_{wieght}$ = W × 0 m = 0 J

The work done by the weight = 0 J

7 0

3 years ago

Water has a specific heat capacity nearly nine times that of iron. Suppose a 50-g pellet of iron at a temperature of 200∘C is dr

miv72 [106K]

Answer:

a. closer to 20∘C

Explanation:

$m_{p}$ = mass of pallet = 50 g = 0.050 kg

$c_{p}$ = specific heat of pallet = specific heat of iron

$T_{pi}$ = Initial temperature of pellet = 200 C

$m_{w}$ = mass of water = 50 g = 0.050 kg

$c_{w}$ = specific heat of water

$T_{wi}$ = Initial temperature of water = 20 C

$T_{e}$ = Final equilibrium temperature

Also given that

$c_{w} = 9 c_{p}$

Using conservation of energy

Energy gained by water = Energy lost by pellet

$m_{w} c_{w} (T_{e} - T_{wi}) = m_{p} c_{p} (T_{pi} - T_{e})\\(0.050) (9) c_{p} (T_{e} - 20) = (0.050) c_{p} (200 - T_{e})\\ (9) (T_{e} - 20) = (200 - T_{e})\\T_{e} = 38 C$

hence the correct choice is

a. closer to 20∘C

4 0

4 years ago

A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.

kodGreya [7K]

Answer:

The z-component of the force is $\= F_z = 0.00141 \ N$

Explanation:

From the question we are told that

The charge on the particle is $q = 7.76 *0^{-8} \ C$

The magnitude of the magnetic field is $B = 0.700\r i \ T$

The velocity of the particle toward the x-direction is $v_x = -1.68*10^{4}\r i \ m/s$

The velocity of the particle toward the y-direction is

$v_y = -2.61*10^{4}\ \r j \ m/s$

The velocity of the particle toward the z-direction is

$v_y = -5.85*10^{4}\ \r k \ m/s$

Generally the force on this particle is mathematically represented as

$\= F = q (\= v X \= B )$

So we have

$\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)$

$\= F = q (v_y B(-\r k) + v_z B\r j)$

substituting values

$\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)$

$\= F= 0.00303\ \r j +0.00141\ \r k$

So the z-component of the force is $\= F_z = 0.00141 \ N$

Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).

7 0

3 years ago

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CaHeK987 [17]

Answer:

Explanation:

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8 0

3 years ago

A squirrel jumps down from a tall tree. Assuming the squirrel is in free fall, how far will it have fallen in 1.5 seconds?

snow_tiger [21]

D = 1/2gt^2 <-- free fall formula

d = (0.5)(9.81)(1.5^2)
d = 11.03625
d = 11.04 m

hope this helps and have a great day :)

7 0

3 years ago

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