At what rate is work done if the 250 Newton object from number six is moved into a hot at 4 m in four seconds

What is the effect of load resistor on ripple factor

Zolol [24]

<span>ripple factor can be reduced by increasing the value of the load resistor (which means reducing the load of the circuit)</span>

7 0

3 years ago

If you walk a distance of 8 blocks and then 3 blocks south from home, what is your position compared to home? What distance did

Harman [31]

The position compared to that of home is a reference to displacement, I believe.

Displacement = x total - x initial

So I believe the answer is 5 blocks due north (if you’re walking linearly from your home), unless the questions is referring to relative displacement, in which then you’d need to use the Pythagorean theorem to find the hypotenuse between both positions. And then you’d have to find theta for the degrees between the south direction and the other unmentioned direction. But I don’t think that’s the case.

Distance refers to x total and doesn’t care for direction, as this refers to a scalar quantity opposed to a vector. Thus the equation is just

d = x

So 8 blocks + 3 blocks = a distance of eleven blocks walked total

7 0

3 years ago

Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

Brrunno [24]

1) $+2.19\mu C$

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$ (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

8 0

3 years ago

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

7nadin3 [17]

Answer:

(a) r = 1.062·R $_E$ = $\frac{531}{500} R_E$

(b) r = $\frac{33}{25} R_E$

(c) Zero

Explanation:

Here we have escape velocity v $_e$ given by

$v_e =\sqrt{\frac{2GM}{R_E} }$ and the maximum height given by

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$

Therefore, when the initial speed is 0.241v $_e$ we have

v = $0.241\times \sqrt{\frac{2GM}{R_E} }$ so that;

v² = $0.058081\times {\frac{2GM}{R_E} }$

v² = ${\frac{0.116162\times GM}{R_E} }$

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$ is then

$\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}$

Which gives

$-\frac{0.941919}{R_E} = -\frac{1}{r}$ or

r = 1.062·R $_E$

(b) Here we have

$K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}$

Therefore we put $\frac{0.241GM}{R_E}$ in the maximum height equation to get

$\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}$

From which we get

r = 1.32·R $_E$

(c) The we have the least initial mechanical energy, ME given by

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

3 0

3 years ago

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A hydrogen atom has a diameter of about 12.7 nm, express this diameter in centimeters.

Andrej [43]

1 meter = 1e9 nm
To get meters, divide nanometers by 1e9: 9.95nm / 1x10^9 = 9.95x10^-9 meters
Answer: 9.95e-9 meters

3 0

2 years ago

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