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Paraphin [41]
2 years ago
10

A small metal ball is given a negative charge, then brought near to end a of the rod (figure 1). What happens to end a of the ro

d when the ball approaches it closely this first time?.
Physics
1 answer:
erma4kov [3.2K]2 years ago
7 0

What happens to end a of the rod when the ball approaches it closely this first time is; It is strongly attracted.

<h3>Electrostatics</h3>

I have attached the image of the rod.

We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.

This means that their fields will cancel.

Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.

This also applies to a strong conducting rod and therefore it is strongly attracted.

Read more about Electrostatics at; brainly.com/question/18108470

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A baseball is thrown vertically up to a height of 30 m on Earth. If the same ball is thrown up on the moon with the same initial
nekit [7.7K]

Answer:

hmax = 181.48m

Explanation:

In order to calculate the maximum height reached by the same ball in the moon, you first calculate the initial velocity of the ball by using the information about the maximum height on the Earth. You use the following formula:

h_{max}=\frac{v_o^2}{2g}     (1)

hmax: maximum height reached by the ball in the Earth = 30m

vo: initial velocity of the ball = ?

g: gravitational acceleration on Earth = 9.8m/s^2

From the equation (1) you solve for vo:

v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(30m)}=24.24\frac{m}{s}

Next, you use the same equation (1) but for the gravitational acceleration of the moon, which is given by:

g' = 1.62m/s^2

h_{max}=\frac{(24.24m/s)^2}{2(1.62m/s^2)}=181.48m

The same ball, with the same initial velocity, will reache a heigth of 181m in the moon.

6 0
3 years ago
The equation for gear ratio
Ymorist [56]
In a gear train with two gears, the gear ratio is defined as follows
R= \frac{\omega _A}{\omega _B} 

where \omega _A is the angular velocity of the input gear while \omega _B is the angular velocity of the output gear. 

This can be rewritten as a function of the number of teeth of the gears. In fact, the angular velocity of a gear is inversely proportional to the radius r of the gear:
\omega = \frac{v}{r}
But the radius is proportional to the number of teeth N of the gear. Therefore we can rewrite the gear ratio also as
R= \frac{\omega _A}{\omega _B} = \frac{r_B}{r_A} = \frac{N_B}{N_A}

4 0
3 years ago
1.<br> Kinetic energy is the energy of _____.
lorasvet [3.4K]

Explanation:

kinetic energy is energy that it possesses due to its motion.

7 0
3 years ago
A father racing his son has 1/2 the kinetic energy of the son, who has 1/3 the mass of the father. The father speeds up by 1.4 m
BaLLatris [955]

Answer: a) 0.78 m/s b) 1.57 m/s

Explanation:

M = father's mass


m = son's mass = M/3


V = father's initial speed


v = son's initial speed

(1/2)MV^2 = (1/2)*(1/2)*m v^2


M*V^2 = (1/2)(M/3)v^2


V^2/v^2 = 1/4


V = v/2

Second equation:


(1/2)M*(V + 1.4)^2 = (1/2)m*v^2


= (1/2)*(M/3)*(3V)^2


cancel out the M's and (1/2)'s


(V + 1.4)^2 = 3V^2


V^2 + 2.8V + 1.96 = 3V^2


V^2 -1.4V -0.98 = 0

V^2 = 0.98/0.4 = 2.45

V = 1.57

3 0
3 years ago
Read 2 more answers
Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.8 x 105 kg on springs that hav
Vikki [24]

Answer:

The force constant is  k =1.316 *10^{7} \  N/m

The energy stored in the spring is  E =  1.68 *10^{7} \ J

Explanation:

From the question we are told that

   The mass of the object is  M  = 4.8*10^{5} \ kg

    The period is T  = 1.2 \ s

The period of the spring oscillation is  mathematically represented as

         T  =2 \pi \sqrt{ \frac{M}{k}}

where  k is the force constant

   So making k the subject

       k = \frac{4 \pi ^2 M }{T^2}

substituting values

       k = \frac{4 (3.142) ^2 (4.8 *10^{5}) }{(1.2)^2}

      k =1.316 *10^{7} \  N/m

The energy stored in the spring is mathematically represented  as

       E =  \frac{1}{2} k x^2

Where x is the spring displacement which is given as

        x =  1.6 \ m

substituting values

      E =  \frac{1}{2} (1.316 *10^{7}) (1.6)^2

       E =  1.68 *10^{7} \ J

   

7 0
3 years ago
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