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ycow [4]
3 years ago
14

The force that moving charged particles exert on one another is called

Physics
1 answer:
Irina-Kira [14]3 years ago
6 0
It should be electromagnetic force
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Please help: I don't know how to do these problems
antiseptic1488 [7]
d =2.55.68m and t = 11.36s
In my opinion
3 0
3 years ago
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
Thank you so muchhhhh
loris [4]

Answer:

I'm taking a wild guess at c

Explanation:

c. winter solstice

6 0
3 years ago
A plane traveling horizontally at 120 ​m/s over flat ground at an elevation of 3610 m must drop an emergency packet on a target
allsm [11]

Answer:

Explanation:

Horizontal displacement

x = 120 t

Vertical position

y = 3610 - 4.9 t²

y = 0 for the ground

0 = 3610 - 4.9 t²

t = 27.14 s

This is the time it will take to reach the ground .

During this period , horizontal displacement

x = 120 x 27.14 m

= 3256.8 m

So packet should be released 3256.8 m before the target.

3 0
3 years ago
Read 2 more answers
A golf club hits a 0.04551 kg golf ball off a golf tee. The club is in contact with the ball for 0.020 s, and the force applied
Juliette [100K]

Answer:

v = 50.5 m/s

Explanation:

F = (m)(^v/^t)

115N = (0.04551kg)(v/(0.020s))

2,526.917161 m/s² = v/(0.020s)

v = 50.53834322 m/s

v = 50.5 m/s

8 0
3 years ago
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