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rodikova [14]
4 years ago
13

A magnetic field applies forces on:a)static chargesb)moving chargesc)water flow​

Physics
2 answers:
lisabon 2012 [21]4 years ago
4 0

Answer:

moving charges b)

I think this is the answer

konstantin123 [22]4 years ago
3 0

Answer:

It is moving charges.

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A ruler vibrates 70 times in 20 seconds.what is the frequency of its vibration
Bad White [126]

Answer:

we know that that frequency= n/t

= 70/20

= 3.5 Hertz

Explanation:

i think that the answer

8 0
3 years ago
If a ball is thrown vertically upward from the roof of 32 foot building with a velocity of 64 ft/sec, its height after t seconds
stepan [7]

Answer:

a) s_{max} = 96\,ft, b) v(4.449\,s) = -78.368\,\frac{ft}{s}

Explanation:

a) The maximum height is obtained with the help of the First and Second Derivative Tests:

First Derivative

v(t) = 64 - 32\cdot t

64 - 32\cdot t = 0

t = 2\,s

Second Derivative

a(t) = -32 (absolute maximum)

The maximum height reached by the ball is:

s (2\,s) = 32 + 64\cdot (2\,s) - 16\cdot (2\,s)^{2}

s_{max} = 96\,ft

b) The time required by the ball to hit the ground is:

32+64\cdot t - 16\cdot t^{2} = 0

-16\cdot (t^{2}-4\cdot t - 2) = 0

t^{2}-4\cdot t - 2 = 0

(t -4.449)\cdot (t+0.449)\approx 0

Just one root offers a solution that is physically reasonable:

t = 4.449\,s

The velocity of the ball when it hits the ground is:

v(4.449\,s) = 64 - 32\cdot (4.449\,s)

v(4.449\,s) = -78.368\,\frac{ft}{s}

6 0
3 years ago
Read 2 more answers
A critical period begins and ends abruptly
STatiana [176]

Answer:

yes

Explanation:

<em>A critical period? </em>

<em>A.begins and ends abruptly </em>

<em>B.begins and ends gradually </em>

<em>C.is unaffected by stimuli </em>

<em>D.is unlikely to impact development</em>

<em>correct answer (A.begins and ends abruptly )</em>

8 0
3 years ago
Please help meeeeeeeeeeeeeeee
maxonik [38]

probabilityAnswer:

2/27

Explanation:

The elk can not be eaten so we remove that from the probablity

so we have x/18

songbird = 4/18

mice = 6/18

4/18*6/18 = 2/27

5 0
2 years ago
A marble runs off the edge of a table that is 1.5 m high and the marble lands 0.50 m from the base of the table. a. How much tim
Free_Kalibri [48]

Answer:

t = 0.55[sg]; v = 0.9[m/s]

Explanation:

In order to solve this problem we must establish the initial conditions with which we can work.

y = initial elevation = - 1.5 [m]

x = landing distance = 0.5 [m]

We set "y" with a negative value, as this height is below the table level.

in the following equation (vy)o is equal to zero because there is no velocity in the y component.

therefore:

y = (v_{y})_{o}*t - \frac{1}{2} *g*t^{2}\\   where:\\(v_{y})_{o}=0[m/s]\\t = time [sg]\\g = gravity = 9.81[\frac{m}{s^{2}}]\\ -1.5 = 0*t -4.905*t^{2} \\t = \sqrt{\frac{1.5}{4.905} } \\t=0.55[s]

Now we can find the initial velocity, It is important to note that the initial velocity has velocity components only in the x-axis.

(v_{x} )_{o} = \frac{x}{t} \\(v_{x} )_{o} = \frac{0.5}{0.55} \\(v_{x} )_{o} =0.9[m/s]

3 0
4 years ago
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