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lions [1.4K]
3 years ago
7

A charge is moving in a magnetic field that points to the left. What direction can the charge move and experience no magnetic fo

rce? Check all that apply A. up B. down C. left D. right E. into the screen F. out of the screen .
What must be known to determine the direction of the magnetic force on a charge? Check all that apply. A. the type of the charge B. the amount of the charge C. the direction of the magnetic field D. the velocity of the charge E. the strength of the magnetic field
Physics
2 answers:
dlinn [17]3 years ago
7 0

its c and d fam \(-u-)/

tiny-mole [99]3 years ago
4 0
The answer is left and right. 
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C po baka namn nakatulong

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2 years ago
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the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity
NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

\int\limits^s_0 {a} \, ds  = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds  = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\

Remember that

v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t=  (5\sqrt{2} ) ln  \frac{| [s + 300 + \sqrt{(s^{2}  + 600s)} ] |}{300} .......2

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

8 0
3 years ago
NewtonsXmeters / time is a unit of which of the following:
GuDViN [60]

Answer:

energy I think I'm not sure of the answer.

7 0
2 years ago
classroom and draw an approxim object Three forces are acting on an object (Figure 1.32) which is in equilibrium. Determine forc
kumpel [21]
I’m not good with physics but I’m good with the theory what goes up must come down
6 0
2 years ago
Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho
77julia77 [94]

Answer:

The time constant is 1.049.

Explanation:

Given that,

Charge q{t}= 0.65 q_{0}

We need to calculate the time constant

Using expression for charging in a RC circuit

q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]

Where, \dfrac{t}{RC} = time constant

Put the value into the formula

0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]

1-e^{-(\dfrac{t}{RC})}=0.65

e^{-(\dfrac{t}{RC})}=0.35

-\dfrac{t}{RC}=ln (0.35)

-\dfrac{t}{RC}=-1.049

\dfrac{t}{RC}=1.049

Hence, The time constant is 1.049.

6 0
3 years ago
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