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hichkok12 [17]
3 years ago
6

Two balloons having the same charge of 1.2 × 10-6 coulombs each are kept 5.0 × 10-1 meters apart. What is the magnitude of the r

epulsive force between them? (k = 9.0 × 109 newton·meter2/coulombs2)
1.2 × 10-2 newtons

3.2 × 10-3 newtons

6.5 × 10-3 newtons

3.0 × 10-4 newtons

5.2 × 10-2 newtons
Physics
1 answer:
ra1l [238]3 years ago
4 0
Steps: 1)Data: q1=charge of first balloon=1.2 × 10^-6 q2=charge of second balloon=1.2 × 10^-6 Condition:q1=q2 then for convenience,we say q instead of q1 and q2. Radius=r=5*10^-1 Radius is the distance between balloons! Magnitude of repulsive force=F=? 2)Solution: F=kq1q2/r^2 As q1=q2=q So F=kqq/r^2 =kq^2/r^2 Putting values: F=((9*10^9)(1.2 × 10^-6)^2)/(5*10^-1)^2 Using calculator: F=0.05184 N
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How many ohms of resistance must be present in a circuit that has 240 volts and a current of 15 amps?
PIT_PIT [208]

Answer:

16 ohms

Explanation:

V= I ⋅ R

where,  V  is the net potential difference in the circuit,  I  is the current in the circuit and  R  is the net resistance of the circuit.  

In this case,  V = 240  volts,  I = 15  amperes.

240 = 15 ⋅ R  

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1) Subatomic particles called muons can be created in the upper atmosphere by collisions of cosmic rays (energetic particles com
Vsevolod [243]

Answer and explanation:

A.

Muon travelled straight down towards the earth. Therefore the tree moves up in the rest frame of muon (option a)

B.

In muon rest frame it travels Zero meters

C.

Distance, d = Velocity, v * Time, s

where, v = 0.9c = 0.9 \times 8 \times 10^8 , s = 2.2 \mu s

d = 0.9 \times 3 \times 10^8 \times 2.2 \times 10^{-6}\\\\d = 594m

D.

Distance from the top of the mountain to the tree is the same as the distance travelled by the tree in the muons rest frame

that is same as in part C which is 594m

E.

Using lorentz contraction

In the rest frame of someone standing on the mountain

the distance is given by

d' = \frac{d}{\gamma} = d\sqrt{1 - \frac{v^2}{c^2}}, where, \frac{1}{\gamma}= \sqrt{1 - \frac{v^2}{c^2}}

d' = 594\sqrt{1 - \frac{(0.9c)^2}{c^2}}

d' = 594\sqrt{1 - 0.81}

d' = 594 \times 0.4359

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F.

in the rest frame of someone standing on the mountain,

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3 years ago
1. Differentiate between speed and velocity.
Mazyrski [523]

Answer:

1. Speed and velocity both involve a numeric rate describing the distance traveled by a body in a unit of time. However, speed describes the rate of a body traveling in any direction in a unit of time, while velocity describes the rate of a body traveling in a particular direction in a unit of time.

2. Answers may vary, but should resemble the following:

Average velocity explains the velocity the body traveled overall, not taking into consideration each spot in the trip. If a car moves at 65 km/h on average, it may have slowed down for some parts and sped up for others. Overall though, it would have made a certain distance of travel within a specified unit of time that totals the average velocity of 65 km/h.

Instantaneous velocity explains the velocity of a body at a particular instant of the trip. The instantaneous velocity of a car stopped at a stop sign would be 0 m/s even if it was moving before and will continue to move after this stop. The velocity at that particular instant is the instantaneous velocity.

Uniform velocity is when the distance being covered is changing uniformly with time. For example, if a car moves 20 km every 30 minutes and continues to do so in the same direction, it's traveling with a uniform velocity.

3. a=v2−v1t

a=20 m/s−60 m/s6 s

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a = –6.7 m/s2

4. v2 = v1 + at

v2 = 14 m/s + (3 m/s2 × 6 s)

v2 = 14 + 18

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5. v=st

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6. First, convert the minutes to seconds. Since there are 60 seconds in one minute, multiply:

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s = v × t

s = 6 m/s × 900 s

s = 5,400 m

7. t=sv

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t = 2.29 hr

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a=50 m/s−15 m/s4 s

a=35 m/s4 s

a = 8.75 m/s2

9. vav=v1+v22

vav=15 m/s+50 m/s2

vav=65 m/s2

vav = 32.5 m/s

10. a=v2−v1t

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a = –3.29 m/s2

Explanation:

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