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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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8 0
3 years ago
An arrow is shot upward from the roof of a building 75 m high with an initial speed of 40 m/s. If g = 10 m/s2 , what total dista
Elenna [48]

Answer:

60m

Explanation:

According to one of the equation of motions, v² = u²+2as where;

S is the distance

u is the initial velocity

v is the final velocity

a is the acceleration

Since the arrow is shot upwards, the body will experience a negative acceleration due to gravity i.e a = -g

Therefore our equation will become;

v² = u² - 2gS

Given u = 40m/s, g = 10m/s², S = 75m

Substituting to get the final velocity of the arrow we will have;

v² = 40²-2(10)(75)

v² = 1600 - 1500

v² = 100

v = √100

v = 10m/s

Total distance traveled is speed of the object × time taken

Total distance traveled = 10 × 6

= 60m

The arrow has therefore traveled 60m after 6seconds

3 0
3 years ago
A _____ space zone is one that is obstructed in some way
telo118 [61]
A closed space zone is one that is obstrutced in some way.

Hope this helps :)
5 0
3 years ago
Read 2 more answers
A 2:2 kg toy train is con ned to roll along a straight, frictionless track parallel to the x-axis. The train starts at the origi
Liula [17]

Answer:

a) 10.51 J

b) 3.48 m/s

Explanation:

Given data :

mass of train ( M ) = 2.2 kg

Given initial velocity ( u ) = 1.6 m/s

<u>a) calculating work done by the force over the journey of the train</u>

F = mx + b  ------ ( 1 )

m = slope  = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m

x = distance travelled on the x axis by the train = 7.5 m

F = force experienced by the train = 2.8 N

x = 0

∴ b = 2.8

hence equation 1 can be written as

F = ( -0.373) x + 2.8   ----- ( 2 )

hence to determine the work done by the force

W   = \int\limits^7_0 { ( -0.373) x + 2.8  )} \, dx     Note:  the limits are actually 7.5 and 0

∴ W ( work done ) = -10.49 + 21 = 10.51 J

<u>b) calculate the speed of the train at the end of its journey</u>

we will apply the work energy theorem

W = 1/2 m*v^2  -  1/2 m*u^2

∴ V^2 = 2 / M ( W + 1/2 M*u^2 )  ( input values into equation )

 V^2 = 12.11

hence V = 3.48 m/s

6 0
3 years ago
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Natalka [10]
Low coefficient of friction

1. flying a plane (friction between air and plane)
2. ice skating (friction between ice and skate blade)
3. swimming (water & skin)
4. rowing a boat (water and boat)

6 0
3 years ago
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