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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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Write the adverbs use in sentences no.1-5 ty​
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Answer:

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5 0
3 years ago
For a short period of time, the frictional driving force acting on the wheels of the 2.5-Mg van is N, where t is in seconds. If
KATRIN_1 [288]

Answer:

15.6m/s

Completed Question;

For a short period of time, the frictional driving force acting on the wheels of the 2.5-Mg van is N= 600t^2 , where t is in seconds. If the van has a speed of 20 km/h when t = 0, determine its speed when t = 5

Explanation:

Mass m = 2500kg

Speed v1 = 20km/h = 20/3.6 m/s = 5.556 m/s

To determine speed v2;

Using the principle of momentum and impulse;

mv1 + ∫₀⁵ F dt = mv2

8 0
3 years ago
Jessie and Jaime complete a 5.0 km race. Each has a mass of 68 kg. Jessie runs the race at 15 km/h; Jaime walks it at 5 km/h. Ho
Leto [7]

The total metabolic energy used by each to complete the course is determined as 656.91 J.

<h3>Kinetic energy of Jessie and Jaime</h3>

The kinetic energy of Jessie and Jaime is calculated as follows;

K.E = ¹/₂mv²

where;

  • m is mass of Jaime
  • v is speed

15 km/h = 4.17 m/s

5 km/h = 1.39 m/s

K.E = ¹/₂(68)(4.17)² + ¹/₂(68)(1.39)²

K.E = 656.91 J

Thus, the total metabolic energy used by each to complete the course is determined as 656.91 J.

Learn more about kinetic energy here: brainly.com/question/25959744

5 0
2 years ago
The speed of an arrow fired from a compound
san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

Y= Y_0 +v_{y0} t - \frac{g}{2} t^2

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, Y=0, then

0= Y_0 +v_{y0} t - \frac{g}{2} t^2

0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2

Caculate the segcond degree equation to obtain the two possible values for t:

t_1= 10.18 \\t_2= -0.04046

But, in physics, time it could not be negative, so we take t_1= 10.18

Caculate now:

X=79m/s*cos(\39)*10.18s= 624.996 m

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

v_0= 79m/s+13m/s= 92m/s

Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

And we obtain:

t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

Finally:

X=92m/s *cos(39)*11.845s=846.887 m

6 0
3 years ago
A rugby player passes the ball 6.8 m across the field, where the ball is caught at the same height as it left his hand. show ans
kherson [118]

Answer:

11.86°

Explanation:

Projectile motion is a form of motion where an object moves in parabolic path (trajectory). Projectile motion only occurs when there is one force applied at the beginning on the trajectory, after which the only interference is from gravity.

The range of an object experiencing projectile motion is given by:

R = u²sin(2θ) / g

where u is the initial velocity, θ is the angle with horizontal and g is the acceleration due to gravity

Given that R = 6.8 m, g = 10 m/s², u = 12 m/s.  

R = u²sin(2θ) / g

substituting:

6.8 = 13²sin(2θ) / 10

13²sin(2θ) = 68

sin(2θ) = 0.4024

2θ = sin⁻¹(0.4024)

2θ = 23.73

θ = 11.86°

3 0
3 years ago
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