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lidiya [134]
2 years ago
5

The magnitude of the electrical force acting between a +2.4 × 10–8 C charge and a +1.8 × 10–6 C charge that are separated by 0.0

08 m is N, rounded to the tenths place.
Physics
2 answers:
solong [7]2 years ago
8 0
To answer this question, we will simply use Coulomb's which is stated as follows:
F = k*q1*q2 / r^2 where:
F is the magnitude of the force between the two charges
k is a constant = 9.00 * 10^9 N.m^2/C^2
q1 =  <span>+2.4 × 10–8 C
q2 = </span><span>+1.8 × 10–6 C
r is the distance between the two charges = </span><span>0.008 m

Substituting in the equation, we get:
F = (9*10^9)(2.4*10^-8)(1.8*10^-6) / (0.008)^2 = 6.075 which is approximately 6.1 Newtons

</span>
brilliants [131]2 years ago
8 0

On Ed. its 6.1 N. Just took the test and got it right soooo.

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Two disks with the same rotational inertia i are spinning about the same frictionless shaft, with the same angular speed ω, but
valentina_108 [34]

Answer:

3. none of these

Explanation:

The rotational kinetic energy of an object is given by:

K=\frac{1}{2}I \omega^2

where

I is the moment of inertia

\omega is the angular speed

In this problem, we have two objects rotating, so the total rotational kinetic energy will be the sum of the rotational energies of each object.

For disk 1:

K_1 = \frac{1}{2}I (\omega)^2 = \frac{1}{2}I\omega^2

For disk 2:

K_2 = \frac{1}{2}I(-\omega)^2 = \frac{1}{2}I\omega^2

so the total energy is

K=K_1 + K_2 = \frac{1}{2}I\omega^2 + \frac{1}{2}I\omega^2 = I\omega^2

So, none of the options is correct.

5 0
2 years ago
A runner runs 300 m at an average speed of 3.0 m/s. She then runs another 300m at an average
Kaylis [27]

Answer:

B. 4 m/s

Explanation:

v=d/t

Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.

3 0
2 years ago
Two balls with charges +Q and +4Q are separated by 3R. Where should you place another charged ball Q0 on the line between the tw
azamat

Answer:

yes independent of the sign or valve of Q

Explanation:

7 0
2 years ago
You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w
Anit [1.1K]

Answer:

v₀ =3.8 m/s

Explanation:

Newton's second law of the box:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Known data

m=2.1 kg  mass of the box

d= 5.4m  length of the roof

θ = 20° angle θ of the roof with respect to the horizontal direction

μk= 0.51 : coefficient of kinetic friction between the box and the roof  

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the box on the roof  and the y-axis in the direction perpendicular to it.

W: Weight of the box  : In vertical direction

N : Normal force : perpendicular to the direction the  roof

fk : Friction force: parallel to the direction to the roof

Calculated of the weight  of the box

W= m*g  =  (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y weight components

Wx= Wsin θ= (20.58)*sin(20)° =7.039 N

Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N

Calculated of the Normal force

∑Fy = m*ay    ay = 0

N-Wy= 0

N=Wy =19.34 N

Calculated of the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx-f = ( 2.1)*a

7.039 - 9.86  = ( 2.1)*a

-2.821 = ( 2.1)*a

a=(-2.821) /( 2.1)

a= -1.34  m/s²

Kinematics of the box

Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:  

d:displacement  = 5.4 m

v₀: initial speed  

vf: final speed  = 0

a : acceleration of the box = -1.34  m/s²

We replace data in the formula (2)

0²=v₀²+2*(-1.34)*(5.4)

2*(1.34)*(5.4)= v₀²

v_{o} =\sqrt{14.472}

v₀ = 3.8 m/s

7 0
2 years ago
Diamond has an index of refraction of 2.419. What is the critical angle for internal reflection inside a diamond that is in liqu
Anestetic [448]

Answer:

B) 55.8°

Explanation:

n₁ = Refractive index of liquid = 2

n₂ = Refractive index of diamond = 2.419

Here it can be seen that the light ray is in the more dense medium and approaching the less dense medium which is a case of total internal reflection.

Critical angle

\theta_r=sin^{-1}\frac{n_1}{n_2}\\\Rightarrow \theta_r=sin^{-1}\frac{2}{2.419}\\\Rightarrow \theta_r=sin^{-1}0.826\\\Rightarrow \theta_r=55.7701^{\circ}

∴ Critical angle is 55.8°

7 0
2 years ago
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