3

The force of attraction between two oppositely charged pith is 5mx 10 to the -6th power newtons. If the charge on the two is 6.7

My name is Ann [436]

Answer:

0.28 m

Explanation:

The following data were obtained from the question:

Force (F) = 5×10¯⁶ N

Charge 1 (q₁) = 6.7×10¯⁹ C

Charge 2 (q₂) = 6.7×10¯⁹ C

Electrical constant (K) = 9×10⁹ Nm²C¯²

Distance apart (r) =?

Thus, the distance between the two charges can be obtained as follow:

F = Kq₁q₂/r²

5×10¯⁶ = 9×10⁹ × 6.7×10¯⁹ × 6.7×10¯⁹/r²

5×10¯⁶ = 4.0401×10¯⁷ / r²

Cross multiply

5×10¯⁶ × r² = 4.0401×10¯⁷

Divide both side by 5×10¯⁶

r² = 4.0401×10¯⁷ / 5×10¯⁶

Take the square root of both side

r = √(4.0401×10¯⁷ / 5×10¯⁶)

r = 0.28 m

Therefore, the distance between the two charges is 0.28 m

4 0

3 years ago

Twins Jody and Taylor are rearranging the furniture in their bedroom and want to move a dresser across the room. The dresser has

xeze [42]

Answer:

yes, They will be able to move the dresser.

Explanation:

sliding force 90N

55N + 38N = 93N

therefore, yes the twins can move the dresser

7 0

3 years ago

If the mass of an object is 10 Kg and it experiences a Net force of 60 Newtons. What is the acceleration?

Solnce55 [7]

Answer:

6 m/s^2

Explanation:

F = ma

60N = (10 kg)a

a = 6m/s^2

3 0

3 years ago

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

Aleks04 [339]

Explanation:

It is given that,

Height of the building, h = 44 m

Initial horizontal velocity, $u_x=ucos\theta=8.6\ m/s$

Initial vertical velocity, $u_y=usin\theta=10.5\ m/s$

It can be assumed to find the maximum height of the projectile and time taken to reach the maximum height.

The formula for the maximum height is given by :

$H=\dfrac{(u\ sin\theta)^2}{2g}$

g is the acceleration due to gravity

$H=\dfrac{(10.5)^2}{2\times 9.8}=5.625\ m$

The total maximum height, h' = h + H

$h'=44+5.625=49.625\ m$

The formula for the time of flight is given by :

$t_{max}=\dfrac{u\ sin\theta}{g}$

$t_{max}=\dfrac{10.5}{9.8}=1.07\ s$

Hence, this is the required solution.

3 0

3 years ago

iv. An object is 14 cm in front of a convex mirror. The image is 5.8 cm behind the mirror. What is the focal length of the mirro

Dvinal [7]

Answer:

C. 9.9 cm

Explanation:

The location of the object, u = 14 cm

The location of the image, v = 5.8 cm

The focal length of the mirror, f = Required

The mirror formula for a convex mirror is given as follows;

$\dfrac{1}{u} -\dfrac{1}{v} = -\dfrac{1}{f}$

Therefore, we get;

$\dfrac{1}{14 \ cm} -\dfrac{1}{5.8 \ cm} =- \dfrac{41}{406 \ cm} = -\dfrac{1}{f}$

Therefore;

$f = \dfrac{406 \ cm}{41} \approx 9.9 \ cm$

The focal length of the mirror, f ≈ 9.9 cm

6 0

3 years ago