Help me bro, thanks all

Two charges that are 1 meter apart repel each other with a force of 2 N. If the distance between the charges is increased to 2 m

Savatey [412]

Answer:

b) 0.5 N

Explanation:

From coulomb's law,

F = kq'q/r².................... Equation 1

Where F =force of repulsion between the charges, q' = first charge, q = second charge, r = distance between the charges, k = proportionality constant.

q'q = Fr²/k........................... Equation 2

Given: F = 2 N, r = 1 m, k = 9.0×10⁹ Nm²/C²

Substituting into equation 2

q'q = 2(1)²/(9.0×10⁹)

q'q = 2/9.0×10⁹ C².

If the distance between the charges is increased to 2 meters,

r = 2 m, q'q = 2/9.0×10⁹ C².

Substitute into equation 1

F = 9.0×10⁹(2/9.0×10⁹)/2²

F = 2/4

F = 1/2 = 0.5 N.

The right option is b) 0.5 N

5 0

4 years ago

PLS THIS IS DUE IN 2 MINUTES

luda_lava [24]

Answer:

The toy car

Explanation:

the real car is parked so yeah but maybe in some way technically the real car has more "momentum"

7 0

3 years ago

A 250 kg engine block is being dragged across the pavement by a sturdy chain with a tension force of 10,000 N, as seen in the fr

Musya8 [376]

Answer: B

Explanation:

The friction force is less than the tension force

4 0

3 years ago

A 50-g ball moving at 10 m/s in the +x direction suddenly collides head-on with a stationary ball of mass 100 g. If the collisio

Savatey [412]

Answer:

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses of the object

u1 and u2 are the velocities

v is the final velocity

Given

m1 = 50g

u1 = 10m/s

m2 = 100g

u2 = 0m/s (stationary ball)

Required

Common velocity v

Substitute

50(10) + 100(0) = (50+100)v

500 = 150v

v = 500/150

v = 3.33m/s

Hence the velocity of each ball immediately after the collision is 3.33m/s

6 0

3 years ago

6. A 15 kg box is given an initial push so that it slides across the floor and comes to a stop.

Illusion [34]

Answer:

(a) Frictional force = 44.145 N

(b) Acceleration = -2.943 m/s²

(c) Distance covered = 1.529 m

Explanation:

Given:

Mass of the box is, $m=15\ kg$

Coefficient of friction is, $\mu =0.30$

Acceleration due to gravity is, $g=9.81\ m/s^2$

Normal force acting on the box is, $N=mg=(15)(9.81) = 147.15\ N$

(a)

Frictional force is given as:

$f=\mu N=0.30\times 147.15=44.145\ N$

Therefore, the frictional force is 44.145 N.

(b)

The acceleration of the box is given using Newton's second law as:

$a=-\frac{f}{m}=-\frac{44.145}{15}=-2.943\ m/s^2$

Therefore, the acceleration of the box is -2.943 m/s².

(c)

Initial velocity is, $u=3.0\ m/s$

Final velocity is, $v=0\ m/s$

Acceleration of the box is, $a=-2.943\ m/s^2$

The displacement of the box can be determined using equation of motion as:

$v^2=u^2+2ad\\0=3^2+2\times (-2.943)d\\0=9-5.886d\\5.886d=9\\d=\frac{9}{5.886}=1.529\ m$

Therefore, the displacement of the box is 1.529 m.

4 0

4 years ago

Help me bro, thanks all​

Help me bro, thanks all