What is the station 2 code for cupids spread the love escape room?

two masses are kept 2 metre apart there is gravitational force of 2 Newton what is the gravitational force when they are kept at

sukhopar [10]

Answer: 0.5N

Explanation:

Gravitational force is calculated using the formula :

F = Gm1m2/r^2

Where G is the gravitational constant (6.67 × 10^-11)

At a distance 'r' of 2metres apart:

Mass of objects are m1 and m2

Gravitational force 'F1' = 2N

Inputting values into the formula :

2 = Gm1m2 / 2^2 - - - - - (1)

At a distance 'r' of 4meters apart:

Mass of objects are m1 and m2

Gravitational force 'F2' = y

Inputting values

F2 = Gm1m2 / 4^2 - - - - - (2)

Dividing equations 1 and 2

2 = Gm1m2 / 2^2 ÷ F2 = Gm1m2 / 4^2

2 / F2 = (Gm1m2 / 4) / (Gm1m2 / 16)

2 / F2 = (Gm1m2 / 4) × (16 / Gm1m2)

2/F2 = 16 / 4

Cross multiply

2 × 4 = 16 × F2

8 = 16F2

F2 = 8/16

F2 = 0.5N

7 0

3 years ago

NEEED HELP!!!

Amanda [17]

Answer:

The correct choice is A. Yes, the more exercise the better.

5 0

2 years ago

Una pelota de golf se lanza con una rapidez de 50 m/seg y un ángulo de elevación de 40º. Calcular:

miss Akunina [59]

Answer:

good morning you are you still

8 0

2 years ago

A car maintains a constant speed v as it traverses the hill and valley as shown below. Both the hill and valley have a radius of

Zigmanuir [339]

Answer:

As given that the car maintains a constant speed v as it traverses the hill and valley where both the valley and hill have a radius of curvature R.

(i) At point C, the normal force acting on the car is largest because the centripetal force is up. gravity is down and normal force is up. net force is up so magnitude of normal force must be greater than the car's weight.

(ii) At point A, the normal force acting on the car is smallest because the centripetal force is down. gravity is down and normal force is up. net force is up so magnitude of normal force must be less than car's weight.

(iii) At point C, the driver will feel heaviest because the driver's apparent weight is the normal force on her body.

(iv) At point A, the driver will feel the lightest.

(v)The car can go that much fast without losing contact with the road at A can be determined as follow:

Fn=0 - lose contact with road

Fg= mv²/r

mg=mv²/r

v=sqrt (gr)

8 0

2 years ago

Two football players are pushing a 60 kg blocking sled across the field at a constant speed of 2.0 m/s. The coefficient of kinet

MArishka [77]

Answer:

The sled slides d=0.155 meters before rest.

Explanation:

m= 60 kg

V= 2 m/s

μ= 0.3

g= 9.8 m/s²

W= m * g

W= 588 N

Fr= μ* W

Fr= 176.4 N

∑F = m * a

a= (W+Fr)/m

a= 12.74m/s²

t= V/a

t= 0.156 s

d= V*t - a*t²/2

d= 0.155 m

7 0

3 years ago