Answer:
Four fundamental forces are gravitational, electromagnetic, strong, and weak.
Explanation:
The gravitational and electromagnetic interactions, which produce significant long-range forces whose effects can be seen directly in everyday life and the strong and weak interactions, which produce forces at minuscule, subatomic distances and govern nuclear interactions.
Answer:
<h2>1116.9 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 438 × 2.55
We have the final answer as
<h3>1116.9 N</h3>
Hope this helps you
Answer: a) vcar= 7 m/s ; b) a train= 0.65 m/s^2
Explanation: By using the kinematic equation for the car and the train we can determine the above values of the car velocity and the acceletarion of the train, respectively.
We have for the car
distance = v car* t, considering the length of train (81.1 m) travel by the car during the first 11.6 s
the v car = distance/time= 81.1 m/11.6s= 7 m/s
In order to calculate the acceleration we have to use the kinematic equation for the train from the rest
distance train = (a* t^2)/2
distance train : distance travel by the car at constant speed
so distance train= (vcar*36.35)m=421 m
the a traiin= (2* 421 m)/(36s)^2=0.65 m/s^2
Answer:
C. 
Explanation:
Let initial charges on both spheres be,

When the sphere C is touched by A, the final charges on both will be,
#Now, when C is touched by B, the final charges on both of them will be:

Now the force between A and B is calculated as:

Hence the electrostatic force becomes 3F/8
Answer:
Time period of the osculation will be 2.1371 sec
Explanation:
We have given mass m = (B+25)
And the spring is stretched by (8.5 A )
Here A = 13 and B = 427
So mass m = 427+25 = 452 gram = 0.452 kg
Spring stretched x= 8.5×13 = 110.5 cm
As there is additional streching of spring by 3 cm
So new x = 110.5+3 = 113.5 = 1.135 m
Now we know that force is given by F = mg
And we also know that F = Kx
So 

Now we know that 
So 

