Let's start with the concept of momentum. What is it? Linear momentum in physics is mathematically written as a product of mass and velocity of an object. Now let us suppose a body of mass m is moving in an inertial frame of reference with velocity v. Consider the fact that no external force is acting on the system. The momentum of this body is given by mv, where m is the mass and v is its velocity. In case of simple real world problems not delving into the realms of relativity, mass is a conserved quantity and it cannot be zero. Hence the velocity of the body must be zero and hence the momentum.
However, photons are considered to have a rest mass zero.
However note the point carefully "rest mass". A body in motion cannot have mass to be zero.
<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em><em> ❤️</em>
Although many characteristics are common<span> throughout the </span>group<span>, the heavier metals such as Ca, Sr, Ba, and Ra are almost as reactive as the </span>Group<span> 1 Alkali Metals. All the </span>elements<span> in </span>Group 2 have two<span> electrons in their valence shells, giving them an oxidation state of +</span><span>2.</span>
The answer is cardiovascular.
The period of the block's mass is changed by a factor of √2 when the mass of the block was doubled.
The time period T of the block with mass M attached to a spring of spring constant K is given by,
T = 2π(√M/K).
Let us say that, when we increased the mass to 2M, the time periods of the block became T', the spring constant is not changed, so, we can write,
T' = 2π(√2M/K)
Putting T = 2π(√M/K) above,
T' =√2T
So, here we can see, if the mass is doubled from it's initial value. The time period of the mass will be changed by a factor of √2.
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Answer:
a) The centripetal acceleration of the car is 0.68 m/s²
b) The force that maintains circular motion is 940.03 N.
c) The minimum coefficient of static friction between the tires and the road is 0.069.
Explanation:
a) The centripetal acceleration of the car can be found using the following equation:
Where:
v: is the velocity of the car = 51.1 km/h
r: is the radius = 2.95x10² m
Hence, the centripetal acceleration of the car is 0.68 m/s².
b) The force that maintains circular motion is the centripetal force:
Where:
m: is the mass of the car
The mass is given by:
Where P is the weight of the car = 13561 N
Now, the centripetal force is:
Then, the force that maintains circular motion is 940.03 N.
c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:
Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.
I hope it helps you!
