Answer:
Explanation:
We shall apply concept of Doppler's effect of apparent frequency to this problem . Here observer is moving sometimes towards and sometimes away from the source . When observer moves towards the source , apparent frequency is more than real frequency and when the observer moves away from the source , apparent frequency is less than real frequency . The apparent frequency depends upon velocity of observer . The formula for apparent frequency when observer is going away is as follows .
f = f₀ ( V - v₀ ) / V , f is apparent , f₀ is real frequency , V is velocity of sound and v is velocity of observer .
f will be lowest when v₀ is highest .
velocity of observer is highest when he is at the equilibrium position or at middle point .
So apparent frequency is lowest when observer is at the middle point and going away from the source while swinging to and from before the source of sound .
Answer:
Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:
lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:
To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):
=![\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]](https://tex.z-dn.net/?f=%5Cfrac%7Bq%7D%7B4%5Cpi%20e_%7B0%7D%20%7D%20%5B%5Cfrac%7B1%7D%7BR%7D%20-%5Cfrac%7Br%5E%7B2%7D-R%5E%7B2%7D%20%20%7D%7B2R%5E%7B3%7D%20%7D%20%5D)
∴NOTE: Graph is attached
Answer:
There is absolutely No relationship between the weight of an object (which is constant) and the frictional force. If a block is sliding on a surface, that surface will be exerting a force on the block. That force can be resolved into a component parallel to the surface (which we call the frictional component), and a component perpendicular to the surface (called the normal component). For many situations, we find experimentally that the frictional component is approximately proportional to the normal component. The frictional component divided by the normal component is defined to be a quantity called the coefficient of kinetic or sliding friction. The coefficient of kinetic friction obviously depends on the nature of the surfaces involved. The normal component on an object can be decreased if you pull in the direction of the normal component (the weight does not change). However pulling this way on the object not only decreases the normal component, but it also decreases the frictional component since they are proportional. This is why it is easier to slide something if you pull up on it while you push it. If you push down, the normal and frictional components increase so it is harder to slide the object. The weight of an object is the downward force exerted by Earth’s gravity on that object, and it does not change no matter how you push or pull on the object.
Answer:
2.10L
Explanation:
Given data
V1= 2.5L
T1= 275K
P1= 2.1atm
P2= 2.7 atm
T2= 298K
V2= ???
Let us apply the gas equation
P1V1/T1= P2V2/T2
substitute into the expression we have
2.1*2.5/275= 2.7*V2/298
5.25/275= 2.7*V2/298
Cross multiply
275*2.7V2= 298*5.25
742.5V2= 1564.5
V2= 1564.5/742.5
V2= 2.10L
Hence the final volume is 2.10L
