Answer:
1) A downward force of magnitude 5 N is exerted on the book by the force of of gravity
2) An upward force of magnitude 5 N is exerted on the book by the table
Explanation:
First of all, any object near the Earth's surface experiences the forces of gravity, which is also called weight of the object. This force always acts downward.
For the book in the problem, the magnitude of the weight is 5 N.
We also know that the book is at rest: this means that the net force acting on it is zero, and there must be another force balancing the weight, in order to give a zero net force. This other force is the reaction force exerted by the table on the book: the magnitude of this force must be equal to the force of gravity (so, 5 N) and its direction is opposite to the weight, therefore upward.
Answer:
Part a)
E = 0
Part b)
Part c)
Electric field inside the conductor is again zero
Part d)
Explanation:
Part a)
conducting sphere is of radius
R = 2 cm
so electric field inside any conductor is always zero
So electric field at r = 1 cm
E = 0
Part b)
Now at r = 3 cm
By Gauss law
Part c)
Again when we use r = 4.50 cm
then we will have
Electric field inside the conductor is again zero
Part d)
Now at r = 7 cm
again by Gauss law
Luminosities
Thanks to this relationship between period and luminosity, a Cepheid provide a practical and accurate method to evaluate their absolute magnitude. Once this is known, it is possible to know the distance of the Cepheid, calculating the difference with respect to the apparent magnitude.
Let's be clear: The plane's "395 km/hr" is speed relative to the
air, and the wind's "55 km/hr" is speed relative to the ground.
Before the wind hits, the plane moves east at 395 km/hr relative
to both the air AND the ground.
After the wind hits, the plane still maintains the same air-speed.
That is, its velocity relative to the air is still 395 km/hr east.
But the wind vector is added to the air-speed vector, and the
plane's velocity <span>relative to the ground drops to 340 km/hr east</span>.
