The initial force of the throw overcomes gravity quite easily. Then, gravity begins to bring it back down to earth, making a curved path.
Answer:
Explanation:
Given:
Specific heat of gold = 0.031cal/°C
Specific heat of silver = 0.057cal/°C
To know the metals that will heat up faster, we must understand the meaning of specific heat capacity.
It is the amount of heat required to raise the temperature of 1g of a substance by 1°C.
Now,
The higher the specific heat capacity the more energy it is required to heat up the substance.
So, Gold with a specific heat capacity of 0.031cal/°C will heat up faster.
Answer:
c)At a distance greater than r
Explanation:
For a satellite in orbit around the Earth, the gravitational force provides the centripetal force that keeps the satellite in motion:
where
G is the gravitational constant
M is the Earth's mass
m is the satellite's mass
r is the distance between the satellite and the Earth's centre
v is the speed of the satellite
Re-arranging the equation, we write
so we see from the equation that when the speed is higher, the distance from the Earth's centre is smaller, and when the speed is lower, the distance from the Earth's centre is larger.
Here, the second satellite orbit the Earth at a speed less than v: this means that its orbit will have a larger radius than the first satellite, so the correct answer is
c)At a distance greater than r
Explanation:
The two postulates of special theory of relativity
Postulate 1: The law of physics are invariant under any of inertial frame of reference.
Postulate 2: The velocity of light is remains same in each ans every frame of reference and independent of relativity.
They are differ from classical mechanics that in classical mechanics there is no change in mass and length in relative velocity but in relativistic mechanics it changes.
These two postulates implements in phenomenon like time dilation , length contraction etc.
Thanks
Answer:
Pressure,P=6×10^3Pa
Explanation:
The gas has an ideal gas behaviour and ideal gas equation
PV=NKT
T= V/N p/K ...eq1
Average transitional kinetic energy Ktr=1.8×10-23J
Ktr=3/2KT
T=2/3Ktr/K....eq2
Equating eq1 and 2
V/N p/K = 2/3Ktr/K
Cancelling K on both sides
P= 2/3N/V( Ktr)
Substituting the value of N/V and dividing by 10^-6 to convert cm^3 to m^3
P = 2/3 (5.0×10^20)/10^-6 × 1.8×10^-23
P= 6 ×10^3Pa