The number of orders that one can see the entire visible speed is 5 orders.
<h3>How to calculate the orders?</h3>
From the information given, it should be noted that the grating spacing will be:
d = (1.00 × 10^-4) / 250
d = 4000nm
Therefore, the number of times that are needed to complete the order will be the same as the number of orders which the long wavelength time will be visible. This will be:
= (4000 × sin 90°)/700
= 5.71
Therefore, the maximum orders will be 5.
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Is this multiple choice because i dont see it
Answer:
The answer is below
Explanation:
Given that:
x(t) = at – bt2+c
a) x(t) = at – bt2+c
Substituting a = 1.4 m/s, b = 0.06 m/s2 and c =50 m gives:
x(t) = 1.4t - 0.06t² + 50
At t = 5, x(5) = 1.4(5) - 0.06(5)² + 50 = 55.5 m
At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m
The average velocity (v) is given as:
b) x(t) = 1.4t - 0.06t² + 50
At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m
At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m
The average velocity (v) is given as:
c) x(t) = 1.4t - 0.06t² + 50
At t = 15, x(5) = 1.4(15) - 0.06(15)² + 50 = 57.5 m
At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m
The average velocity (v) is given as:
The total displacement is 4.0 m east.
Answer:
Concepts and Principles
1- Kinetic Energy: The kinetic energy of an object is:
K=1/2*m*v^2 (1)
where m is the object's mass and v is its speed relative to the chosen coordinate system.
2- Gravitational potential energy of a system consisting of Earth and any object is:
U_g = -Gm_E*m_o/r*E-o (2)
where m_E is the mass of Earth (5.97x 10^24 kg), m_o is the mass of the object, and G = 6.67 x 10^-11 N m^2/kg^2 is Newton's gravitational constant.
Solution
The argument:
My friend thinks that escape speed should be greater for more massive objects than for less massive objects because the gravitational pull on a more massive object is greater than the gravitational pull for a less massive object and therefore the more massive object needs more speed to escape this gravitational pull.
The counterargument:
We provide a mathematical counterargument. Consider a projectile of mass m, leaving the surface of a planet with escape speed v. The projectile has a kinetic energy K given by Equation (1):
K=1/2*m*v^2 (1)
and a gravitational potential energy Ug given by Equation (2):
Ug = -G*Mm/R
where M is the mass of the planet and R is its radius. When the projectile reaches infinity, it stops and thus has no kinetic energy. It also has no potential energy because an infinite separation between two bodies is our zero-potential-energy configuration. Therefore, its total energy at infinity is zero. Applying the principle of energy consersation, we see that the total energy at the planet's surface must also have been zero:
K+U=0
1/2*m*v^2 + (-G*Mm/R) = 0
1/2*m*v^2 = G*Mm/R
1/2*v^2 = G*M/R
solving for v we get
v = √2G*M/R
so we see v does not depend on the mass of the projectile
