Answer:
- magnitude : 1635.43 m
- Angle: 130°28'20'' north of east
Explanation:
First, we will find the Cartesian Representation of the 
and 
vectors. We can do this, using the formula
where 
its the magnitude of the vector and θ the angle. For we have:
where the unit vector 
points east, and 
points north. Now, the will be:
Now, taking the sum:
This is
Now, for the magnitude, we just have to take its length:
For its angle, as the vector lays in the second quadrant, we can use:
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
Answer:
14 m/s
Explanation:
Step One:
Given data:
mass of automobile m= 1700kg
Force F= 170,000 N
time t= 0.14s
v= ??
Required
The velocity of the car
Step two:
From the expression given below, we can find the velocity
Ft= mv
make v subject of the formula
v= Ft/m
v= 170000*0.14/1700
v=23800/1700
v=14 m/s
This is a tough one.
In A, B, and C, there are various routes to get from one end to the other end through 2 or 3 capacitors.
D is the only configuration where it's possible to get from one end to the other through only 1 of them.
I'm not totally confident, but I think D is the one that's not like the others.
Answer:
Speed at bottom of the hill (v) = 11.74 m/s
Explanation:
Given:
Combined mass = 48.8 kg
Height h = 7.05 m
Find:
Speed at bottom of the hill (v)
Computation:
v² = 2gh
v = √2 x 9.8 x 7.05
v = √138.18
v = 11.74 m/s
Speed at bottom of the hill (v) = 11.74 m/s
