To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:
Where,
m = Mass
v = Velocity
Replacing we have that the Total Kinetic Energy is
On the other hand the required Energy to heat up t melting point is
Where,
m = Mass
Specific Heat
Change at temperature
Latent heat of fussion
Heat required to heat up to melting point,
The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.
Answer:
a) (0, -33, 12)
b) area of the triangle : 17.55 units of area
Explanation:
<h2>
a) </h2>
We know that the cross product of linearly independent vectors 
and 
gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.
Luckily for us, we know that vectors 
and 
are living in the plane through the points P, Q, and R, and are linearly independent.
We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).
If they weren't linearly independent, we will obtain vector zero as the result of the cross product.
So, for our problem:
<h2>B)</h2>
We know that and are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:
so:
K=1400*V^2/2
K=20000*25^2/2. => 1400*V^2/2=20000*25^2/2 <=> 1400*V^2=20000*25^2
14*V^2=200*225
v^2=100*225/7
v=250/7^(1/2)
Answer: 250*7^(1/2)/7
Answer:
Vi = 8.28 m/s
Explanation:
This problem is related to the projectile motion.
As we know there are two components of motion associated with this, the horizontal component and vertical component.
The horizontal distance covered by the ball is
Vx*t = x
Vx*t = 5.3
Vx = 5.3/t eq. 1
Also we know that
Vx = Vicos(60)
Vx = Vi*0.5 eq. 2
equate eq. 1 and eq. 2
5.3/t = Vi*0.5
5.3/0.5 = Vi*t
Vi*t = 10.6 eq. 3
The vertical distance is
Vy = y1 + Vyi*t - 0.5gt²
also we know that
Vyi = Visin(60)
Vyi = Vi*0.866
It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance
3 = 1.9 + Vi*0.866*t - 0.5gt²
3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
1.1 = 0.866(Vi*t) - 4.9t²
0.866(Vi*t) = 4.9t² + 1.1
substitute Vi*t = 10.6 in above equation
0.866(10.6) = 4.9t² + 1.1
9.18 = 4.9t² + 1.1
4.9t² = 8.08
t² = 8.08/4.9
t² = 1.648
t = 1.28 sec
Finally, initial speed can be found by substituting the value of t into eq. 3
Vi*t = 10.6
Vi = 10.6/t
Vi = 10.6/1.28
Vi = 8.28 m/s
Answer:
There is less gravitational force in space.
Explanation:
Gravity doesn't exist as it does on Earth. Earth's gravity is as 6 times stronger as it is on the moon.
