1.igneous rocks are formed when magma or lava cools and hardens
2. the cooling rate of the rock
The lithosphere because it includes the outer region of the earth including the crust and outer mantle
Answer:
You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.
Explanation:
Given Values:
L = 50 cm = 0.5 m
H = 170 j/s
To find the diameter of the rod, we have to find the area of the rod using the following formula.
Here Tc = 100.0° C
k = 50.2
H = k × A × ![\frac{[T_{H -}T_{C} ] }{L}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BT_%7BH%20-%7DT_%7BC%7D%20%5D%20%7D%7BL%7D)
Solving for A
A = ![\frac{H * L }{k * [ T_{H}- T_{C} ] }](https://tex.z-dn.net/?f=%5Cfrac%7BH%20%2A%20L%20%7D%7Bk%20%2A%20%5B%20T_%7BH%7D-%20T_%7BC%7D%20%5D%20%7D)
A = ![\frac{170 * 0.5}{50.2 * [ 350 - 100 ]}](https://tex.z-dn.net/?f=%5Cfrac%7B170%20%2A%200.5%7D%7B50.2%20%2A%20%5B%20350%20-%20100%20%5D%7D)
A = 
= 6.77 ×
m²
Now Area of cylinder is :
A = 
d²
solving for d:
d = 
d = 9.28 cm
Answer:
i) 3.514 s, ii) 5.692 m/s
Explanation:
i) We can use Newton's second law of motion to find out how long does it take for the Eagle to touch down.
as the equation says for free-falling
h = ut +0.5gt^2
Here, h = 10 m, g = acceleration due to gravity = 1.62 m/s^2( on moon surface)
initial velocity u = 0
10 = 0.5×1.62t^2
t = 3.514 seconds
Therefore, it takes t = 3.514 seconds for the Eagle to touch down.
ii) use Newton's 1st equation of motion to calculate the velocity of the lunar module when it hits the surface of the moon
v = u + gt
v = 0+ 1.62×3.514
v= 5.692 m/s
Answer:
The correct option is;
3 times of X
Explanation:
The algebraic properties of a vector are;
Multiplicative identity for real numbers 1
1P = P for each P
Scalar associative property = r(sP) = (rs)P
Scalar distributive property (r + s)P = rP + sP
Vector distributive property r(P + Q) = rP + rQ
Additive inverse of a vector such that a vector P has an inverse -P such that we have;
P + (-P) = 0
Vector associative property (P + Q) + R = P + (Q + R)
Vector commutative property P + Q = Q + P.
