Answer:
25.6 m/s
Explanation:
Draw a free body diagram of the sled. There are two forces acting on the sled:
Normal force pushing perpendicular to the hill
Weight force pulling straight down
Take sum of the forces parallel to the hill:
∑F = ma
mg sin θ = ma
a = g sin θ
a = (9.8 m/s²) (sin 38.0°)
a = 6.03 m/s²
Given:
v₀ = 0 m/s
a = 6.03 m/s²
t = 4.24 s
Find: v
v = at + v₀
v = (6.03 m/s²) (4.24 s) + (0 m/s)
v = 25.6 m/s
The value of the c will be (0 i+0 j+0 k). c is a vector that is along the positive x-axis and has the condition a(b+c)=0.
<h3>What is a vector?</h3>
A vector is a quantity or phenomena with magnitude and direction that are independent of one another. The phrase also refers to a quantity's mathematical or geometrical representation.
Given ;
a=3i-2j+k
b=-i-4j+3k
Given property:
a(b+c)=0
-i-4j+3k ((-i-4j+3k)+c)=0
(3+8+3)(-i-4j+3k)c=0
14(-i-4j+3k)c=0
c=0 i+0 j+0 k
Hence, the value of the c will be (0 i+0 j+0 k).
To learn more about the vector refer to the link;
brainly.com/question/13322477
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Answer:
RE of Hydrogen = 6.47 x RE of Krypton
Explanation:
Actually the correct formula for comparing rate of effusion (RE) of two gases is:
RE of Gas A
------------------- = √ ( Molar mass of B / Molar mass of A)
RE of Gas B
You can designate which of the two gases you have (hydrogen and krypton) will be your gas A and gas B. So for this particular problem, let us make hydrogen as gas A and Krypton as gas B. So the equation becomes:
RE of Hydrogen
------------------------- = √ (Molar mass of Krypton / Molar mass of Hydrogen)
RE of Krypton
Get the molar masses of Hydrogen and Krypton in the periodi table:
RE of Hydrogen
------------------------- = √ (83.798 g/mol / 2 g/mol)
RE of Krypton
RE of Hydrogen
------------------------- = 6.47 ====> this can also be written as:
RE of Krypton
RE of Hydrogen = 6.47 x RE of Krypton
It means that the rate of effusion of Hydrogen gas will be 6.47 faster than the rate of effusion of Krypton gas. With the type of question you have, it doesn't matter which gases goes on your numerator and denominator. What's important is that you show the rate of effusion of a gas with respect to the other. But if that's concerns you the most, then take the gas which was stated first as your gas A and the latter as your gas B unless the problem tells you which one will be on top and which is in the bottom.
