Is "Noble gas" A compound or an element? I'm confused :/
2 answers:
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Answer:
a) the first vector has magnitude 58 cm and the angle is 15 measured clockwise from the positive side of the x-axis
b) the second vector, the magnitude is 55.7 cm and the angle is 35 half from the negative side of the x-axis in a clockwise direction
c) the magnitude is 54.2 cm with an angle of 18 measured counterclockwise from the x-axis
Explanation:
For this exercise we draw a Cartesian coordinate system in this system: East coincides with the positive part of the x-axis and North with the positive part of the y-axis.
a) the first vector has magnitude 58 cm and the angle is 15 measured clockwise from the positive side of the x-axis
b) the second vector, the magnitude is 55.7 cm and the angle is 35 half from the negative side of the x-axis in a clockwise direction
c) the magnitude is 54.2 cm with an angle of 18 measured counterclockwise from the x-axis
In the attachment we can see the representation of the three vectors
Answer:
The speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.
Explanation:
Given;
mass of block, m = 4 kg
coefficient of kinetic friction, μk = 0.25
angle of inclination, θ = 30°
initial speed of the block, u = 5 m/s
From Newton's second law of motion;
F = ma
a = F/m
Net horizontal force;
∑F = mgsinθ + μkmgcosθ
At the top of the ramp, energy is conserved;
Kinetic energy = potential energy
¹/₂mv² = mgh
¹/₂ v² = gh
¹/₂ x 5² = 9.8h
12.5 = 9.8h
h = 12.5/9.8
h = 1.28 m
Height of the ramp is 1.28 m
Now, calculate the speed of the block (in m/s) when it has returned to the bottom of the ramp;
v² = u² + 2ah
v² = 5² + 2 x 7.022 x 1.28
v² = 25 + 17.976
v² = 42.976
v = √42.976
v = 6.56 m/s
Therefore, the speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.