Answer:
A) If you want to achieve the SMALLEST possible resistance, you should attach the leads to the opposite faces that measure b) 5 cm by 8 cm.
B) If you want to achieve the LARGEST possible resistance, you should attach the leads to the opposite faces that measure a) 3 cm × 5 cm
Explanation:
Resistivity is directly proportional to lenght and inversely properly to cross sectional area.
For the first case, 5 cm by 8 cm gives the largest area and leave 3 cm as the lenght. The resistivity of the metal will be smallest in these dimensions.
For the second case, 3 cm by 5 cm gives the smallest area, leaving 8 cm as the lenght. This is the maximum arrangement that can give the largest resistance possible.
Answer:
26.83 N.
Explanation:
If the angle between two vector is 90°, to get the resultant, we use Pythagoras theorem.
a² = b²+c²......................... Equation 1
Where a = R = Resultant, b = 12 N, c = 24 N.
Substitute these values into equation 1
R² = 12²+24²
R² = 144+576
R² = 720
√R² = √720
R = 26.83 N.
Hence, the result of the two force is 26.83 N.
Answer: perpendicular to it oscillations.
Explanation: A transverse wave is a wave whose oscillations is perpendicular to the direction of the wave.
By perpendicular, we mean that the wave is oscillating on the vertical axis (y) of a Cartesian plane and the vibration is along the horizontal axis (x) of the plane.
Examples of transverse waves includes wave in a string, water wave and light.
Let us take a wave in a string for example, you tie one end of a string to a fixed point and the other end is free with you holding it.
If you move the rope vertically ( that's up and down) you will notice a kind of wave traveling away from you ( horizontally) to the fixed point.
Since the oscillations is perpendicular to the direction of wave, it is a transverse wave
The largest resultant amplitude would be that created by constructive interference, basically when the two waves are of the same phase, so it would be 0.36m+0.22m= 0.58 m.
Answer:
v = 15.8 m/s
Explanation:
Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement. So the variation of mechanical energy is equal to the work of the fictional force
= ΔEm = 
-Em₀
Let's write the mechanical energy at each point
Initial
Em₀ = Ke = ½ k x²
Final
= K + U = ½ m v² + mg y
Let's use Hooke's law to find compression
F = - k x
x = -F / k
x = 4400/1100
x = - 4 m
Let's write the energy equation
fr d = ½ m v² + mgy - ½ k x²
Let's clear the speed
v² = (fr d + ½ kx² - mg y) 2 / m
v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50) 2/60.0
v² = (160 + 8800 - 1470) / 30
v = √ (229.66)
v = 15.8 m/s
