Answer:
a) 0.64 b) 2.17m/s^2 c) 8.668joules
Explanation:
The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,
Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move
Frictional force = mgsin20o + 5N = 6.71+5N = 11.71
The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44
Coefficient of static friction = 11.71/18.44= 0.64
Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)
b) coefficient of kinetic friction = frictional force/ normal force
Fr = 0.4* mgcos 20o = 7.375N
F due to motion = ma = total force - frictional force
Ma = 11.71 - 7.375 = 4.335
a= 4.335/2(mass of the block) = 2.17m/s^2
C) work done = net force *distance = 4.335*2= 8.67Joules
Answer:
Explanation:
Given:
Horizontal distance = 30 m
Note that Wavelength, lambda is the distance between two consecutive crests/troughs. Since,
One boat is at trough, the other is at crest.
The distance between a crest and a trough next to it = lambda/2
Complete cycles = 3 cycles
Time taken for the 3 cycles = 15 s
Vertical distance = 3.8 m
Wavelength, lambda = 2 × horizontal distance
= 2 × 30
= 60 m
Amplitude = vertical distance from the extreme loint to the mid point
= y/2
= 3.8/2
A = 1.9 m
In one cycle = 18/3
= 6 s/cycle
frequency, f = 1/T
= 1/6 = 0.17 Hz
speed,v = lambda × frequency
= 60 × 0.17
= 10 m/s
Answer:
Hipparchus was an ancient Greek who classified stars based on the brightness in 129 B.C. He grouped the brightest stars and ranked them 1 (first magnitude) and dimmest stars as 6 (sixth magnitude). Thus, the smaller numbers indicated brighter stars. Now, the scale extends in negative axis as well. More the negative number, brighter is the star. For example, Sun has magnitude -26.74.
This the apparent magnitude which means the classification is based on the brightness of the star as it appears from the Earth.
Answer:
z = 93.2 m
Explanation:
We can appreciate that this expression is equivalent to the linear motion equation with constant acceleration
v² = v₀² + 2 a d
If we make a term-to-term comparison with the expression obtained, they are equivalent
u² = v² + 2 a z
From here we can clear the position
2 a z = u² –v²
z = (u² –v²) / 2 a
Let's calculate
For the speed to reduce the acceleration must be negative
z = (0 - 21.8²) / 2(- 2.55)
z = 93.2 m