Answer:
Explanation:
Given
Mass of crate is m=574 kg
The crate is raised up to a height of 1.3 m
time taken t=2 s
Work done to raise the crate is equal to the change in its potential energy
Answer:
I believe 0
Explanation:
this is due to the fact that the box will not move anywhere the forces are canceling themselves out.
Answer:
<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong):</em>
a) a×b = 34.27k
b) a·b = 128.43
c) (a + b)·b = 305.17
d) The component of a along the direction of b = 9.66
Explanation:
<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong)</em> we can proceed as follows:
a) The vectorial product, a×b is:
b) The escalar product a·b is:
c) <u>Asumming (a</u><u> </u><u>+ b)·b</u> <em>instead a+b·b</em> we have:
d) The component of a along the direction of b is:
I hope it helps you!
Answer:
Please find attached, the required wave drawn with MS Excel
Explanation:
Functions that represent waves is given as follows
A general form of the wave equation is A·sin(B·x) + D
Where;
B = 2·π/T
T = The period of the wave = 1/f
D = The vertical shift of the wave = 0
A = The amplitude of the wave = 1 for sine wave
v = The wave velocity
λ = The wavelength of the wave
f = The frequency of the wave
v = f·λ
At constant <em>v</em>, λ ∝ 1/f
∴ λ ∝ T
Where T = 3, we have;
B = 2·π/T
∴ B = 2·π/3
Therefore, we have the wave with an amplitude of 1 cm, and wavelength, 3 cm, given as follows
y = sin((2·π/3)·x)
Plotting the above wave with MS Excel, we can get the attached wave
Answer:
Explanation:
Given that
Height of the tree is 3.7m
Therefore yo=3.7m
yox= 0m
The tiger lands 4.8m from the tree
Then, Range x=4.8m
Since she flies horizontally and she lands away from the bottom of the tree, her path will be trajectory as shown in the attachment
Let know the time of flight, using the equation of motion
voy is the initial velocity of the vertical motion of the is the tiger, which is zero at the beginning.
y = y0 + voy*t + ½*g*t²
Given that,
y=3.7m
yo=0m
voy=0m/s
g=9.81m/s²
y = y0 + voy*t + ½*g*t²
3.7=0+0•t+½×9.81×t²
3.7=0+0+4.905t²
3.7=4.905t²
t²=3.7/4.905
t²=0.7543
t=√0.7543
t=0.87sec
Time to reach the ground is
Now to know the initial velocity of the horizontal motion, using equation of motion
x=xo+Voxt
Vox is the horizontal initial velocity of the tiger.
x=xo+Voxt
4.8=0+Vox×0.87
4.8=0+0.87Vox.
4.8=0.87Vox.
Then, Vox=4.8/0.87
Vox=5.52m/s
Then, her initial velocity Vo
Vo=√voy²+vox²
Vo=√0²+5.52²
Vo=√5.52²
Vo=5.52m/s
The initial velocity is 5.52m/s