Answer:
Force applied to smaller cross section is
= 82.63 N
Explanation:
As we know
where 
signifies the weight of the two chair in a hydraulic-lift system
And 
signifies the area of the two respective chairs in a hydraulic-lift system
Given -
N
Square centimeter
Square centimeter
Substituting the given values in above equation, we get -
Force applied to smaller cross section is
= 82.63 N
Answer:
the wave length becomes doubled or becomes two times the initial wavelength = 240 cm
Explanation:
From wave,
v = λf................ Equation 1
Where v = velocity of the wave, λ = wavelength of the wave, f = frequency of the wave.
Given: f = 1200 Hz, λ = 120 cm = 1.2 m
Substitute into equation 1
v = 1200(1.2)
v = 1440 m/s.
When the ship sent out a 600 Hz sound wave,
make λ the subject of formula in equation 1
λ = v/f............. Equation 2
Given: f = 600 Hz, v = 1440 m/s
Substitute into equation 2
λ = 1440/600
λ = 2.4 m or 240 cm.
When the ship sent out a 600 Hz sound wave instead, the wave length becomes doubled or becomes two times the initial wavelength = 240 cm
Answer:
43.58 m
Explanation:
If you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees
Using trigonometry ratio
Sin 5 = opposite/hypothenus
Where the hypothenus = 500m
Opposite = height h
Sin 5 = h/500
Cross multiply
500 × sin 5 = h
h = 500 × 0.08715
h = 43.58m
Therefore, the height above the starting point is equal to 43.58m
A simple electromagnet consisting of a coil of wire wrapped around an iron core. <u><em>A core of ferromagnetic material like iron serves to increase the magnetic field created.</em></u> The strength of magnetic field generated is proportional to the amount of current through the winding.
your answer is b :)
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On comparing values , we see that student which has the largest percent error is <u>A. Student 4: 9.61 m/s2
.</u>
<u>Explanation:</u>
Here, we have Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Let's calculate which student’s measurement has the largest percent error :
<u>A. Student 4: 9.61 m/s2
</u>
Percentage of error = 
%.
<u>B. Student 3: 9.88 m/s2
</u>
Percentage of error = 
%.
<u>C. Student 2: 9.79 m/s2
</u>
Percentage of error = 
% .
<u>D. Student 1: 9.78 m/s2</u>
Percentage of error = 
% .
On comparing values , we see that student which has the largest percent error is <u>A. Student 4: 9.61 m/s2
.</u>
