Answer:
You should use second Newton law
Explanation:
F=ma=>m=F/a
Answer:
1. The answer is that there will be more force with more speed.
2. less speed means less force.
3. And last, but certainly not the least, the less the mass has, the less the force will be, unless the object is moving at a high speed.
Explanation:
Answer: The mean of the samples is 32.325.
Proportion of the defective unit is =![\frac{2}{40}=\frac{1}{20}=0.05](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B40%7D%3D%5Cfrac%7B1%7D%7B20%7D%3D0.05)
Explanation:
The point estimate of the mean of the sample:
First calculate the mean of the 40 samples:
![\frac{\text{Sum of all the samples}}{\text{Number of observations}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BSum%20of%20all%20the%20samples%7D%7D%7B%5Ctext%7BNumber%20of%20observations%7D%7D)
![\frac{1,293}{40}=32.325](https://tex.z-dn.net/?f=%5Cfrac%7B1%2C293%7D%7B40%7D%3D32.325)
The mean of the samples is 32.325.
Defected pieces in the samples whose life span is less than the 26 days = 2
Proportion of the defective unit is =![\frac{2}{40}=\frac{1}{20}=0.05](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B40%7D%3D%5Cfrac%7B1%7D%7B20%7D%3D0.05)
Answer:
As b ∝ (L/r²) and
the distance of the sun from the earth is 0.00001581 light years
and
the distance of the Sirius from the earth is 8.6 light years
hence,
the Sun appear brighter in the sky
Explanation:
The brightness (b) is directly proportional to the Luminosity of the star (L) and inversely proportional to the square of the distance between the star and the observer (r).
thus, mathematically,
b ∝ (L/r²)
now,
given
L for sirius is 23 times more than the sun i.e 23L
now,
the distance of the sun from the earth is 0.00001581 light years
and
the distance of the Sirius from the earth is 8.6 light years
thus,
using the the relation between conclude that the value of brightness for the Sirius comes very very low as compared to the value for brightness for the Sun.
hence, the sun appears brighter
Answer:
The work required to pump the liquid out of the spout is 2.6 × 10⁶ J
Explanation:
<u>Step 1:</u> consider a spherical tank, sliced into two equal part, since it is half filled.
The new circular surface has a radius, if we construct a right angled-triangle on the surface. The initial radius of the spherical tank becomes the hypotenuse of the triangle and the radius is calculated as follows;
![a^2 +b^2 =3^2\\\\a^2 = 9-b^2](https://tex.z-dn.net/?f=a%5E2%20%2Bb%5E2%20%3D3%5E2%5C%5C%5C%5Ca%5E2%20%3D%209-b%5E2)
volume (v)
![= \pi (9-b^2) \delta b](https://tex.z-dn.net/?f=%3D%20%5Cpi%20%289-b%5E2%29%20%5Cdelta%20b)
<u>Step 2:</u> Total height, d, in which the liquid is pumped out;
d = b + 3 + 1
= b + 4
<u>Step 3:</u> Force required to pump the liquid out:
F = mg
But, m = density (ρ) x volume (v)
F = ρvg
Given;
ρ = 900 kg/m³ and g = 9.8 m/s²
![= \pi (9-b^2) \delta b](https://tex.z-dn.net/?f=%3D%20%5Cpi%20%289-b%5E2%29%20%5Cdelta%20b)
![F = (900 X9.8) \pi (9-b^2) \delta b =\pi (8820)(9-b^2) \delta b](https://tex.z-dn.net/?f=F%20%3D%20%28900%20X9.8%29%20%5Cpi%20%289-b%5E2%29%20%5Cdelta%20b%20%3D%5Cpi%20%288820%29%289-b%5E2%29%20%5Cdelta%20b)
<u>Step 4:</u> The work done in pumping the liquid out of the spout
W = F x d
![W' = \pi (8820)(9-b^2) \delta b *(d)\\\\W' = \pi (8820)(9-b^2) \delta b *(b + 4)\\\\W' = \pi (8820)(9-b^2)(b + 4) \delta b\\\\W = \int\limits^3_0 { \pi (8820)(9-b^2)(b + 4)}\ \delta b \\\\W = [(\frac{-b^4}{4} -\frac{4b^3}{3} +\frac{9b^2}{2} +36b)(8820\pi )]^3_0 \\\\W = [(\frac{-(3)^4}{4} -\frac{4(3)^3}{3} +\frac{9(3)^2}{2} +36(3))(8820\pi )] - (0)\\\\W = 2.6*10^6J](https://tex.z-dn.net/?f=W%27%20%3D%20%5Cpi%20%288820%29%289-b%5E2%29%20%5Cdelta%20b%20%2A%28d%29%5C%5C%5C%5CW%27%20%3D%20%5Cpi%20%288820%29%289-b%5E2%29%20%5Cdelta%20b%20%2A%28b%20%2B%204%29%5C%5C%5C%5CW%27%20%3D%20%5Cpi%20%288820%29%289-b%5E2%29%28b%20%2B%204%29%20%5Cdelta%20b%5C%5C%5C%5CW%20%3D%20%5Cint%5Climits%5E3_0%20%7B%20%5Cpi%20%288820%29%289-b%5E2%29%28b%20%2B%204%29%7D%5C%20%5Cdelta%20b%20%5C%5C%5C%5CW%20%3D%20%5B%28%5Cfrac%7B-b%5E4%7D%7B4%7D%20-%5Cfrac%7B4b%5E3%7D%7B3%7D%20%2B%5Cfrac%7B9b%5E2%7D%7B2%7D%20%2B36b%29%288820%5Cpi%20%29%5D%5E3_0%20%5C%5C%5C%5CW%20%3D%20%20%5B%28%5Cfrac%7B-%283%29%5E4%7D%7B4%7D%20-%5Cfrac%7B4%283%29%5E3%7D%7B3%7D%20%2B%5Cfrac%7B9%283%29%5E2%7D%7B2%7D%20%2B36%283%29%29%288820%5Cpi%20%29%5D%20-%20%280%29%5C%5C%5C%5CW%20%3D%202.6%2A10%5E6J)
Therefore, the work required to pump the liquid out of the spout is 2.6 × 10⁶ J