The question is incomplete as it does not have the options which have been provided in the attachment.
Answer:
Option-D
Explanation:
In the given question, the effect of the sunlight on the growth of the plant has been studied. The values provided in the Option-D can be considered correct as the values are measured in the decimal value up to two decimal value.
The values are measured after the first week, second week, and the initial readings. The difference in the values provided in Option-D does not show much difference as well as are up to two decimal places.
Thus, Option-D is the correct answer.
Answer: If a force on an object is aimed in the direction of the object’s velocity, the force does positive work
(b).
Explanation:
Hi, the answer is oprion b. positive work.
If a force on an object is aimed in the direction of the object’s velocity, the force does positive work.
An object's kinetic energy will only change if the force acting on the object changes the object's speed. This will only happen if there is a component of the force in the direction that the object moves.
So, A force will do work only if the force has a component in the direction that the object moves.
Given:
rod of circular cross section is subjected to uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]
From given above, area of cross section = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=force/area
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)
(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183
= 0.0318 (to three significant figures)
(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm (to three significant figures)
Answer:
Magnitude of the force is 2.135N and the direction is 41.8° below negative y-axis
Explanation:
The stiff wire 50.0cm long bent at a right angle in the middle
One section lies along the z axis and the other is along the line y=2x in the xy plane
tan θ = 2
Therefore,
slope m = tan θ = y / x
Then length of each section is 25.0cm
so, length vector of the wire is
And magnetic field is B = (0.318T)i
Therefore,
![\bar F = (20.0)[(0.112m)i +(0.224m)j-(0.250m)k \times 90.318T)i]](https://tex.z-dn.net/?f=%5Cbar%20F%20%3D%20%2820.0%29%5B%280.112m%29i%20%2B%280.224m%29j-%280.250m%29k%20%5Ctimes%2090.318T%29i%5D)
![= (20.0)(i(0)+j(-0.250)(0.318T)+k[0-(0.224m)(0.318T)]\\\\=(20.0)(-0.250)(0.318)j-(20.0)(0.224)(0.318T)\\\\=-(1.59N)j-(1.425N)k](https://tex.z-dn.net/?f=%3D%20%2820.0%29%28i%280%29%2Bj%28-0.250%29%280.318T%29%2Bk%5B0-%280.224m%29%280.318T%29%5D%5C%5C%5C%5C%3D%2820.0%29%28-0.250%29%280.318%29j-%2820.0%29%280.224%29%280.318T%29%5C%5C%5C%5C%3D-%281.59N%29j-%281.425N%29k)
Magnitude of the force is
Direction is
Magnitude of the force is 2.135N and the direction is 41.8° below negative y-axis
