Answer: f=150cm in water and f=60cm in air.
Explanation: Focal length is a measurement of how strong light is converged or diverged by a system. To find the variable, it can be used the formula:
= (nglass - ni)(
-
).
nglass is the index of refraction of the glass;
ni is the index of refraction of the medium you want, water in this case;
R1 is the curvature through which light enters the lens;
R2 is the curvature of the surface which it exits the lens;
Substituting and calculating for water (nwater = 1.3):
= (1.5 - 1.3)(
-
)
= 0.2(
)
f = = 150
For air (nair = 1):
= (1.5 - 1)(
-
)
f = = 60
In water, the focal length of the lens is f = 150cm.
In air, f = 60cm.
Answer:
Part a) When collision is perfectly inelastic
Part b) When collision is perfectly elastic
Explanation:
Part a)
As we know that collision is perfectly inelastic
so here we will have
so we have
now we know that in order to complete the circle we will have
now we have
Part b)
Now we know that collision is perfectly elastic
so we will have
now we have
Answer:
130.5
Explanation:
According to the stemplot attached (Which I think it is, and if not, then you only need to replace the procedure with your data and you should be fine), you need to calculate first the points of all ten students. In that plot, we can easily calculate the points.
The first number in the colum represents the centen of the point, while the numbers of the second column, represents the units of that centen, for example if you see:
16 | 8 5 6
This means that the point for the students are 168, 165 and 166. Three students, three points.
If you watch the stemplot, the points for the students are:
116, 118, 121, 124, 128, 133, 137, 142, 146 and 179.
The median can be calculated using the mean between the two values in the middle of the sequence.
In this case, half of ten is 5, so, the numbers from the middle in this sequence are 128 and 133, therefore:
Median = 128 + 133 / 2 = 130.5
Answer:
- the forces on the left hand side is 1.038 kN
- the forces on the right hand side is 1.483 kN
Explanation:
Given the data in the question, as illustrated in the image below;
Length of the scaffold = 3 m
weight of the scaffold = 395 N
Weight of Bob = 805 N and stands 1 m from the left end
weight of washing equipment = 500N and on sits 2 m from the left end
Weight of Joe = 820 N and stand 0.500 m from the right end
so the force on the left cable will be;
=
[ (805 N)( (3-1) m) + ( 395 N )(
) + ( 500 N )(1m ) + ( 820 N)( 0.500m ) ]
=
[ 1610 + 592.5 + 500 + 410 ]
=
[ 3112.5 ]
= 1037.5 N
= 1.038 kN
Therefore, the forces on the left hand side is 1.038 kN
On the right hand side;
= ( 805 N + 395 N + 500 N + 820 N ) - 1037.5 N
= 2520 N - 1037.5 N
= 1482.5 N
= 1.483 kN
Therefore, the forces on the right hand side is 1.483 kN
