During the pyramid-building era the ancient Egyptians used two units of length, the palm and the ell. An ell was 7 palms and a p
1 answer:
Knowing that the units of the ancient Egyptians were palm and ell, where 1 ell was 7 palms and 1 palm was 3 inches, the speed of a hawk tracked as 3 ells/s is equal to 63 in/s. The given options are not the correct ones.
We have that:
1 ell = 7 palms
1 palm = 3 inches
v: is the speed of the hawk = 3 ells/s
To find the <u>speed of the hawk</u> in inches per second, we need to convert 3 ells/s to in/s, as follows:
Therefore, the speed of the hawk is 63 inches per second. The given options are not the correct ones.
You can find another example of speed here:
I hope it helps you!
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Answer:
Check the explanation
Explanation:
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
The image is virtual
The image is upright
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
Kindly check the diagram in the attached image below.
Answer: 5166.347
Explanation:
The specific gravity of a solid (also called relative density) is the ratio of the density of that solid
to the density of water
(normally at
):
(1)
On the other hand, the density of the rock is calculated by:
(2)
Where:
is the mass of the rock
is the volume of the rock, since is spherical
Well, we already know the value of , but we need to find
in order to find the rock's specific gravity; and in order to do this, we firsly have to find
and then calculate
:
In the case of the mass of the rock, we can calclate it by the following equation:
(3)
Where:
is the weight if the rock in mars
is the acceleration due gravity in Mars
Isolating :
(4)
(5)
To find we can use the following equation of the potential gravitational energy
:
(6)
Where:
is the potential energy
is the height at which the rock has the mentioned potential energy
Isolating :
(7)
(8)
(9)
Substituting (9) in (5):
(10)
(11)
Substituting (11) in (2):
(12) At this point we only need to find the volume of the rock, knowing its diameter is
, hence its radius is
(13)
(14)
Substituting (14) in (12):
(15)
(16)
Substituting (16) in (1):
(17)
Finally we obtain the specific gravity of the rock: