Answer:
The COP of the system is = 4.6
Explanation:
Given data
Higher pressure = 1.8 M pa
Lower pressure = 0.12 M pa
Now we have to find out high & ow temperatures at these pressure limits.
Higher temperature corresponding to pressure 1.8 M pa
°c = 335.9 K
Lower temperature corresponding to pressure 0.2 M pa
°c = 262.9 K
COP of the system is given by


COP = 4.6
Therefore the COP of the system is = 4.6
Answer: 2.5 m/s and 6.25 m
Explanation:
u = 0
a = 0.5 m/s²
t = 5 s
v = u + at
= 0 + 0.5 × 5
= <u>2.5 m/s</u>
s = ut + 1/2 at²
= 1/2 × 2.5 × 5
=<u> 6.25 m</u>
Correction
A student measures the mass <em><u>8cm3</u></em> block of brown sugar to be 12.9g. what is the density of the brown sugar
Answer:

Explanation:
Density is defined as mass per unit volume of an object expressed as
where
is the density, m is the mass of sugar and v is the volume of the sugar. Considering that the volume is given as 8cm3 for sugar then we substitute this for v and mass of 12.9 g we substitute for g then the density will be

Answer:
The correct option is;
a- sea surface temperature anomaly, in degrees Celsius
Explanation:
From the diagram related to the question we have two graphs super imposed of Sea surface temperature anomaly, in degrees Celsius and cholera incidence anomaly (%) both plotted against time in years.
On the left the y-axis represents the sea surface temperature anomaly while on the right, the y-axis represents the cholera incidence anomaly (%).
The display of the graph shows the sea surface temperature anomaly in blue.
Answer:
I would love to be a gardener :)
Explanation:
Plants are kool!!!