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vladimir1956 [14]
2 years ago
12

How did Robert Whittaker change classification?

Physics
2 answers:
larisa [96]2 years ago
7 0
He reorganized life into five kingdoms
lorasvet [3.4K]2 years ago
6 0

Answer:

d

Explanation:

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A ventilating fan is operated by 0.5hp electric motor. How much work in joules can the fan do in 3 hours? (I need the answer asa
shtirl [24]

Answer:

Power= 0.5hp = 375W

T = 3hrs = 10800s

Power = Work done/ Time

Work = Power * Time = 375 * 10800 = 4050000J

3 0
3 years ago
during a journey, a car travels at 40 km in 2.5 hours, next 62 km in 3 hours, then took a break for 30 minutes, again travelled
wlad13 [49]
45mph is the answer if you do the math right
6 0
3 years ago
During metamorphism, what is the major effect of chemically active fluids?
xz_007 [3.2K]

Answer:

Option b

Explanation:

Metamorphism is the process where the variation of the geological texture resulting from the different arrangement of the minerals or the variation of minerals in protoliths, i.e., pre- existing rocks take place such that there occurs no change in state of the protolith, i.e., it does not melt into magma.

The change takes place as a result of the presence of chemically active fluids, heat and pressure.

There is a reaction between the chemically active fluid and the rock through which it passes and promotes the movement of the dissolved ions of silicate and promotes the growth of the mineral grains.

7 0
2 years ago
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
What is the average velocity of the particle from rest to 9 seconds?
geniusboy [140]
Answer: v = 2[m/s]Explanation:This avarage velocity can be found with the ... B. 2 meters/ second. C. 3 meters/second. D. 4 meters/second. 1.
7 0
2 years ago
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