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motikmotik
2 years ago
6

A weather forecaster uses a computational model on a Monday to predict the weather on Friday. Why might that forecast change? (1

point)
A. The forecaster may have done the calculations a second time because he made a mistake in the calculation.

B. Someone may have reported the weather incorrectly before the first computation.

C. The forecaster may have had the computer model do the calculations a second time, and he found that the prediction changed.


D. An area of low pressure might move more quickly on Tuesday and Wednesday than expected.
Physics
1 answer:
vodomira [7]2 years ago
6 0

Answer:

D. An area of low pressure might move more quickly on Tuesday and Wednesday than expected.

Explanation:

I took the test and got it wrong :/

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Glass bottle holds 500g of liquid at 40oC and only 300g at 70oC. Find the real cubic
ahrayia [7]

Answer:

The real cubic expansivity is 0.024 per kelvin

Explanation:

check attachment

7 0
3 years ago
A swimmer is capable of swimming at 1.4m/s in still water. a. How far downstream will he land if he swims directly across a 180m
ozzi

Answer:

t = 180 / 1.4 = 129 sec   (time to swim horizontally across river)

S = 129 sec * V     where V is speed of current and S is the distance he will be carried downstream

The problem does not specify V the speed of the river

8 0
3 years ago
Read 2 more answers
A triangle ∆P QR has vertices P(3, 2, −4), Q(1, 0, −4), R(2, 1, 1). Use the distance formula to decide which one of the followin
777dan777 [17]

Answer:

a. FALSE

b.TRUE

C. FALSE

Explanation:

The formula fot the distance between two points is given as

d=\sqrt{(x_{2}-x_{1})^{2} +(y_{2}-y_{1})^{2} +(z_{2}-z_{1})^{2}}\\

hence we determine the distances between all the points

a.P(3,2,-4), Q(1,0,-4), R(2,1,1)

PQ=\sqrt{(1-3)^{2} +(0-2)^{2} +(-4-(-4))^{2}}\\PQ=\sqrt{4+4+0}\\ PQ=\sqrt{8}

For point PR

we have

PR=\sqrt{(2-3)^{2} +(1-2)^{2} +(1-(-4))^{2}}\\PR=\sqrt{1+1+9}\\ PR=\sqrt{11}\\

|PQ|\neq |PR|

B. For point RP

RP=\sqrt{(3-2)^{2} +(2-1)^{2} +(-4-1)^{2}}\\RP=\sqrt{1+1+25}\\ RP=\sqrt{27}

for point RQ  we have

RQ=\sqrt{(1-2)^{2} +(0-1)^{2} +(-4-1)^{2}}\\RQ=\sqrt{1+1+25}\\ RQ=\sqrt{27}

|RP|=|R Q|

C.

QP=\sqrt{(3-1)^{2} +(2-0)^{2} +(-4+4)^{2}}\\QP=\sqrt{4+4+0}\\ QP=\sqrt{8}

For point Q R

QR=\sqrt{(2-1)^{2} +(1-0)^{2} +(1-(-4))^{2}}\\QR=\sqrt{1+1+9}\\ QR=\sqrt{11}\\

QP\neq QR

6 0
3 years ago
A positive and a negative charge are released from rest in vacuum. They move toward each other. As they do: A positive and a neg
Pachacha [2.7K]

Answer:

A negative potential energy becomes more negative

Explanation:

Let the charges be - Q₁ and Q₂ . Let the distance between them be d .

Potential energy = k -Q₁x Q₂ / R

= - KQ₁ Q₂ / R

Now if the magnitude of R decreases , the magnitude of potential energy increases . So we see that the negative  potential energy becomes more negative .  

4 0
4 years ago
How deep is the outer core beneath the surface
sukhopar [10]
Precisely around 1,800 miles below.
6 0
3 years ago
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