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ra1l [238]
3 years ago
9

On hot sunny days or days with thin clouds, the temperature over the land is warmer than the temperature over the water. The war

mer air rises and the cooler air moves over the land, producing sea breezes. This means that,
A: Water takes longer to heat up than the land
B: Soil warms up more slowly than the water
C: The air above the land becomes cool and rises
D: Land takes longer to heat up tan the water
Physics
1 answer:
Luden [163]3 years ago
4 0

Answer:

i think it would be a but i am not 100% sure on that

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What is the most likely reason it is dangerous to break open a compressed gas tank?
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How much power is needed to lift the 200-N object to a height of 10 m in 4 s?
harkovskaia [24]

Answer:

500 watts

Explanation:

Recall that the definition of power is the amount of energy delivered per unit of time.

In our case, the energy delivered is potential energy which we can estimate as the product of the weight of the object times the distance it is lifted above ground:

200 N x 10 m = 2000  Nm

then the power is the quotient of this potential energy divided the time it took to lift the object to that position:

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6 0
2 years ago
A force acts on a 9.90 kg mobile object that moves from an initial position of to a final position of in 5.40 s. Find (a) the wo
horrorfan [7]

Given that,

Mass of object = 9.90 kg

Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

We need to calculate the displacement

Using formula of displacement

s=r_{2}-r_{1}

Where, r_{1} = initial position

r_{2} = final position

Put the value into the formula

s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)

s= -6.80i+6.20j-0.1k

(a). We need to calculate the work done on the object

Using formula of work done

W=F\cdot s

Put the value into the formula

W=(2.00i + 9.00j + 5.30k)\cdot (-6.80i+6.20j-0.1k)

W=-13.6+55.8-0.53

W=41.67\ J

(b). We need to calculate the average power due to the force during that interval

Using formula of power

P=\dfrac{W}{t}

Where, P = power

W = work

t = time

Put the value into the formula

P=\dfrac{41.67}{5.40}

P=7.71\ Watt

(c). We need to calculate the angle between vectors

Using formula of angle

\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})

Put the value into the formula

\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})

\theta=79.7^{\circ}

Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.

(b). The average power due to the force during that interval is 7.71 Watt.

(c).  The angle between vectors is 79.7°

7 0
3 years ago
Find the mass of a sample of water if its temperature dropped 24.8 °C when it lost 870 J of
rosijanka [135]

Answer:

8.35 × 10^{-3} kg

Explanation:

Remember the formula for heat:

Q=mc\Delta T

You just need to solve for m, doing so you're left with:

m=\dfrac{Q}{c\Delta T}

Replace the variables with numbers and that's it.

7 0
2 years ago
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