All categories

2 years ago

Which can cause physical weathering of rock? A acid B oxygen C sun D wind

Physics

1 answer:

zepelin [54]2 years ago

3 0

Answer:

D wind

Explanation:

Heavy winds may cause much damage to the Earth along with rain and extreme temperature changes.

You might be interested in

What is cosmic ray spallation

love history [14]

The Cosmic Ray is a natural way for nuclear fission and nucleosynthesis to occur. It refers to the formation of chemical elements from the impact of cosmic rays on an object.

6 0

2 years ago

Read 2 more answers

A swimming pool measures 5.0 m long x 4.0 m wide x 3.0 mdeep.

QveST [7]

Answer:

a) 588,000 N

b) 294000 N

Explanation:

Given that

Density of water = 1000kg/m3

(g) = 9.8m/s2

volume is given as (V)= 5m*4m*3m

a) force will be equal to weight of water

$W = mg = \rho g V = 1000\times 9.8 \times (5*4*3) = 588,000 kg m/s^2$

b) at either end

$P = \frac{F}{A} = P_o + \rho gh$

$dF = PdA$

$dF = \rho g w \int h dh$

$F = \rho g w \frac{h^2}{2}$

$F = \rho g A \frac{h}{2}$ [A = wh]

$F = 1000\times 9.8 \times 5\times 4 \times\frac{3}{2}$

F = 294000 N

3 0

3 years ago

Nutritionists use the ______ to measure how much energy we get from foods.

Natali [406]

<span>It is measured in calories </span>

6 0

2 years ago

Read 2 more answers

A.complete the following sentences with your own result or condition

ludmilkaskok [199]

1. If I hadn't bought that car yesterday.

2. I would have never bought the tickets for today..

3. She wouldn't have gotten suspended.

4. We could've worked on our project.

5. but I was in the shower.

6 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

2 years ago