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3 years ago

What is the 5 sources of chemical weathering

1 answer:

6 0

The 5 sources of chemical weathering is Hydration, oxidation, carbonation, hydrolysis and dissolution are forms of chemical weathering.
Hope this helps.

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A robot is exploring charon, the dwarf planet pluto's largest moon. Gravity on charon is 0.278 meters per second [m divided b

DochEvi [55]

In order to find the efficiency first we will find the Change in Potential energy of the small stone that robot picked up

First we will find the mass of the stone

As it is given that stone is spherical in shape so first we will find its volume

$V = \frac{4}{3}\pi r^3$

$V = \frac{4}{3}\pi *(\frac{0.06}{2})^3$

$V = 1.13 * 10^{-4} m^3$

Now it is given that it's specific gravity is 10.8

So density of rock is

$\rho = 10.8 * 10^3 kg/m^3$

mass of the stone will be

$m = \rho V$

$m = 10.8* 10^3 * 1.13 * 10^{-4}$

$m = 1.22 kg$

now change in potential energy is given as

$\Delta U = mgH$

here

g = gravity on planet = 0.278 m/s^2

H = height lifted upwards = 15 cm

$\Delta U = 1.22* 0.278 * 0.15$

$\Delta U = 0.051 J$

Now energy supplied by internal circuit of robot is given by

$E = Vit$

V = voltage supplied = 10 V

i = current = 1.83 mA

t = time = 12 s

$E = 10* 1.83 * 10^{-3} * 12$

$E = 0.22 J$

Now efficiency is defined as the ratio of output work with given amount of energy used

$\eta = \frac{\Delta U}{E}*100$

$\eta = \frac{0.051}{0.22} = 0.23$

so efficiency will be 23 %

5 0

2 years ago

A 70 kg driver gets into an empty taptap to start the day's work. The springs compress 2.3×10−2 m . What is the effective spring

ANEK [815]

Answer:

29856.521 N/m

Explanation:

$m=Mass\ of\ diver=70\ kg\\x=Length\ compressed\ by\ spring=2.3\times 10^{-2}\ m\\a=Acceleration\ due\ to\ gravity=9.81\ m/s^2\\F=Force\ exerted\ by\ diver=m\times a\\\Rightarrow F=70\times 9.81\\\Rightarrow F=686.7\ N\\k=spring\ constant\\F=k\times x\\\Rightarrow k=\frac {F}{x}\\\Rightarrow k=\frac {686.7}{2.3\times 10^{-2}}\\\therefore k=29856.521\ N/m$

5 0

3 years ago

What happens in a tug of war if the net forces are balanced and why?

FinnZ [79.3K]

Answer:

Balanced forces are responsible for unchanging motion. Balanced forces are forces where the effect of one force is cancelled out by another. A tug of war, where each team is pulling equally on the rope, is an example of balanced forces. The forces exerted on the rope are equal in size and opposite in direction.

Explanation:

6 0

3 years ago

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

2 years ago

Starting from rest, a solid sphere rolls without slipping down an incline plane. At the bottom of the incline, what does the ang

Marrrta [24]

Answer:

2/R*sqrt (g*s*sin(θ)) = w

Explanation:

Assume:

- The cylinder with mass m

- The radius of cylinder R

- Distance traveled down the slope is s

- The angular speed at bottom of slope w

- The slope of the plane θ

- Frictionless surface.

Solution:

- Using energy principle at top and bottom of the slope. The exchange of gravitational potential energy at height h, and kinetic energy at the bottom of slope.

ΔPE = ΔKE

- The change in gravitational potential energy is given as m*g*h.

- The kinetic energy of the cylinder at the bottom is given as rotational motion: 0.5*I*w^2

- Where I is the moment of inertia of the cylinder I = 0.5*m*R^2

We have:

m*g*s*sin(θ) = 0.25*m*R^2*w^2

2/R*sqrt (g*s*sin(θ)) = w

- The angular velocity depends on plane geometry θ , distance travelled down slope s, Radius of the cylinder R , and gravitational acceleration g

3 0

3 years ago