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Roman55 [17]
3 years ago
12

What are the main causes of crack growth in rocks overtime

Physics
2 answers:
garik1379 [7]3 years ago
5 0
Erosion and weathering. They both make rocks crack more.
lana [24]3 years ago
4 0
Ice Wedging and Plant Growth
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Is the Earth getting larger as more sea floor is added? explain
Natasha_Volkova [10]

The seafloor is only expanding in one area, which is the Atlantic Ocean. It's just Earth's seafloor recylcing itself.

4 0
3 years ago
The distance between two electric pole is 800m then express this in terms of mm​
Lana71 [14]

Answer:

Mm stands for milimetres

Explanation:

6 0
3 years ago
Read 2 more answers
We are going to make an imaginary engine using water. We are going to heat 100 grams of water to 120 C from its initial temperat
Svetach [21]

Answer:

The work done by this engine is 800 cal

Explanation:

Given:

100 g of water

120°C final temperature

22°C initial temperature

30°C is the temperature of condensed steam

Cw = specific heat of water = 1 cal/g °C

Cg = specific heat of steam = 0.48 cal/g °C

Lw = latent heat of vaporization = 540 cal/g

Question: How much work can be done using this engine, W = ?

First, you need to calculate the heat that it is necessary to change water to steam:

Q_{1} =m_{w} C_{w} (100-22)+m_{w}L_{w}+m_{w}C_{g}(120-100)

Here, mw is the mass of water

Q_{1} =(100*1*78)+(100*540)+(100*0.48*20)=62760cal

Now, you need to calculate the heat released by the steam:

Q_{2} =m_{w}C_{g}(120-100)+m_{w}L_{w}+m_{w}C_{w}(100-30)=(100*0.48*20)+(100*540)+(100*1*70)=61960cal

The work done by this engine is the difference between both heats:

W=Q_{1}-Q_{2}=62760-61960=800cal

8 0
3 years ago
If the effort distance = 8 meters and the resistance distance = 1/2 meter. How much force does this man have to apply to lift th
Delicious77 [7]

Answer:

Explanation:

Effort x effort distance = load x resistance  distance

effort distance = 8 m ,

load = 2000N

resistance distance = 1/2 m = 0.5 m

Putting the values in the equation above

effort x 8m = 2000N x .5

effort = 2000 x 0.5 / 8

= 125 N

force required = 125 N .

5 0
3 years ago
A 23.7 kg kid slides down a
anyanavicka [17]

Answer:

\text { The acceleration of the kid is } 7.18 \mathrm{m} / \mathrm{s}^{2}

Explanation:

Mass of the kid 23.7 kg.

\text { The kid is accelerating down at an angle is } 47.2^{\circ} .

^{\prime \prime} \mathrm{g}^{\prime \prime} \text { acceleration due to gravity is } 9.8 \mathrm{m} / \mathrm{s}^{2}

We need to find the acceleration of the kid,

We know that, Parallel force acted on the kid at an angle is

F = m × g × sinθ (F = ma)

m × a = m × g × sinθ

Now, substitute the given values in the above formula to find acceleration of the kid,

23.7 \times a=23.7 \times 9.8 \times \sin 47.2^{\circ}

23.7 × a = 232.26 × 0.733

23.7 × a = 170.24

a=\frac{170.24}{23.7}

a=7.18 \mathrm{m} / \mathrm{s}^{2}

\text { Therefore, acceleration of the kid is } 7.18 \mathrm{m} / \mathrm{s}^{2}

6 0
3 years ago
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