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MariettaO [177]

3 years ago

13

swelling magma chambers below the surface can cause the crust above to uplift and form a large flat regions higher than the surr

ounding land known as

1 answer:

klemol [59]3 years ago

3 0

Answer: I belive the answer is 666

Explanation:

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A 5000kg freight car moving at 2 m/s East collides with a 10,000kg freight car at rest. Upon collision, they got stuck and moved

mr_godi [17]

Answer:

A

Explanation:

6 0

3 years ago

Given this measurement 2642 ft, how many significant figures are there?

Stolb23 [73]

4 sig figs. Just count

8 0

3 years ago

State two factors that affect the rate of diffusion of a substance

Ronch [10]

Explanation:

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<em>Diffusion</em><em> </em><em>of</em><em> </em><em>substance</em><em> </em><em>plays</em><em> </em><em>an</em><em> </em><em>important</em><em> </em><em>role</em><em> </em><em>on</em><em> </em><em>cellular</em><em> </em><em>transport</em><em> </em><em>in</em><em> </em><em>plants</em><em>.</em><em> </em>
<em>Diffusion</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>passive</em><em> </em><em>movement</em><em> </em><em>of</em><em> </em><em>substance</em><em> </em><em>from</em><em> </em><em>a</em><em> </em><em>region</em><em> </em><em>of</em><em> </em><em>higher</em><em> </em><em>concentration</em><em> </em><em>to</em><em> </em><em>a</em><em> </em><em>region</em><em> </em><em>of</em><em> </em><em>lower</em><em> </em><em>concentration</em><em>. </em>

4 0

3 years ago

A student places a block on a table and hangs one mass from the block. The student lets the block go and observes the block move

NeTakaya

Magnitude of acceleration

Explanation:

We know that acceleration can increase depending in the force applied on an object, any object with a greater mass will apply a greater force. F = M(a).

7 0

3 years ago

A Foucault pendulum consists of a brass sphere with a diameter of 31.0 cm suspended from a steel cable 11.0 m long (both measure

kozerog [31]

Answer:

43.7 °C

Explanation:

$\alpha_b$ = Coefficient of linear expansion of brass = $18\times 10^{-6}\ ^{\circ}C$

$\alpha_s$ = Coefficient of linear expansion of steel = $11\times 10^{-6}\ ^{\circ}C$

$L_{0b}$ = Initial length of brass = 31 cm

$L_{0s}$ = Initial length of steel = 11 m

$\Delta L$ = Total change in length = 3 mm

Total change in length would be

$\Delta L=\Delta L_b+\Delta L_s\\\Rightarrow \Delta L=L_{0b}\alpha_b\Delta T+L_{0s}\alpha_b\Delta T\\\Rightarrow \Delta T=\frac{\Delta L}{L_{0b}\alpha_b+L_{0s}\alpha_b}\\\Rightarrow \Delta T=\frac{0.003}{0.31\times 18\times 10^{-6}+11\times 10^{-6}\times 11}\\\Rightarrow \Delta T=23.7\ ^{\circ}C$

$\Delta T=23.7\\\Rightarrow T_f-T_i=23.7\\\Rightarrow T_f=23.7+T_i\\\Rightarrow T_f=23.7+20\\\Rightarrow T_f=43.7\ ^{\circ}C$

The final temperature is 43.7 °C

6 0

4 years ago