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2 years ago

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Mr Smith is working in a muddy garden. When he picks up a paving stone his feet sink deeper into the mud. Explain why his feet s

ink into the mud more when he picks up the paving stone. Write about pressure in your answer

1 answer:

Anni [7]2 years ago

3 0

because he is carrying more mass and as the ground is muddy his feet goes in due to the pull of gravity

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The phosphorus cycle is important to ecosystems. Choose all of these statements that are true concerning the phosphorus cycle.

Lesechka [4]

Answer:

The largest reservoir of phosphorous is sedimentary rock.

Major sources of phosphorous to aquatic ecosystems are fertilizer runoff, sewage leaks, and industrial wastes.

Eccess phosphorous can lead to eutrophication

Explanation:

Phosphorus come from different sources such as aquatic ecosystems and fertilizers used for plants. When these substances containing phosphorus and those from industrial wastes find their way into water bodies, they tend to cause eutrophication, which is the natural enrichment of water bodies.

Also, it is known that a very small portion of phosphoric acid contribute to acid rain in the atmosphere.

7 0

2 years ago

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Which of the following terms refers to the shortest length from a starting

shutvik [7]

Answer:

A) Distance

Explanation:

The shortest distance that is represented as the measurement from one point to another is called distance.

It should not be confused with displacement as it represents the sum of the distances travelled by a body or person.

3 0

2 years ago

a rocket is launched with a constant acceleration straight up. exactly 4.00 seconds after lift off, a bolt falls off the side of

Papessa [141]

The initial speed of the bolt is not 58.86 m/s.

Let a be the acceleration of the rocket.

During the 4 sec lift off, the rocket has reached a height of

h = (1/2)*a*t^2

with t=4,

h = (1/2)*a^16

h = 8*a

Its velocity at 4 sec is

v = t*a

v = 4*a

The initial velocity of the bolt is thus 4*a.

During the 6 sec fall, the bolt has the initial velocity V0=-4*a and it drops a total height of h=8*a. From the equation of motion,

h = (1/2)*g*t^2 + V0*t

Substituting h0=8*a, t=6 and V0=-4*a into it,

8*a = (1/2)*g*36 - 4*a*6

Solving for a

a = 5.52 m/s^2

6 0

2 years ago

A puck of mass 0.70 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the movi

natali 33 [55]

Answer:

$v_1 = \ 5.196 \frac{m}{s}$

$v_2 = 3 \frac{m}{s}$

Explanation:

For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

$\vec{p}_i = \vec{p}_f$

where the suffix i means initial, and the suffix f means final.

The initial momentum will be:

$\vec{p}_i = m_1 \ \vec{v}_{1_i} + m_2 \ \vec{v}_{2_i}$

as the second puck is initially at rest:

$\vec{v}_{2_i} = 0$

Using the unit vector $\vec{i}$ pointing in the original line of motion:

$\vec{v}_{1_i} = 6.0 \frac{m}{s} \hat{i}$

$\vec{p}_i = 0.70 \ kg \ 6.0 \frac{m}{s} \ \hat{i} + 0.70 \ kg \ 0$

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

So:

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i} = \vec{p}_f$

$\vec{p}_f = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

$\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

So, our velocity vectors will be:

$\vec{v}_{1_f} = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )$

$\vec{v}_{2_f} = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

We got

$\vec{p}_f = 0.7 \ kg \ \vec{v}_{1_f} + 0.7 \ kg \ \vec{v}_{2_f}$

$4.2 \ \frac{kg \ m}{s} \ \hat{i} = 0.7 \ kg \ v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ ) + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

So, we got the equations:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ v_1 \ cos(30 \°) + 0.7 \ kg \ v_2 \ cos(-60 \°)$

and

$0 = 0.7 \ kg \ v_1 \ sin(30 \°) + 0.7 \ kg \ v_2 \ sin(-60 \°)$ .

From the last one, we get:

$0 = 0.7 \ kg \ ( v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°) )$

$0 = v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°)$

$v_1 \ sin(30 \°) = - \ v_2 \ sin(-60 \°)$

$v_1 = \ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) }$

and, for the first one:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ ( v_1 \ cos(30 \°) + v_2 \ cos(60 \°) )$

$\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} = v_1 \ cos(30 \°) + v_2 \ cos(60 \°)$

$\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} = v_1 \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = (\ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = v_2 (\ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + cos(60 \°)$

$6 \ \frac{m}{s} = v_2 * 2$

so:

$v_2 = 6 \ \frac{m}{s} / 2 = 3 \frac{m}{s}$

and

$v_1 = \ 3 \frac{m}{s} \ \frac{sin(60 \°)}{ sin(30 \°) }$

$v_1 = \ 5.196 \frac{m}{s}$

3 0

2 years ago

2 H2O2-> 2H20+02

Luden [163]

Answer:

This reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen. Add 2 molecules of hydrogen peroxide and 2 molecules of water. Since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

3]Explanation: This reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen. Add 2 molecules of hydrogen peroxide and 2 molecules of water. Since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

4]Two moles of hydrogen peroxide H2O2 decomposes to produce two moles of water H2O and one mole of oxygen gas O2(g) , which then bubbles off

6 0

2 years ago