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gogolik [260]
3 years ago
15

What is sludge dumping?HELP ASAP PLEASE

Physics
1 answer:
yulyashka [42]3 years ago
5 0

Answer:

What is sludge dumping? Sludge is the solid waste in raw sewage. Sludge dumping is discharging that waste into the ocean.

Explanation:

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The origin of the universe remains a question. true false
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True ..............................
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Which statement best define percent error? check all that apply.
Art [367]

Percent error is the difference between the experimental value and theoretical value and measures the accuracy of the result found. The larger the error, lesser is the accuracy and vice versa.

Solution:

It is a mathematical way of showing accuracy

The higher the percent error, the less accurate the data set,

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1 year ago
What geological feature can be created by a divergent boundary
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5 0
3 years ago
Read 2 more answers
A 3.9 g dart is fired into a block of wood with a mass of 24.6 g. The wood block is initially at rest on a 1.5 m tall post. Afte
Galina-37 [17]

Answer:

46.48m/s

Explanation:

The problem is a combination of the principle of conservation of linear momentum and projectile motion.

The principle of conservation of linear momentum states that in a closed system, the total momentum of colliding bodies before impact is equal to the total momentum after impact. The masses stated in the problem experienced an inelastic collision. In an inelastic collision, the bodies involved stick together after the collision and move with a common velocity.

For two bodies of masses m_1 and m_2 moving with velocities u_1 and u_2 before impact, if they experience inelastic collision, the conservation of their momenta is as stated in equation (1);

m_1u_1+m_2u_2=(m_1+m_2)v..................(1)

were v is their common velocity after impact. If the second mass m_2 was at rest before the impact, then its initial velocity u_2=0m/s. therefore m_2u_2=0. Equation (1) then becomes;

m_1u_1=(m_1+m_2)v..............(2)

In the problem stated, the second mass taken as the mass of the wooden block was at rest before the impact and the collision was inelastic since both the wood and the dart stuck together and moved with a common velocity after the impact. Therefore we can use equation (2) for the problem.

Given;

m_1=3.9g=0.0039kg\\u_1=?\\m_2=24.6g=0.0246kg\\v=?

Substituting these values into (2), we get the following;

0.0039*u_1=(0.0039+0.024)v\\0.0039u_1=0.0285v.........(3)

Their common v velocity after impact now makes both the wooden block and the dart (as a single body) to fall vertically through a height h of 1.5m over a range R of 3.5m as stated by the problem; hence by the principle of projectile motion for a body projected horizontally, the following relationship holds;

R= vt............(4)

were t is the time taken to fall through the height h. To obtain t we use the second equation of free fall under gravity;

h=\frac{1}{2}gt^2...........(5)

were g is acceleration due to gravity taken as 9.8m/s^2. Therefore;

1.5=\frac{1}{2}*9.8*t^2\\1.5=4.9t^2\\t^2=\frac{1.5}{4.9}=0.306\\t=\sqrt{0.306} =0.55s

We then substitute R and t into equation (4) to obtain v.

3.5=v*0.55\\v=\frac{3.5}{0.55}\\v=6.36m/s

We now further substitute this value of v into (3) to obtain u_1;

u_1=\frac{0.0285v}{0.0039}\\\\u_1=\frac{0.0285*6.36}{0.0039}\\\\u_1=\frac{0.18126}{0.0039}\\\\u_1=46.48m/s

4 0
4 years ago
A girl drops a stone from the top a tower 45m tall. At the same time, a boy standing at the base of the tower, projects another
Advocard [28]

Answer:

(i) The stones meet at 1.8 second

(ii) The point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.

Explanation:

(i)

First we consider the stone dropped by the girl. We have data:

Vi = Initial Velocity of Stone = 0 m/s   (Since the stone was initially at rest)

t = Time Period

g = 10 m/s²

s₁ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₁ = Vi t + (0.5)gt²

s₁ = (0)(t) + (0.5)(10)t²

s₁ = 5t²   ----- equation (1)

Now, we consider the stone throne vertically upward by the boy. We have data:

Vi = Initial Velocity of Stone = 25 m/s

t = Time Period

g = - 10 m/s²   (negative sign due to upward motion)

s₂ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₂ = Vi t + (0.5)gt²

s₂ = (25)(t) + (0.5)(-10)t²

s₂ = 25t - 5t²   ----- equation (2)

At, the point where both the stones meet, the sum of distances covered by both stones must be equal to the height of building (i.e 45 m).

s₁ + s₂ = 45

using values from equation (1) and equation (2)

5t² + 25t - 5t² = 45

25t = 45

t = 45/25

<u>t =  1.8 sec</u>

(ii)

using this value of of t in equation (2)

s₂ = (25)(1.8) - (5)(1.8)²

<u>s₂ = 28.8 m</u>

using this value of of t in equation (1)

s₁ = (5)(1.8)²

<u>s₁ = 16.2 m</u>

<u>Hence, the point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.</u>

6 0
3 years ago
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