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Soloha48 [4]
3 years ago
14

Describe how water may be treated before people use it

Physics
2 answers:
kramer3 years ago
7 0
They purify the water so that people can use it for things and drinking. They chloride, boil, or use ultraviolet light to kill bacterias that can harm us
brilliants [131]3 years ago
3 0
Water is treated by purifying it by adding slaked lime or potash alum

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Help pleasseeee URGENT
Zina [86]

Answer:

The speed of the 8-ball is 2.125 m/s after the collision.

Explanation:

<u>Law Of Conservation Of Linear Momentum</u>

The total momentum of a system of masses is conserved unless an external force is applied. The momentum of a body with mass m and velocity v is calculated as follows:

P=mv

If we have a system of masses, then the total momentum is the sum of all the individual momentums:

P=m_1v_1+m_2v_2+...+m_nv_n

When a collision occurs, the velocities change to v' and the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

In a system of two masses, the law of conservation of linear momentum is simplified to:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

The m1=0.16 Kg 8-ball is initially at rest v1=0. It is hit by an m2=0.17 Kg cue ball that was moving at v2=2 m/s.

After the collision, the cue ball comes to rest v2'=0. It's required to find the final speed v1' after the collision.

The above equation is solved for v1':

\displaystyle v'_1=\frac{m_1v_1+m_2v_2-m_2v'_2}{m_1}

\displaystyle v'_1=\frac{0.16*0+0.17*2-0.17*0}{0.16}

\displaystyle v'_1=\frac{0.34}{0.16}

v'_1=2.125\ m/s

The speed of the 8-ball is 2.125 m/s after the collision.

8 0
3 years ago
What is the tangential speed of a point on a wheel of a car if the point is located 0.114 m from the axis of rotation; and the w
Svetach [21]

Answer:

Tangential speed = R * w  where w is the angular speed

For a wheel   v = .114 m * 6.53 / sec = .744 m/s

Note that this is the tangential speed - If the wheel were moving at speed V then as viewed from the ground you have

top V + v

bottom   V - v

That is the wheel also has speed due to the speed of the axle (axis of rotation)

3 0
3 years ago
A 15 m ladder with a mass of 51 kg is leaning against a frictionless wall, which makes an angle of 60 degrees with the horizonta
timofeeve [1]

Answer:

Explanation:

Given that, .

Mass of ladder is 51kg

Then, it weight is

WL = mg = 51 × 9.81 = 500.31N

This weight will act at the midpoint of the ladder

Length of ladder is 15m

The ladder makes an angle 55°C with the horizontal

An object whose mass is 81kg is at 4m from the bottom of the ladder

Then, weight of object

Wo = mg = 81 × 9.81 = 794.61 N

Using newton second law

Check attachment

Ng is normal force on the ground

Ff is the horizontal frictional force

Nw Is the normal force on the wall

ΣFy = 0

Ng = Wo + WL

Ng = 794.61 + 500.31

Ng = 1294.92 N

Also

ΣFx = 0

Ff — Nw = 0

Then,

Ff = Nw

Now taking moment about point A.

Check attachment

using the principle of equilibrium

Sum of clockwise moment equals to sum of anti-clockwise moment

Also note that the Normal force on the wall is not perpendicular to the ladder, so we will resolve that and also the weights of ladder and weight of object

Clockwise = Anticlockwise

Wo•Cos60 × 4 + WL•Cos60 × 7.5 = Nw•Sin60 × 15

794.61Cos60 × 4 + 500.31Cos60 × 7.5 = Nw × Sin60 × 15

1589.22 + 1876.163 = 12.99•Nw

3465.383 = 12.99•Nw

Nw = 3465.383 / 12.99

Nw = 266.77 N

Since, Nw = Ff

Then, Ff = 266.77N

the horizontal force exerted by the ground on the ladder is 266.77 N

8 0
3 years ago
Read 2 more answers
A girl lifts a 160-N load to a height of 1 min 0.5 s. How much power is used to lift the load?
Aleonysh [2.5K]

Answer: 320 W

Explanation:

P = W/t

W = F * d

F = 160 N

d = 1 m

t = 0.5 s

W = 160 * 1

W = 160 J

P = 160 J / 0.5 s

P = 320 W

8 0
3 years ago
A 700 g can of beans is dropped from a shelf that is 1.5 m high. What is the gravitational potential energy of this can? Round y
OLga [1]
<span>To find the gravitational potential energy of an object, we can use this equation: GPE = mgh m is the mass of the object in kg g = 9.80 m/s^2 h is the height of the object in meters GPE = mgh GPE = (0.700 kg) (9.80 m/s^2) (1.5 m) GPE = 10.3 J The gravitational potential energy of this can is 10.3 J</span>
4 0
3 years ago
Read 2 more answers
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