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hammer [34]
2 years ago
6

What hazardous wastes are common in ordinary households? What can you do to reduce the impact you make on the environment from t

he use of hazardous wastes?
Physics
1 answer:
labwork [276]2 years ago
3 0

Answer:

The most common products include aerosols, anti-freeze, asbestos, fertilizers, motor oil, paint supplies, photo chemicals, poisons, and solvents, cleaning supplies.

Explanation:

Use homemade cleaners

You can find local retailers to take rechargeable batteries from laptop computers, cordless power tools, cellular and cordless phones, and camcorders at the Rechargeable Battery Recycling Corporation’s website

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A car is travelling to the right with a speed of 40 m/s when the driver slams on the brakes. The car skids for 4 s with constant
vazorg [7]

Answer:

10m/s^2

Explanation:

Given data

velocity= 40m/s

time= 4 seconds

Acceleration a =????

We know that

a= velocity/time

substitute

a=40/4

a= 10m/s^2

Hence the acceleration will be 10m/s^2

3 0
2 years ago
In the diagram, q1= -2.60*10^-9 C and
Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

E_{net}=90.37\: N/c

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}

Where:

  • k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
  • q1 is the first charge
  • d1 is the distance from q1 to P

|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}

|E_{1}|=80.84\: N/C

And E2 will be:

|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}

|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}

|E_{2}|=40.39\: N/C

Finally, we need to use the  Pythagoras theorem to find the magnitude of the net electric field.

E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}

E_{net}=\sqrt{80.84^{2}+40.39^{2}}

E_{net}=90.37\: N/c

I hope it helps you!

7 0
3 years ago
the earth has a radius of 6.38×10^16 meter and turns around once on its axis in 24 hour.what is the radial acceleration of perso
Scrat [10]

337493603.8m/s²

Explanation:

Radius of the earth = 6.38 x 10¹⁶m

time = 24hr (86400s)

Unknown:

Centripetal acceleration = ?

Solution:

The centripetal acceleration is directed inward to keep the body from falling off the surface of the earth.

     centripetal acceleration  = \frac{v^{2} }{r}

  where v is the velocity and r is the radius

   also;

            v  = wr

  where w is the angular velocity

substituting in the equation for centripetal acceleration gives;

                 

          a = w²r

 also w = \frac{2 x pi}{T}

      therefore;

                 a = \frac{4 \pi  ^{2} r }{T^{2} }

 a = \frac{ 4 x 3.142^{2}  x 6.38 x 10^{16} }{86400^{2} }

 a = 337493603.8m/s²

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

5 0
3 years ago
What is the minimal mass of helium (density 0.18 kg/m3) needed to lift a balloon carrying two people in a basket, if the total m
Sergio039 [100]

Answer:   
 M[min] = M[basket+people+ balloon, not gas] * ΔR/R[b] 
 Î”R is the difference in density between the gas inside and surrounding the balloon. 
 R[b] is the density of gas inside the baloon.   
 ==================================== 
 Let V be the volume of helium required. 
 Upthrust on helium = Weight of the volume of air displaced = Density of air * g * Volume of helium = 1.225 * g * V   
 U = 1.225gV newtons 
 ---- 
 Weight of Helium = Volume of Helium * Density of Helium * g 
 W[h] = 0.18gV N   
 Net Upward force produced by helium, F = Upthrust - Weight = (1.225-0.18) gV = 1.045gV N  -----

 
 Weight of 260kg = 2549.7 N 
 Then to lift the whole thing, F > 2549.7 
 So minimal F would be 2549.7 
 ---- 
 1.045gV = 2549.7 
 V = 248.8 m^3   
 Mass of helium required = V * Density of Helium = 248.8 * 0.18 = 44.8kg (3sf)   
 =====   
 Let the density of the surroundings be R 
 Then U-W = (1-0.9)RgV = 0.1RgV   
 So 0.1RgV = 2549.7 N 
 V = 2549.7 / 0.1Rg   
 Assuming that R is again 1.255, V = 2071.7 m^3 
 Then mass of hot air required = 230.2 * 0.9R = 2340 kg   
 Notice from this that M = 2549.7/0.9Rg * 0.1R so   
 M[min] = Weight of basket * (difference in density between balloon's gas and surroundings / density of gas in balloon)   
 M[min] = M[basket] * ΔR/R[b]
3 0
3 years ago
Suppose that the radius of the circular path is r when the speed of the rocket is v and the acceleration of the rocket has magni
SCORPION-xisa [38]

Answer:

A3

Explanation:

4 0
3 years ago
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