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jeka57 [31]
3 years ago
6

What Type Of Technologies are helping reduce ozone?

Physics
1 answer:
atroni [7]3 years ago
4 0

Sector-specific technological measures – vapor recovery nozzles at gasoline pumps, cleaner-burning fuels, strong vehicle inspection programs and strict emission limits for refineries, industrial emissions and combustion sources – all can help control ozone pollution and have contributed to lower emissions over the last three decades

hope this helps

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How does charging by conduction compare with charging by induction?
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Oppositely charged objects form in induction but not in conduction.
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If you push twice as hard against a stationary brick wall, the amount of work you do
Alecsey [184]

Answer:

Zero

Explanation:

The work done on an object is given by:

W=Fd cos \theta

where

F is the force applied on the object

d is the displacement of the object

\theta is the angle between the direction of the force and the displacement

In this problem, you are pushing again a stationary wall: this means that the walls does not move. As a result, the displacement is zero: d=0. Therefore, the work done is also zero: W=0.

8 0
3 years ago
assume that the initial speed is 25 m/s and the angle of projection is 53 degree above the hroizontal. the cannon ball leaves th
finlep [7]

Answer:

A.  xmax = 131.49 m

B.  t = 8.74 s

C.  ymax = 220.33 m

Explanation:

A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:

y=y_o+v_osin\theta-\frac{1}{2}gt^2      (1)

yo: height from the projectile is fired = 200m

vo: initial velocity of the projectile = 25m/s

g: gravitational acceleration = 9.8 m/s^2

θ: angle between the direction of the initial motion of the ball and the horizontal = 53°

t: time

You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.

When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:

0=200+(25)sin53\°t-\frac{1}{2}(9.8)t^2\\\\0=200+19.96t-4.9t^2 (2)

You use the quadratic formula to obtain the value of t:

t_{1,2}=\frac{-19.96\pm\sqrt{(19.96)^2-4(-4.9)(200)}}{2(-4.9)}\\\\t_{1,2}=\frac{-19.96\pm65.71}{-9.8}\\\\t_1=8.74s\\\\t_2=-4.66s

You use the positive value because it has physical meaning.

Now, you can calculate the horizontal range of the projectile by using the following formula:

x_{max}=v_ocos\theta t      

x_{max}=(25m/s)(cos53\°)(8.74s)=131.49m

The cannon ball travels a horizontal distance of 131.49 m

B. The cannon ball reaches the canon for t = 8.74s

C. The maximum height is obtained by using the following formula:

y_{max}=y_o+\frac{v_o^2sin^2\theta}{2g}     (3)

By replacing in the equation (3) the values of all parameters you obtain:

y_{max}=200m+\frac{(25m/s)^2(sin53\°)^2}{2(9.8m/s^2)}\\\\y_{mac}=200m+20.33m=220.33m

The maximum height reached by the cannon ball is 220.33m

3 0
3 years ago
4000 J of heat energy is applied to a 70 g sample of water initially at 35oC. What is the final temperature of the sample?
BARSIC [14]

Answer:

48.7°C

Explanation:

Step one:

given data

Quantity of heat = 4000J

mass of sample= 70g= 0.07kg

initial temperature T1=35°C

Water has a specific heat capacity of 4182 J/kg°C

Required

The final temperature T2

Step two:

Q= mcΔT

4000= 0.07*4182(T2-35)

4000=292.74(T2-35)

4000=292.74T2-10245.9

collect like terms

4000+10245.9=292.74T2

14245.9= 292.74T2

divide both sides by 292.74

T2= 14245.9/ 292.74

T2=48.7°C

8 0
3 years ago
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