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Stels [109]
3 years ago
7

Madison is a responsible homeowner. She often thinks about the environment when working on home projects. What should Madison do

to help slow pollution as she works?
1 Measure carefully and more than once when cutting lumber to reduce waste.
2 Purchase tools that have multiple uses rather than buying many specialized tools.
3 Pour unused paint down the drain instead of taking it to the recycling center.
4 Plant vegetation and trees on hilly areas to hold nutrient-rich soil in place.
5 When given a choice, pick the most expensive tool or material for the job.
Physics
2 answers:
uranmaximum [27]3 years ago
4 0

2 and or 4. 2 would be the most sensible option, however. because it would cut back on her personal carbon footprint

Alex3 years ago
3 0

Answer:

4. Plant vegetation and trees on hilly areas to hold nutrient-rich soil in place.

Explanation:

This is something that Madison could do in order to be a responsible homeowner. When remodelling a home, it is important to think of what the impact of our actions is. One way in which Madison could minimize the negative impact of her changes would be by planting vegetation and trees on hilly areas in order to hold nutrient-rich soil in place. This will maintain the quality of the soil, as well as reduce damaging erosion.

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Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with an
Alona [7]

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

F=\dfrac{kQ^2}{r^2} ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}.....(2)

Dividing equation (1) and (2), we get :

\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}

Hence, the correct option is (d) i.e. " 4F/9"

7 0
3 years ago
The maximum distance particles of the medium move when a wave passes through them is wave.
joja [24]

Answer:

amplitude

Explanation:

6 0
2 years ago
) The radius of sphere measured repeated values 5.63 m, 5.54 m and 5.53 m. Determine the most
Nadusha1986 [10]

Answer:

v_{ average} = 5.57

Explanation:

The most probable value of a measure is

       v_average = \frac{1}{N} ∑ x_i

where N is the number of measurements

in tes case N = 3

         v_{average} = ⅓ (5.63 +5.54 + 5.53)

         V_{average} = 5,567

The number of significant figures must be equal to the number of figures that have the least in the readings.

          v_{ average} = 5.57

3 0
2 years ago
Determine in symbols an expression for the magnetic force exerted on the falling bar (and determine the direction of that force)
Katarina [22]

Answer:

magnetic force on falling Bar F = B*i*L*sin(90) = B*(B*L*v/R)*L = B^2*L^2*v/R

direction of the force is vertically upwards

Explanation:

8 0
3 years ago
A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr
valina [46]
The  spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
T=2 t = 2 \cdot 0.100 s = 0.200 s
Which means that the frequency is
f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz
and the angular frequency is
\omega=2 \pi f = 2 \pi (5 Hz)=31.4 rad/s

In a spring-mass system, the maximum velocity of the object is given by
v_{max} = A \omega
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s
6 0
3 years ago
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