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kogti [31]
3 years ago
7

A pesticide was applied to a population of roaches, and it was determined that the LD50 was 55mgkg. If the average mass of a roa

ch was 0.02kg, which of the following approaches will determine the dose in mg per roach?
Physics
1 answer:
zhannawk [14.2K]3 years ago
0 0

Answer:

55mg/kg * 0.02kg which gives 1.1 mg pesticides

Explanation:

(Options are missing)

If the average mass or weight of a roach is 0.02kg

And 55mg is needed per 1 kg roaches

The amount or size of pesticide required for 0.02kg roach is 55mg/kg * 0.02kg

Size = 55mg/kg * 0.02 kg

Size = 1.1mgkg/kg

Size = 1.1 mg

So, the size of pesticide required is 1.1mg

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How many spoonfuls of water did it take for your sponge to be 100% saturated?
adell [148]

Answer:

19

Explanation:

I legit did this and it took 19.

7 0
2 years ago
A concrete piling of 50 kg is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. How much will the wire stretch?
pentagon [3]

Explanation:

It is given that,

Mass of concrete pilling, m = 50 kg

Diameter of wire, d = 1 mm

Radius of wire, r = 0.0005 m

Length of wire, L = 11.2

Young modulus of steel, Y=20\times 10^{10}\ N/m^2

The young modulus of a wire is given by :

Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}

Y=\dfrac{F.L}{A\Delta L}

\Delta L=\dfrac{F.L}{A.Y}

\Delta L=\dfrac{50\ kg\times 9.8\ m/s^2\times 11.2\ m}{\pi (0.0005\ m)^2\times 20\times 10^{10}\ N/m^2}

\Delta L=0.034\ m

So, the wire will stretch 0.034 meters. Hence, this is the required solution.

8 0
3 years ago
Help please an observer at the North Pole sees the moon phase shown below. What Moon phase will be observed approximately one we
ziro4ka [17]

Answer:

it should be a new moon

Explanation:

4 0
3 years ago
Find the mass of a flying discus that has a net force of 1.05 newtons and accelerates at 3.5 m/s^2
Ilya [14]
F=ma
m=1.05/3.5= 0.3kg
8 0
3 years ago
A square wave has amplitude 0 V for the low voltage and 4 V for the high voltage. Calculate the average voltage by integrating o
Margarita [4]

Answer:

V_{average} = \frac{1}{2}  V_o  ,     V_{average} = 2 V

Explanation:

he average or effective voltage of a wave is the value of the wave in a period

            V_average = ∫ V dt

in this case the given volage is a square wave that can be described by the function

           V (t) = \left \{ {{V=V_o \ \ \  t<  \tau /2} \atop {V=0 \ \  \ \  t> \tau /2 }   } \right.

to substitute in the equation let us separate the into two pairs

             V_average = \int\limits^{1/2}_0 {V_o} \, dt + \int\limits^1_{1/2} {0} \, dt

             V_average = V_o \ \int\limits^{1/2}_0 {} \, dt

             V_{average} = \frac{1}{2}  V_o

we evaluate  V₀ = 4 V

             V_{average} = 4 / 2)

             V_{average} = 2 V

6 0
3 years ago
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