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2 years ago

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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?

1 answer:

zhannawk [14.2K]2 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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A policeman starts giving chase 60 seconds after a stolen car zooms by at 108 km/hr. At what minimum speed should he drive if he

Iteru [2.4K]

Answer:

30.93 m/s

Explanation:

Given that, the speed of stolen car is,

$v_{s} =108km/hr\\v_{s} =108\times \frac{5}{18}m/s\\ v_{s} =30m/s$

As policeman start chasing the stolen car after 60 seconds.

Now suppose the speed of policeman car is, $v_{p}$

The policeman catches the stolen car at a distance of,

$S=60km\\S=60000m$

Now the distance covered by the policeman in time t is $v_{p}\times t$

And the distane cover by the thief in stolen car in time(t+60s) is $v_{s}\times (t+60sec)$ .

And these distances are equal and they are equal to 60000 m.

Therefore,

$v_{p}\times t=v_{s}\times (t+60sec)=60000m$

Therfore,

$v_{s}\times (t+60sec)=60000m\\30m/s\times (t+60sec)=60000m\\(t+60s)=2000s\\t=1940s$

Now use this value to solve for minimum speed of policeman's car.

$v_{p}\times 1940=60000\\v_{p}=30.93 m/s$

Therefore minimum speed of policeman's car is 30.93 m/s.

6 0

2 years ago

The drag force, fd, imposed by the surrounding air on a vehicle moving with velocity v is given by fd = cdaρv 2/2 where cd is a

kiruha [24]

Answer: 191 N

Explanation:

Given that,

Velocity v = 80 km/hr

Area $a = 2.3\ m^{2}$

Air density $\rho = 1.2\ kg/m^{3}$

Constant cd = 0.28

We know the drag force,

$F_{d} = cda\dfrac{1}{2}\rho v^{2}$

$F_{d} = 0.28\times 2.3\ m^{2}\times\dfrac{1}{2}\times1.2\ kg/m^{3} \times (\dfrac{800}{36}\ m/s)^{2}$

$F_{d} = 190.8 N$

$F_{d} = 191 N$

Hence, the drag force is 191 N.

7 0

2 years ago

A small town has decided to forego the use of electrical power and send energy through town via mechanical waves on ropes. They

mojhsa [17]

Answer:

the required frequency of waves is 2.066 Hz

Explanation:

Given the data in the question;

μ = 1.50 kg/m

T = 6000 N

Amplitude A = 0.500 m

P = 2.00 kW = 2000 W

we know that, the average power transmit through the rope can be expressed as;

p = $\frac{1}{2}$ vμω²A²

p = $\frac{1}{2}$ √(T/μ)μω²A²

so we solve for ω

ω² = 2P / √(T/μ)μA²

we substitute

ω² = 2(2000) / √(6000/1.5)(1.5)(0.500)²

ω² = 4000 / 23.71708

ω² = 168.65

(2πf)² = ω²

so

(2πf)² = 168.65

4π²f² = 168.65

f² = 168.65 / 4π²

f² = 4.27195

f = √4.27195

f = 2.066 Hz

Therefore, the required frequency of waves is 2.066 Hz

3 0

2 years ago

A machine is currently set to a feed rate of 5.921 inches per minute (IPM). Te machinist changes this setting to 6.088 IPM. By h

lukranit [14]

Answer:

By 16.7% or 0.167 IPM

Explanation:

Substracting the final IPM (6.088) to the initial IPM (5.921) gives us the net difference, which is how much did it increase in IPM. Multiplying this number by 100 gives us the percentual increase in the feed rate.

4 0

2 years ago

Read 2 more answers

A turntable rotates counterclockwise at 76 rpm . A speck of dust on the turntable is at 0.47 rad at t=0. What is the angle of th

icang [17]

To solve this exercise it is necessary to apply the kinematic equations of angular motion.

By definition we know that the displacement when there is constant angular velocity is

$\theta= \theta_0 +\omega t$

From our given data we know that,

$\omega = 76\frac{rev}{min}$

$\omega = 76\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1 min}{60s})$

$\omega = 7.958rad/s$

Moreover we know that

$\theta_0 = 0.47 rad$

Therefore for time t=8.1s we have,

$\theta= \theta_0+ \omega t$

$\theta= 0.47+(7.958)(8.1)$

$\theta = 64.9298rad$

That number in revolution is:

$\theta = 64.9298rad(\frac{1rev}{2\pi})$

$\theta = 15.108 Revolutions$

Here, we see that there are 15 complete revolutions

And 0.108 revolutions i not complete, so the tunable rotation is

$\theta_{net} = 0.108*2\pi=0.216\pi$

Therefore the angle of the speck at a time 8.1s is $0.216\pi$

4 0

2 years ago