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mafiozo
3 months 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:
Yuliya22 [3.3K]3 months ago
0 0

Answer:

To calculate the pesticide required, you perform 55mg/kg multiplied by 0.02kg, which results in 1.1 mg of pesticides

Explanation:

(Options are missing)

Given that the typical weight of a cockroach is 0.02kg

And the requirement is 55mg for every 1 kg of roaches

The pesticide quantity needed for a 0.02kg roach is calculated as 55mg/kg times 0.02kg

Thus, the amount = 55mg/kg multiplied by 0.02 kg

Amount = 55mg/kg times 0.02 kg

Amount = 1.1mg

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A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4
Yuliya22 [3333]

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

R=\sqrt{P^2+Q^2+2PQCos\theta}

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

\theta= 90^o

R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70

(Cos90^o=0),(sin 90^o=1)

\alpha =tan^{-1}\frac{Qsin\theta}{P+Qcos\theta}=tan^{-1}\frac{4.5 sin 90^o}{3.5+4.5 cos90^o}=tan^{-1}\frac{4.5}{3.5}=52.12^o

Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

8 0
3 months ago
Ability of the muscles to function effectively and efficiently without undue fatigue
Keith_Richards [3271]

Response:

Physical well-being

Clarification:

7 0
3 months ago
A person who weighs 625 N is riding a 98-N mountain bike. Suppose that the entire weight of the rider and bike is supported equa
Keith_Richards [3271]

Answer:

A = 4.76 x 10⁻⁴ m²

Explanation:

Given data:

Person's weight = 625 N

Bike's weight = 98 N

Pressure per tire = 7.60 x 10⁵ Pa

Find: Contact area per tire

Total system weight = 625 + 98 = 723 N

Let F represent the force supported by each tire

2F = 723 N

Therefore, F = 361.5 N

Using the formula F = P × A

A = \dfrac{F}{P}

A = \dfrac{361.5}{7.60 \times 10^5}

Contact area, A = 4.76 x 10⁻⁴ m²

7 0
3 months ago
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