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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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find the time taken, if the speed of a train increased from 72 km/hr to 90 km/hr for 234 km. leave your answer in seconds
Airida [17]

Answer:

Time taken = 10400 s

Explanation:

Given:

Initial speed of the train, u=72\textrm{ km/h}=72\times \frac{5}{18}=20\textrm{ m/s}

Final speed of the train, v=90\textrm{ km/h}=90\times \frac{5}{18}=25\textrm{ m/s}

Displacement of the train, S=234\textrm{ km}=234\times 1000=234000\textrm{ m}

Using Newton's equation of motion,

v - u = at\\a=\frac{v-u}{t}

Now, using Newton's equation of motion for displacement,

v^{2}-u^{2}=2aS

Now, plug in the value of a=\frac{v-u}{t} in the above equation. This gives,

v^{2}-u^{2}=2\times \frac{v-u}{t}\times S\\(v+u)(v-u)=\frac{2(v-u)S}{t}\\t=\frac{2(v-u)S}{(v+u)(v-u)}\\t=\frac{2S}{v+u}

Now, plug in 234000 m for S, 25 m/s for v and 20 m/s for u. Solve for t.

t=\frac{2S}{v+u}\\t=\frac{2\times 234000}{25+20}\\t=\frac{468000}{45}=10400\textrm{ s}

Therefore, the time taken by the train is 10400 s.

3 0
4 years ago
The density of air is 1.3 kg/m^3. What mass of air is contained in a room measuring 2.5m x 4m x 10m? Give your answer in kg. Bra
krek1111 [17]
Mass= density x volume
1.3 kg/m^3 x ( 2.5x4x10) m^3
= 130 kg
7 0
4 years ago
How will the electrostatic force between two electric charges change if the first charge is doubled and the second charge is onl
Vinvika [58]

Answer:

B) \frac{2}{3}

Explanation:

The electric force between charges can be determined by;

F = \frac{kq_{1} q_{2} }{r^{2} }

Where: F is the force, k is the Coulomb's constant, q_{1} is the value of the first charge, q_{2} is the value of the second charge, r is the distance between the centers of the charges.

Let the original charge be represented by q, so that;

q_{1} = 2q

q_{2} = \frac{q}{3}

So that,

F = q_{1}q_{2} x \frac{k}{r^{2} }

  = 2q x \frac{q}{3} x \frac{k}{r^{2} }

  = \frac{2q^{2} }{3} x \frac{k}{r^{2} }

  = \frac{2}{3} x \frac{kq}{r^{2} }

F = \frac{2}{3} x \frac{kq}{r^{2} }

The electric force between the given charges would change by \frac{2}{3}.

4 0
3 years ago
Carmen is helping load furniture and boxes onto a moving truck. She picks up boxes of her things, places them on a cart, and pus
AleksandrR [38]

Answer:

B because of the friction from the wheels

6 0
3 years ago
Read 2 more answers
Pleaseeeee HELPPPP THIS IS TIMED ALSO,
TEA [102]

Answer:

Friction, normal force, and weight

Explanation:

If the book slows down, it means that there must be friction acting in the opposite direction of the direction the book is moving in.

Weight is caused by the gravitational pull of the Earth on the book, and normal force is the table pushing the book up because the book is pushing down on the table (3rd law.)

Note that weight and normal force is not the 3rd law action-reaction pair. The pair is the force of the book on the table and the force of the table on the book.

8 0
2 years ago
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