Ask question

Ask question

All categories

4 years ago

12

A scientist observed the formation of hundreds of maggots and many flies on a rotting meat slice exposed to the environment. Wha

t most likely formed the maggots? Water vapor present in the surrounding air Eggs of flies in the surroundings Nutrient crystals in the meat Oxygen in the environment

1 answer:

Phantasy [73]4 years ago

8 0

Female flies are the one that are saddle with the responsibility of laying eggs during reproduction. The female usually look for a warm and moist place that have abundance food source for laying the eggs. The eggs laid by by the flies transformed into larva, which is also called maggots and the maggots turn into flies.

You might be interested in

A major component of gasoline is octane (C8H18). When octane is burned in air, it chemically reacts with oxygen gas (O2) to prod

stira [4]

Answer:

0.342g of octane

Explanation:

First let us generate a balanced equation for the reaction this is illustrated below:

2C8H18 + 25O2 —> 16CO2 + 18H20

Next, we'll calculate the molar mass and masses of C8H18 and O2

Molar Mass of C8H18 = (12x8) + (18x1 ) = 96 + 18 = 114g/mol

Mass of C8H18 from the balanced equation = 2 x 114 = 228g

Molar Mass of O2 = 16 x 2 = 32g/mol

Mass of O2 from the balanced equation = 25 x 32 = 800g

From the equation,

228g of octane(C8H18) were consumed by 800g of O2.

Therefore, Xg of octane will be consumed by 1.2g of O2 i.e

Xg of octane = (228 x 1.2)/800 = 0.342g

Therefore 0.342g of octane will be consumed

4 0

3 years ago

Which of the following is a true statement?

forsale [732]

C. Electromagnetic waves do not always need a medium to travel.

Since they are transverse waves they do not need a material medium.

4 0

3 years ago

Read 2 more answers

A rider on a bike with the combined mass of 100 kg attains a terminal speed of 15 m/s on a 12% slope. Assuming that the only for

vlabodo [156]

Answer:

0·95

Explanation:

Given the combined mass of the rider and the bike = 100 kg

Percent slope = 12%

∴ Slope = 0·12

Terminal speed = 15 m/s

Frontal area = 0·9 m²

Let the slope angle be β

tanβ = 0·12

As it attains the terminal speed, the forces acting on the combined rider and the bike must be balanced and therefore the rider must be moving download as the directions of one of the component of weight and drag force will be in opposite directions

The other component of weight will get balance by the normal reaction and you can see the figure which is in the file attached

From the diagram m × g × sinβ = drag force

Drag force = 0·5 × d × $C_{D}$ × v² × A

where d is the density of the fluid through which it flows

$C_{D}$ is the drag coefficient

v is the speed of the object relative to the fluid

A is the cross sectional area

As tanβ = 0·12

∴ sinβ = 0·119

Let the fluid in this case be air and density of air d = 1·21 kg/m³

m × g × sinβ = 0·5 × d × $C_{D}$ × v² × A

100 × 9·8 ×0·119 = 0·5 × 1·21 × $C_{D}$ × 15² × 0·9

∴ $C_{D}$ ≈ 0·95

∴ Drag coefficient is approximately 0·95

4 0

3 years ago

Read 2 more answers

A makeshift sign hangs by a wire that is extended over an ideal pulley and is wrapped around a large potted plant on the

ella [17]

Answer:I know the answer for B cus I’m doing the same problem. For B, you would only take the coefficient of friction given and then multiply it by the Normal Force, which in this case is the same as the Gravitational Force.

Explanation:

4 0

3 years ago

A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.58 m/s. The pulling force is 110 N parallel to th

kari74 [83]

Given that,

Mass = 9.2 kg

Force = 110 N

Angle = 20.2°

Distance = 5.10 m

Speed = 1.58 m/s

(A). We need to calculate the work done by the gravitational force

Using formula of work done

$W_{g}=mgd\sin\theta$

Where, w = work

m = mass

g = acceleration due to gravity

d = distance

Put the value into the formula

$W_{g}=9.2\times(-9.8)\times5.10\sin20.2$

$W_{g}=-158.8\ J$

(B). We need to calculate the increase in internal energy of the crate-incline system owing to friction

Using formula of potential energy

$\Delta U=-W$

Put the value into the formula

$\Delta U=-(-158.8)\ J$

$\Delta U=158.8\ J$

(C). We need to calculate the work done by 100 N force on the crate

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=100\times5.10$

$W=510\ J$

We need to calculate the work done by frictional force

Using formula of work done

$W=-f\times d$

$W=-\mu mg\cos\theta\times d$

Put the value into the formula

$W=-0.4\times9.2\times9.8\cos20.2\times5.10$

$W=-172.5\ J$

We need to calculate the change in kinetic energy of the crate

Using formula for change in kinetic energy

$\Delta k=W_{g}+W_{f}+W_{F}$

Put the value into the formula

$\Delta k=-158.8-172.5+510$

$\Delta k=178.7\ J$

(E). We need to calculate the speed of the crate after being pulled 5.00m

Using formula of change in kinetic energy

$\Delta k=\dfrac{1}{2}m(v_{2}^2-v_{1}^{2})$

$v_{2}^2=\dfrac{2\times\Delta k}{m}+v_{1}^2$

Put the value into the formula

$v_{2}^2=\dfrac{2\times178.7}{9.2}+1.58$

$v_{2}=\sqrt{\dfrac{2\times178.7}{9.2}+1.58}$

$v_{2}=6.35\ m/s$

Hence, (A). The work done by the gravitational force is -158.8 J.

(B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.

(C). The work done by 100 N force on the crate is 510 J.

(D). The change in kinetic energy of the crate is 178.7 J.

(E). The speed of the crate after being pulled 5.00m is 6.35 m/s

8 0

3 years ago