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3 years ago

How much energy is passed from one trophic level to the next

1 answer:

6 0

The amount of energy at each trophic level decreases as it moves through an ecosystem. As little as 10 percent of the energy at any trophic level is transferred to the next level; the rest is lost largely through metabolic processes as heat.

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15. A body moving with a velocity of 20 m/s begins to accelerate at 3 m/s2. How far does the body move in 5 seconds? A. 137.5 m

Rudik [331]

Answer is B. According to the equation of motion s = vt + 1/2 at2 Where s is distance covered, v is velocity, a is acceleration and t is time taken. So, by putting all the values, we get s = (20)(5) + 1/2 (3)(5)2 s = 100 + 1/2 (3)(25) s = 100 + 1/2 75 s = 100 + 37.5 s = 137.5 meters

7 0

3 years ago

Read 2 more answers

A brick of gold is 0.1 m wide, 0.1 m high, and 0.2 m long. The density of gold is 19,300 kg/m3. What pressure does the brick exe

PtichkaEL [24]

Answer:

The pressure exerted by the brick on the table is 18,933.3 N/m².

Explanation:

Given;

height of the brick, h = 0.1 m

density of the brick, ρ = 19,300 kg/m³

acceleration due to gravity, g = 9.81 m/s²

The pressure exerted by the brick on the table is calculated as;

P = ρgh

P = (19,300)(9.81)(0.1)

P = 18,933.3 N/m²

Therefore, the pressure exerted by the brick on the table is 18,933.3 N/m².

4 0

3 years ago

What does it mean for a reaction to be endothermic? choose 2 of the following

Rainbow [258]

Answer: I would choose options C and D

Explanation:

3 0

2 years ago

Read 2 more answers

A graduated cylinder.measures 15.3 mL. Convert this measurement to DaL

ololo11 [35]

Answer:

0.000153DaL

Explanation:

We have been given:

15.3mL to convert to DaL

DaL is a unit of volume which indicates a decaliter.

This implies that;

1 Da L = 1 x 10²L

So:

1 mL = 1 x 10⁻³L

So 15.3mL will give 15.3 x 10⁻³L

So;

1 x 10²L = 1 DaL

15.3 x 10⁻³L will give $\frac{15.3 x 10^{-3} }{1 x 10^{2} }$ = 15.3 x 10⁻⁵DaL

Therefore, this is 0.000153DaL

5 0

3 years ago

Two crates, one with mass 5.4 kg and the other with mass 8.2 kg, connected by a light rope. The coefficient of kinetic friction

Gnom [1K]

Answer:

R= 2.5 :ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks

Explanation:

We apply Newton's second law:

∑F=m*a

velocity is constant ,then , a=0

Nomenclature

W: weight

m: mass

N : normal force

Ff: Friction force

μk: coefficient of kinetic friction

T: tension force in the rope

F: applied horizontal force

g: acceleration due to gravity.

Force Calculation

W₁=m₁*g=5.4 kg *9.8m/s²=52.92 N

W₂=m₂*g=8.2 kg *9.8m/s²= 80.36N

∑Fy=0

N₁-W₁=0 , N₁=W₁ = 52.92 N

N₂-W₂=0, N₂=W₂=80.36N

Ff₁= μk* N₁=0.4*52.92 N = 21.16N

Ff₂= μk* N₂=0.4*80.36N = 32.14N

Look at the attached graphic

Free-Body diagram m₁=5.4 kg

∑Fx=0

T- Ff₁=0 , T= Ff₁ , T= 21.16N

Free-Body diagram m₂=8.2 kg

∑Fx=0

F-T- Ff₂=0 , F=T+Ff₂= 21.16N+32.14N=53.3N

Ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks (R)

R= F/T= 53.3N/21.16N = 2.5

3 0

3 years ago