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3 years ago

HELP ASAP!!!!!!!!!

1 answer:

6 0

B) The total energy always decreases in an energy conversion,

This is because: Energy decreases as it moves up trophic levels because energy is lost as metabolic heat when the organisms from one trophic level are consumed by organisms from the next level. Trophic level transfer efficiency (TLTE) measures the amount of energy that is transferred between trophic levels

Hope this helps! ✨

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A runner runs 2.0 km in 5.0 minutes

Nadya [2.5K]

Answer: What is this supposed to be converted into?

Explanation:

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3 years ago

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1. A block is pulled to the right at constant velocity by a 20N force acting at 30o above the horizontal. If the coefficient of

DiKsa [7]

Answer:

44.6 N

Explanation:

Draw a free body diagram of the block. There are four forces on the block:

Weight force mg pulling down,

Normal force N pushing up,

Friction force Nμ pushing left,

and applied force F pulling right 30° above horizontal.

Sum of forces in the y direction:

∑F = ma

N + F sin 30° − mg = 0

N = mg − F sin 30°

Sum of forces in the x direction:

∑F = ma

F cos 30° − Nμ = 0

F cos 30° = Nμ

N = F cos 30° / μ

Substitute:

mg − F sin 30° = F cos 30° / μ

mg = F sin 30° + (F cos 30° / μ)

Plug in values:

mg = 20 N sin 30° + (20 N cos 30° / 0.5)

mg = 44.6 N

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3 years ago

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A projectile is fired vertically from Earth's surface with an initial speed v0. Neglecting air drag, how far above the surface o

lidiya [134]

This may helpv^2=u^2+2as. v=0 at top of flight. a=acceleration of gravity(vo^2)/2a=s.

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4 years ago

Calculate the pressure exerted by a 4000N camel on the sand. The camel’s feet have a

Harrizon [31]

Answer:

The pressure exerted by camel feet is <u>2000 N/m²</u>.

Step-by-step explanation:

<h3><u>Solution</u> :</h3>

Here, we have given that ;

Force applied on camel feet = 4000 N
Total area of camel feet = 2 m²

We need to find the pressure exerted by camel feet.

As we know that :

${\longrightarrow{\pmb{\sf{Pressure= \dfrac{Area}{Force}}}}}$

Substituting all the given values in the formula to find the pressure exerted by camel feet.

$\begin{gathered} \begin{array}{l} {\longrightarrow{\sf{Pressure= \dfrac{Area}{Force}}}} \\ \\ {\longrightarrow{\sf{Pressure= \dfrac{4000}{2}}}} \\ \\ {\longrightarrow{\sf{Pressure= \cancel{\dfrac{4000}{2}}}}} \\ \\ {\longrightarrow{\sf{Pressure= 2000 \: N/{m}^{2}}}} \\ \\\star \: \small\underline{\boxed{\sf{\purple{Pressure= 2000 \: N/{m}^{2}}}}} \end{array}\end{gathered}$

Hence, the pressure exerted by camel feet is 2000 N/m².

$\rule{300}{2.5}$

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2 years ago

Two ladybugs sit on a rotating disk that is slowing down at a constant rate. The ladybugs are at rest with respect to the surfac

d1i1m1o1n [39]

The two ladybugs have same rotational (angular) speed

Explanation:

The rotational (angular) speed of an object in circular motion is defined as:

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval considered

Here we have two ladybugs, which are located at two different distances from the axis. In particular, ladybug 1 is halfway between ladybug 2 and the axis of rotation. However, since they rotate together with the disk, and the disk is a rigid body, every point of the disk cover the same angle $\theta$ in the same time $t$ : this means that every point along the disk has the same angular speed, and therefore the two ladybugs also have the same angular speed.

On the other hand, the linear speed of the two ladybugs is different, because it follows the equation:

$v=\omega r$

where r is the distance from the axis: and since the two ladybugs are located at different $r$ , they have different linear speed.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

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3 years ago