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sineoko [7]
3 years ago
11

Immature forms of many animals, like Guinea worms, before they morph into adults

Physics
2 answers:
AveGali [126]3 years ago
8 0

The immature forms of animals which they are before metamorphosis are the larva.

<u>Explanation:</u>

The larva in plural form is known as the larvae. The eggs are hatch into the form of larva. These larvae grow in different stages which are marked according to the stages as I or II and so on.

After the larval development, when they are ready to be converted into adults, metamorphosis occurs. This process changes the larva into adults. Here the immature form gets converted into mature adult form.

liberstina [14]3 years ago
6 0

Answer:

True

Explanation: If this is a true or false question it is *T*

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Vitek1552 [10]
Choice 'b' is one possible way to state
Newton's second law of motion.

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8 0
3 years ago
QUESTION 10
Elena L [17]

The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

\theta =37.01^{\circ}

We can apply the first Newton's law in x and y-direction.

If we do a free body diagram of the system we will have:

x-direction

All the forces acting in this direction are:

T_{1}sin(\theta)-T_{2}sin(\theta)=0    (1)

Where:

  • T(1) is the tension due to the rope 1
  • T(2) is the tension due to the rope 2

Here we just conclude that T(1) = T(2)

y-direction

The forces in this direction are:

T_{1}cos(\theta)+T_{2}cos(\theta)-W=0   (2)

Here W is the weight of the steel beam.

We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.

Knowing that T(1) = T(2) and W = mg, we have:

T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)=m_{steel}g

T(1) must be equal to 5479 N, so we have:

cos(\theta)=\frac{m_{steel}g}{2T_{1}}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=0.80

Therefore, the maximum angle allowed is θ = 37.01°.

You can learn more about tension here:

brainly.com/question/12797227

I hope it helps you!

8 0
3 years ago
If you were standing at the center of curvature in front of a concave mirror, what image would be projected?
matrenka [14]
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6 0
3 years ago
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A spring with a spring constant of 50 N/m is stretched 15cm. What is the force and energy associated with this stretching?
Olenka [21]
Data:
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m

Formula:
F = k*x

Solving: 
F = k*x
F = 50*0.15
\boxed{\boxed{F = 7.5\:N}}\end{array}}\qquad\quad\checkmark

Data:
E (energy) = ? (joule)
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Formula:
E = \frac{k*x^2}{2}

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E = \frac{k*x^2}{2}
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E =  \frac{50*0.0225}{2}
E =  \frac{1.125}{2}
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What is the speed of a helicopter that traveled 1200 miles in 7 hours
zlopas [31]
1200
-------=171 miles per hour
7

8 0
3 years ago
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