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NARA [144]
3 years ago
5

The hard external covering of some invertebrates is called a what

Physics
2 answers:
IRISSAK [1]3 years ago
6 0
That would be the exoskeleton!
Alexandra [31]3 years ago
4 0
The answer would be exoskeleton.
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How are pupae, larvae, and nymphs similar? *Answer Fast!*
Goshia [24]

Answer:

They are all a cycle!

Explanation:

They just are all cycles.

5 0
3 years ago
If 1.34 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved throu
ioda

Given :

Number of operations move through a pocket calculator during a full day's operation , n=1.34 \times 10^{20} .

To Find :

How many coulombs of charge moved through it .

Solution :

We know , charge in  one electron is :

e^-=-1.6\times 10^{-19}\ coulombs

So , charge on n electron is :

C=e^-\times n\\C=-1.6\times 10^{-19}\times 1.34\times 10^{20} \ C\\C=-21.44\ C

Therefore , -21.44 coulombs of charge is moved through it .

Hence , this is the required solution .

3 0
3 years ago
What are the strengths and limitations of the doppler and transit methods? What kind of planets are easiest to detect with each
Arturiano [62]

\huge\mathfrak\red{✔Answer:-}

Strength: able to detect planets in a wide range of orbits, as long as orbits aren't face on

Limitations: yield only planet's mass and orbital properties

3 0
3 years ago
A 10 m long uniform beam weighing 100 N is supported by two ropes at the ends. If a 400 N person sits at 2.0 m from one end of t
Naddik [55]

Answer:

T1 = 130N, T2 = 370N

Explanation:

In order for the system to be at rest, the sum of all forces must be zero and the torque around a point on the beam must be zero.

1. forces:

Let tension in rope 1 be T1 and in rope 2 be T2:

ma = T1 + T2 - 100N - 400N = 0

(1) T1 + T2 = 500N

2. torque around the center point of the beam:

τ = r x F = 5*T1 + 3*400N - 5*T2 = 0

(2) T1 - T2 = -240N

Solving both equations:

T1 = 130N

T2 = 370N

3 0
4 years ago
6. A car, 1110 kg, is traveling down a horizontal road at 20.0 m/s when it locks up its brakes. The coefficient of friction betw
USPshnik [31]

Answer:

x=22.65m

Explanation:

We have an uniformly accelerated motion, with a negative acceleration. Thus, we use the kinematic equations to calculate the distance will it take to bring the car to a stop:

v_f^2=v_0^2-2ax\\\frac{0^2-v_0^2}{-2a}=x\\x=\frac{v_0^2}{2a}

The acceleration can be calculated using Newton's second law:

\sum F_x:F_f=ma\\\sum F_y:N=mg

Recall that the maximum force of friction is defined as F_f=\mu N. So, replacing this:

\mu N=ma\\\mu mg=ma\\a=\mu g\\a=0.901(9.8\frac{m}{s^2})\\a=8.83\frac{m}{s^2}

Now, we calculate the distance:

x=\frac{v_0^2}{2a}\\x=\frac{(20\frac{m}{s})^2}{2(8.83\frac{m}{s^2})}\\x=22.65m

7 0
3 years ago
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