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3 years ago

Which of the following is an example of a Cnidarian?

Physics

2 answers:

siniylev [52]3 years ago

5 0

The answer to this is C jellyfish i am positive

hope this helps

Agata [3.3K]3 years ago

3 0

Jellyfish are an example of cnidarian

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PLEASE HELP ME WITH THIS ONE QUESTION

Ipatiy [6.2K]

k = $\dfrac{ (\dfrac{h}{ \lambda} )^{2} }{2m}$

k = (6.626×10-¹⁹/590 × 10-⁹ )^{2} /2 × 1.673 × 10-²⁷

k = (1.12 × 10-³⁰)^2/3.346×10-²⁷

k = 1.25 × 10-⁶⁰ /3.346×10-²⁷

k = 0

ldk why, my answer is coming this :(

4 0

2 years ago

A 45.0 kg skier testing a new competitive ski wax drops off a ledge, goes into a crouch and goes straight down a slope of 10.0◦

Neko [114]

Answer:

v = 21.03 m/s

Explanation:

given,

mass of skier = 45 kg

the slope of the snow = 10.0◦

coefficient of friction = 0.114

distance traveled = 300 m

speed = ?

Acceleration = g sin θ - µ g Cos θ

= 9.8 × Sin (10°) - 0.10 × 9.8 × Cos(10°)

= 0.737 m/s²

using equation of motion

v² = u² + 2 a s

v² = 0 + 2 × 0.737 × 300

v = 21.03 m/s

Speed of skier's after travelling 300 m speed is equal to 21.03 m/s

6 0

3 years ago

Many organ system serve to help maintain a healthy body. The excretory system is an organ system which is composed of all sweat

Cerrena [4.2K]

Answer: D

Explanation:

Cause removing excess waste from the body helps to maintain a steady health

3 0

3 years ago

Consider two air-filled parallel-plate capacitors with circular plates. Capacitor 1 has a distance between plates d and plate ra

sergey [27]

Answer:

1/8 x C

Explanation:

The capacitance of parallel plate capacitor

= ε₀ A /d where A is area of plate and d is distance between plate.

for capacitor 1

C = ε₀ A /d

For capacitor 2

radius = R/2

Area = A / 4

Capacitance

= ε₀ (A/4) x ( 1 / 2d )

= ( 1 / 8) x (ε₀ A /d)

= 1/8 x C

4 0

3 years ago

A rectangular loop of wire with width w = 5 cm, length L = 10cm, mass m = 40 g, and resistance R = 20 mΩ has an initial velocity

zhuklara [117]

Answer:

The loop penetrate 4 cm into the magnetic field.

Explanation:

Given that,

Width w= 5 cm

Length L= 10 cm

mass m = 40 g

Resistance R = 20 mΩ

Initial velocity = 1 m/s

Magnetic field = 2 T

We need to calculate the induced emf

Using formula of emf

$\epsilon=v_{0}Bw$

Put the value into the formula

$\epsilon =1\times2\times5\times10^{-2}$

$\epsilon =10\times10^{-2}\ volt$

We need to calculate the current

Using Lenz's formula

$i=\dfrac{\epsilon}{R}$

$i=\dfrac{10\times10^{-2}}{20\times10^{-3}}$

$i=5\ A$

We need to calculate the force

Using formula of force

$F=i(\vec{w}\times\vec{B})$

$F=iwB$

Put the value into the formula

$F=5\times5\times10^{-2}\times2$

$F=0.5\ N$

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{F}{m}$

Put the value in to the formula

$a=\dfrac{0.5}{40\times10^{-3}}$

$a=12.5\ m/s^2$

We need to calculate the distance

Using equation of motion

$v^2=u^2+2as$

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{0-1^2}{2\times(-12.5)}$

$s=0.04\ m$

$s=4\ cm$

Hence, The loop penetrate 4 cm into the magnetic field.

5 0

3 years ago