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3 years ago

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In which stage in the life cycle of a cnidarian is the organism anchored to the bitten of the ocean or lake where it lives

1 answer:

Semmy [17]3 years ago

7 0

<span>In which stage in the life cycle of a cnidarian is the organism anchored to the bitten of the ocean or lake where it lives</span>

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List 3 to 5 ways to reduce friction

pychu [463]

<span>Make the surfaces smoother. Rough surfaces produce more friction and smooth surfaces reduce friction
Lubrication is another way to make a surface smoother
Make the object more streamlined
Reduce the forces acting on the surfaces
<span>Reduce the contact between the surfaces.</span></span>

8 0

3 years ago

Read 2 more answers

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

$H_{max}= \dfrac{v^2 sin^2(\theta)}{g}$

now,

$H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}$

$H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}$

now, equating both the equations

$\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}$

$\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}$

v₂² = 31061.79

v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

8 0

3 years ago

Martin has severe myopia, with a far point on only 17 cm. He wants to get glasses that he'll wear while using his computer whose

marusya05 [52]

Answer:

Explanation:

Far point = 17 cm . That means he can not see beyond this distance .

He wants to see at an object at 65 cm away . That means object placed at 65 has image at 17 cm by concave lens . Using lens formula

1 / v - 1 / u = 1 / f

1 / - 17 - 1 / - 65 = 1 / f

= 1 / 65 - 1 / 17

= - .0434 = 1 / f

power = - 100 / f

= - 100 x .0434

= - 4.34 D .

6 0

4 years ago

Two blocks, A and B, are connected by a rope. A second rope is connected to block B and a steady, horizontal tension force of T

Oksanka [162]

Answer:

40 N

Explanation:

We are given that

Speed of system is constant

Therefore, acceleration=a=0

Tension applied on block B=T=50 N

Friction force=f=10 N

We have to find the friction force acting on block A.

Let T' be the tension in string connecting block A and block B and friction force on block A be f'.

For Block B

$T-f-T'=m_Ba$

Where $m_B$ =Mass of block B

Substitute the values

$50-10-T'=m_B\times 0=0$

$T'==40 N$

For block A

$T'-f'=m_Aa$

Where $m_A=$ Mass of block A

Substitute the values

$40-f'=m_A\times 0=0$

$f'=40 N$

Hence, the friction force acting on block A=40 N

3 0

3 years ago

I WILL GIVE BRAILYEST!!! What is the mass of an object moving at a velocity of 5 m/s if the momentum of the object is 50 kg•m/s?

lianna [129]

Answer:

a. 250kg I think it's the right answer. hope it helps:)

7 0

3 years ago

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