Ask question

Ask question

All categories

4 years ago

9

Ferns that eject spores generally do so in pairs, with two spores flying off in opposite directions. The structure from which th

e spores are launched is quite lightweight. If it takes a certain amount of energy to eject each spore, explain how launching the spores in pairs provides for the greatest initial launch speed for each spore.

1 answer:

Furkat [3]4 years ago

5 0

Answer:

It cancels recoil.

Explanation:

For each action there is an equal an opposite reaction.

The principle of conservation of momentum tell us that if a single spore were ejected the fern would suffer a recoil from it. This recoil would take energy and speed from the spore. But if they are ejected in pairs the recoil is canceled and all the energy is transferred to the spores resulting in higher speeds.

You might be interested in

Two sacks contain the same number of identical apples and are separated by a distance r. The two

andrezito [222]

Answer:

The appropriate response will be " $F_2=\frac{3}{4}F$ ". A further explanation is given below.

Explanation:

According to the question,

⇒ $F_1=\frac{G(m_1 m_2)}{r^2}$ ....(equation 1)

and,

⇒ $F_2=\frac{G(m_1 m_2)}{r^2}$ ....(equation 2)

Now,

On dividing both the equations, we get

⇒ $\frac{F_1}{F_2}=\frac{\frac{G(2)(2)}{(1)^2}}{\frac{G(1)(3)}{(1)^2}}$

⇒ $\frac{F_1}{F_2}= \frac{4}{3}$

⇒ $F_2=\frac{3}{4}F$

4 0

3 years ago

What is the center of our solar system?

Yakvenalex [24]

The sun is the centre of our solar system

8 0

4 years ago

PLEASE HELP ASAP! i will really appreciate it

Setler79 [48]

What you’re doing is so easy pay attention more I think

4 0

3 years ago

A gyroscope flywheel of radius 3.21 cm is accelerated from rest at 13.2 rad/s2 until its angular speed is 2450 rev/min.

SIZIF [17.4K]

Answer:

Part a)

$a_t = 0.423 m/s^2$

Part b)

$a_c = 2113 m/s^2$

Part c)

$d = 80 m$

Explanation:

Part a)

as we know that angular acceleration of the wheel is given as

$\alpha = 13.2 rad/s^2$

now the radius of the wheel is given as

R = 3.21 cm

so the tangential acceleration is given as

$a_t = R\alpha$

$a_t = (0.0321)(13.2)$

$a_t = 0.423 m/s^2$

Part b)

frequency of the wheel at maximum speed is given as

$f = 2450 rev/min$

$f = \frac{2450}{60} = 40.8 rev/s$

now we know that

$\omega = 2\pi f = 2\pi(40.8) = 256.56 rad/s$

now radial acceleration is given as

$a_c = \omega^2 r$

$a_c = (256.56)^2(0.0321) = 2113 m/s^2$

Part c)

total angular displacement of the point on rim is given as

$\Delta \theta = \omega_0 t + \frac{1}{2}\alpha t^2$

here we know that

$\omega = \omega_0 + \alpha t$

$256.56 = 0 + 13.2 t$

$t = 19.4 s$

now angular displacement will be

$\Delta \theta = 0 + \frac{1}{2}(13.2)(19.4)^2$

$\Delta \theta = 2493.3 rad$

now the distance moved by the point on the rim is given as

$d = R\theta$

$d = (0.0321)(2493.3)$

$d = 80 m$

7 0

3 years ago

A mercury thermometer has a glass bulb of interior volume 0.100 cm3 at 10°c. the glass capillary 10) tube above the bulb has an

Nadya [2.5K]

Initial volume of mercury is
V = 0.1 cm³

The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.

Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³

The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm

Answer: 4.5 cm

7 0

4 years ago

Read 2 more answers