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enot [183]
2 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.
Physics
1 answer:
Furkat [3]2 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.

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A car drives 16 miles south and then 12 miles west. What is the magnitude of the car’s displacement? 4 miles 16 miles 20 miles 2
Ostrovityanka [42]
For this case, what we can do is use the Pythagorean theorem to find the magnitude of the displacement of the car.
 We have then
 d ^ 2 = 16 ^ 2 + 12 ^ 2

 From here, we clear the value of d.
 We have then:
 d =  \sqrt{16 ^ 2 + 12 ^ 2} 

 Rewriting:
 d = \sqrt{256 + 144}
 d = \sqrt{400}
 d = 20 miles

 Answer:
 
The magnitude of the car's displacement is:
 
d = 20 miles
7 0
2 years ago
Read 2 more answers
A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.
Sedaia [141]

Answer:

335°C

Explanation:

Heat gained or lost is:

q = m C ΔT

where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Heat gained by the water = heat lost by the copper

mw Cw ΔTw = mc Cc ΔTc

The water and copper reach the same final temperature, so:

mw Cw (T - Tw) = mc Cc (Tc - T)

Given:

mw = 390 g

Cw = 4.186 J/g/°C

Tw = 22.6°C

mc = 248 g

Cc = 0.386 J/g/°C

T = 39.9°C

Find: Tc

(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)

Tc = 335

7 0
2 years ago
Suppose that a sound has initial intensity β1 measured in decibels. This sound now increases in intensity by a factor f. What is
topjm [15]

Answer:

β2= β1+10*f

Explanation:

comparing β2 and β1, it is said that β2 is increased by a factor of f.

for each factor of f, there is a 10*f dB increase.

therefore if the β1 is increases by an intensity of factor f

the new intensity would be β1+ 10*f

4 0
2 years ago
A 5.00 kg crate is on a 21.0 degree hill. Using X-Y axes tilted down the plane, what is the y-component of the normal force?
Rama09 [41]

Answer:

The y-component of the normal force is 45.74 N.

Explanation:

Given that,

Mass of the crate, m = 5 kg

Angle with hill, \theta=21^{\circ}

We need to find the y component of the normal force. We know that the y component of the normal force is given by :

F_y=F\ \cos\theta\\\\F_y=mg\ \cos\theta\\\\F_y=5\times 9.8\ \cos(21)\\\\F_y=45.74\ N

So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.

7 0
1 year ago
A girl and a boy are riding on a merry-go-round that is turning at a constant rate. The girl is near the outer edge, and the boy
skad [1K]

(a) Both the girl and the boy have the same nonzero angular displacement.

Explanation:

The angular displacement of an object moving in uniform circular motion, as the boy and the girl on the merry-go-round, is given by

\theta= \omega t

where

\omega is the angular speed

t is the time interval

For a uniform object in uniform circular motion, all the points of the object have same angular speed. This means that the value of \omega is the same for the boy and the girl.

Therefore, if we consider the same time interval t, the boy and the girl will also have same nonzero angular displacement.

(b) The girl has greater linear speed.

Explanation:

The linear (tangential) speed of a point along the merry-go-round is given by

v=\omega r

where

\omega is the angular speed

r is the distance of the point from the centre of the merry-go-round

In this problem, the girl is near the outer edge, while the boy is closer to the centre: since the value of \omega is the same for both, this means that the value of r is larger for the girl, so the girl will also have a greater linear speed.

3 0
2 years ago
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