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tangare [24]
3 years ago
11

Compare and contrast reproduction and fertilization.

Physics
1 answer:
Anastaziya [24]3 years ago
6 0

Answer:

ew so the difference is.. Pollination and Fertilization occur in plants during sexual reproduction.

the difference is Pollination occurs from anthers of stamens to stigma of the ovary and It is a physical process.

fertilizateion is It is the fusion of female and male gametes and It is a genetic and biochemical process.

Explanation:

this was so hard to write out

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Interestingly, there have been several studies using cadavers to determine the moment of inertia of human body parts by letting
loris [4]

Answer:

0.08735 kgm²

Explanation:

m = Mass of lower leg = 5 kg

L = Length of leg = 18 cm

g = Acceleration due to gravity = 9.81 m/s²

f = Frequency = 1.6 Hz

I = Moment of inertia

Time period is given by

T=2\pi\sqrt{\dfrac{I}{mgL}}

Also

T=\dfrac{1}{f}

So,

I=\dfrac{mgL}{(2\pi f)^2}\\\Rightarrow I=\dfrac{5\times 9.81\times 0.18}{(2\pi 1.6)^2}\\\Rightarrow I=0.08735\ kgm^2

The moment of inertia of the lower leg is 0.08735 kgm²

8 0
4 years ago
Is energy transformation from potential to kinetic 100%?
Arada [10]
Yes. take a bow for instance. while pulling back the string you have potential energy. when you let the string go and the arrow flies towards your target the string is filled with kinetic energy.
8 0
3 years ago
Woman pulls a 6.87 kg suitcase,
GenaCL600 [577]

Answer:

2.24 m/s

Explanation:

resolving force of 29.2 N in x component

Fx = 29.2 cos 57.7

Fx = 15.6N

as force of friction is 12.7 N hence net force which produces acceleration is

15.6-12.7=2.9 N

by Newton 's law a=f/m

a= 2.9/6.87=0.422 m/s^2

now equation of motion is

v^2= U^2+2as

 = 0^2+2(.422)(5.93)

v^2=5.00

v=2.24 m/s

4 0
3 years ago
2. Two charges at fixed locations produce an electric field as shown
vagabundo [1.1K]

Answer:

sum of the two forces as both point to the right is a force that points to the right,

Explanation:

The force on the cast load at point Y is given by

        F = q_y  E

force is a vector magnitude so its result is

        ∑ F = Fₐ + F_b

indicate that the charge at y is negative, we analyze the direction of the force created by each charge

Charge A

as the electric field is incoming the charge is negative and as the test charge is negative both repel each other, consequently the force points to the right

Charge B

in this case the electric field lines are salient, therefore the charge is positive, consequently the force on the charge at y is attractive and points to the right

the sum of the two forces as both point to the right is a force that points to the right, that is, in the direction of the charge located at B

3 0
3 years ago
A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal
Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration (g=9.8 m/s^2) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

v=18.9 m/s

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

v_y^2 - u_y^2 = 2ah (1)

where

u_y = 16.4 m/s is the initial vertical velocity

v_y is the vertical velocity at height h

a=g=-9.8 m/s^2 is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, v_y = 0, so we can use the equation to find the maximum height

h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m

So, at half of the maximum height,

h = \frac{13.7}{2}=6.9 m

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s

And so, the speed of the stone at half of the maximum height is

v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

v_1 = 18.9 m/s

while the speed at half of the maximum height (part b) is

v_2 = 22.2 m/s

So the difference is

\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s

And so, in percentage,

\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%

So, the stone in part (b) is moving 17.4% faster than in part (a).

4 0
4 years ago
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