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3 years ago

A fertilized human egg cannot grow into a crocodile duck or fish because

Physics

2 answers:

Shalnov [3]3 years ago

7 0

Answer:

A fertilized human egg cannot grow into a crocodile, duck or fish specifically because of: genetic code. is a stage in reproduction whereby an egg and a sperm fuse to create a single cell.

Rashid [163]3 years ago

5 0

Explanation:

Because of ; Genetic Code

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A block slides forward 4.73 m while being pulled back by a 39.7 N force at 142° from the direction of motion. How much work does

lubasha [3.4K]

Answer:

work done by the force, W = -147.97 J

Given:

displacement of the box, s = 4.73 m

angle between force and displacement, $\theta = 142^{\circ}$

Force, F = 39.7 N

Solution:

Work done by the force can be expressed as the dot product of Force, F and displacement 's' is given by:

W = Fscos $\theta$

W = 39.7 $\times 4.73\times cos142^{\circ}$

W = -147.97 J

where negative sign indicates that work is done on the block

6 0

4 years ago

A playground carousel is rotating counterclockwise about its center on frictionless bearings. A person standing still on the gro

Shtirlitz [24]

Answer:

1.94601 rad/s

Explanation:

$I_1$ = Moment of inertia of carousel = 124 kgm²

$\omega_1$ = Angular speed of carousel = 3.5 rad/s

$\omega_2$ = Angular speed of person

r = Radius of carousel = 1.53 m

m = Mass of person = 42.3 kg

Moment of inertia of person

$I_2=mr^2\\\Rightarrow I_2=42.3\times 1.53^2\\\Rightarrow I_2=99.02007\ kgm^2$

As the angular momentum is conserved in the system

$I_1\omega_1=(I_1+I_2)\omega_2\\\Rightarrow \omega=\frac{I_1\omega_1}{(I_1+I_2)}\\\Rightarrow \omega_2=\frac{124\times 3.5}{(124+99.02007)}\\\Rightarrow \omega_2=1.94601\ rad/s$

The angular speed of the carousel after the person climbs aboard is 1.94601 rad/s

5 0

3 years ago

A particle initially located at the origin has an acceleration of a=3jm/s^2 and an initial velocity of Vi=5im/s.

-BARSIC- [3]

Given the particle's acceleration is

$\vec a(t) = \left(3\dfrac{\rm m}{\mathrm s^2}\right)\vec\jmath$

with initial velocity

$\vec v(0) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath$

and starting at the origin, so that

$\vec r(0) = \vec 0$

you can compute the velocity and position functions by applying the fundamental theorem of calculus:

$\vec v(t) = \vec v(0) + \displaystyle \int_0^t \vec a(u)\,\mathrm du$

$\vec r(t) = \vec r(0) + \displaystyle \int_0^t \vec v(u)\,\mathrm du$

We have

• velocity at time <em>t</em> :

$\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \displaystyle \int_0^t \left(3\dfrac{\rm m}{\mathrm s^2}\right)\,\vec\jmath\,\mathrm du \\\\ \vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath \\\\ \boxed{\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath}$

• position at time <em>t</em> :

$\vec r(t) = \displaystyle \int_0^t \left(\left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)u\,\vec\jmath\right) \,\mathrm du \\\\ \boxed{\vec r(t) = \left(5\dfrac{\rm m}{\rm s}\right)t\,\vec\imath + \frac12 \left(3\frac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath}$