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2 years ago

.Golden Rice is an example of genetically modified organism. true or false

Physics

2 answers:

Rashid [163]2 years ago

8 0

Answer:

true it would be because rice is a different breed no but ye it's true I don't have an explanation

MissTica2 years ago

7 0

Answer:

True

Explanation:

Golden Rice is an example of genetically modified organism.

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Please help me what defines the above collision as being elastic?

Kryger [21]

A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision. ... Momentum is conserved in inelastic collisions, but one cannot track the kinetic energy through the collision since some of it is converted to other forms of energy.

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2 years ago

What is the best approximate value for the elastic potential energy (EPE) of the spring elongated by 3.0 meters?

DIA [1.3K]

The elastic potential energy of the spring is 6.8 J

Explanation:

The elastic potential energy of a compressed/stretched spring is given by the equation:

$E=\frac{1}{2}kx^2$

where

k is the spring constant

x is the elongation of the spring

The spring constant of the spring in this problem can be found by keeping in mind the relationship between force (F) and elongation (x) (Hooke's law):

$F=kx$

By looking at the graph and comparing it with the formula, we realize that the slope of the force-elongation graph corresponds to the spring constant. Therefore in this case,

$k=\frac{15.0-0}{10.0-0}=1.5 N/m$

Therefore when the spring has a elongation of $x=3.0 m$ , its potential energy is

$E=\frac{1}{2}(1.5)(3.0)^2=6.8 J$

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2 years ago

What is the total moved without regard to direction

soldi70 [24.7K]

The entire motion of an object, regardless of direction.

In physics, motion is that the phenomenon in which an object changes its position with respect to time. Motion is mathematically described in terms of displacement, distance, velocity, acceleration, speed, and frame of regard to an observer and measuring the change in position of the body relative to that frame with change in time. The branch of physics describing the motion of objects without regard to their cause is called kinematics, while the branch studying forces and their effect on motion are named dynamics.

If an object isn't changing relative to a given frame of reference, the thing is said to be at rest, motionless, immobile, stationary, or to possess a constant or time-invariant position with reference to its surroundings. Modern physics holds that, as there's no absolute frame of reference, Newton's concept of absolute motion can't be determined

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1 year ago

(7)Figure 4 shows three charges: Q₁, Q₂ and Q3 . Determine the net force (Fnet) acting on Q3. (Hint: Draw a free body diagram of

NISA [10]

Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is

$F = \dfrac{kq_1q_2}{r^2}$

where k ≈ 8,98 × 10⁹ kg•m³/(s²•C²) is Coulomb's constant.

8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.

8.2. First convert everything to base SI units:

0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C

0,03 µC = 3 × 10⁻⁸ C

0,04 µC = 4 × 10⁻⁸ C

300 mm = 300 × 10⁻³ m = 0,3 m

600 mm = 0,6 m

Force due to Q₁ :

$F_{Q_2/Q_1} = \dfrac{k (6 \times 10^{-16} \,\mathrm C)}{(0,3 \, \mathrm m)^2} \approx \boxed{6,0 \times 10^{-5} \,\mathrm N} = 0,06 \,\mathrm{mN}$

Force due to Q₃ :

$F_{Q_2/Q_3} = \dfrac{k (12 \times 10^{-16} \,\mathrm C)}{(0,6 \, \mathrm m)^2} \approx \boxed{3,0 \times 10^{-5} \,\mathrm N} = 0,03 \,\mathrm{mN}$

8.3. The net force on the particle at Q₂ is the vector

$\vec F = F_{Q_2/Q_1} \, \vec\imath + F_{Q_2/Q_3} \,\vec\jmath = \left(-0,06\,\vec\imath - 0,03\,\vec\jmath\right) \,\mathrm{mN}$

Its magnitude is

$\|\vec F\| = \sqrt{\left(-0,06\,\mathrm{mN}\right)^2 + \left(-0,03\,\mathrm{mN}\right)^2} \approx 0,07 \,\mathrm{mN} = \boxed{7,0 \times 10^{-5} \,\mathrm N}$

and makes an angle θ with the positive horizontal axis (pointing to the right) such that

$\tan(\theta) = \dfrac{-0,03}{-0,06} \implies \theta = \tan^{-1}\left(\dfrac12\right) - 180^\circ \approx \boxed{-153^\circ}$

where we subtract 180° because $\vec F$ terminates in the third quadrant, but the inverse tangent function can only return angles between -90° and 90°. We use the fact that tan(x) has a period of 180° to get the angle that ends in the right quadrant.