What can you say about the speed of the car?

Describe 1 way to reduce the shortage of food for a growing human population.

alexira [117]

The answer is; producing genetically modified organism

GMOs have been shown to produce varieties of organisms that have increased yields and are resistant to the upsets of climate change and diseases. GMOs are made by introducing genes into an organism that will confer particular desired traits such as disease and drought resistance.

5 0

3 years ago

Which statements best describe half-lives of radioactive isotopes? Check all that apply.

victus00 [196]

Answer:

A

Explanation:

half life is the time taken for a radioactive isotopes to dissociate

7 0

2 years ago

A strip of copper 190 µm thick and 4.20 mm wide is placed in a uniform magnetic field of magnitude B = 0.78 T, that is perpendic

Veronika [31]

Answer:

V = 9.682 × 10^(-6) V

Explanation:

Given data

thick = 190 µm

wide = 4.20 mm

magnitude B = 0.78 T

current i = 32 A

to find out

Calculate V

solution

we know v formula that is

V = magnitude× current / (no of charge carriers ×thickness × e

here we know that number of charge carriers/unit volume for copper = 8.47 x 10^28 electrons/m³

so put all value we get

V = magnitude× current / (no of charge carriers ×thickness × e

V = 0.78 × 32 / (8.47 x 10^28 × 190 × 1.602 x 10^(-19)

V = 9.682 × 10^(-6) V

3 0

3 years ago

Two long, parallel wires carry currents of different magnitudes. If the amount of current in one of the wires is doubled, what h

Ulleksa [173]

Answer:

The magnitude of the force that each wire exerts on the other will increase by a factor of two.

Explanation:

force on parallel current carrying wire, F = BILsinθ

where;

B is the strength of the magnetic field

L is the length of the wire

I is the magnitude of current on the wire

θ is the angle of inclination of the wire

Assuming B, L and θ is constant, then F ∝ I

F = kI

$\frac{F_1}{I_1} = \frac{F_2}{I_2}$

When the amount of current is doubled in one of the wires, lets say the second wire;

$\frac{F_1}{I_1} = \frac{F_2}{2I_1} \\\\F_2 = \frac{2F_1I_1}{I_1} \\\\F_2 =2F_1$

Also, if will double the amount of current on the first wire, then

F₁ = 2F₂

Therefore, the magnitude of the force that each wire exerts on the other will increase by a factor of two.

3 0

3 years ago

Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.

Ksivusya [100]

$|v| =\sqrt{ G \cdot M / r}$ , where

$M$ the mass of the planet, and
$G$ the universal gravitation constant.

Explanation:

Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

Equation 1 (see below) relates net force the object experiences, $\Sigma F$ to its orbit velocity $v$ and its mass $m$ required for it to stay in orbit :

$\Sigma F = m \cdot v^{2} / r$ (equation 1)

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force $\Sigma F$ acting on the soccer ball shall equal to its weight, $W = m \cdot g$ where $g$ the gravitational acceleration constant. Thus

$\Sigma F = W = m \cdot g$ (equation 2)

Substitute equation 2 to the left hand side of equation 1 and solve for $v$ ; note how the mass of the soccer ball, $m$ , cancels out:

$m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0)$ (equation 3)

Equation 4 gives the value of gravitational acceleration, $g$ , a point of negligible mass experiences at a distance $r$ from a planet of mass $M$ (assuming no other stellar object were present)

$g = G \cdot M/ r^{2}$ (equation 4)

where the universal gravitation constant $G = 6.67408 \times 10^{-11} \cdot \text{m}^{3} \cdot \text{kg}^{-1} \cdot \text{s}^{-2}$

Thus

$\begin{array}{lll}|v| &=& \sqrt{g \cdot r}\\ & =&\sqrt{ G \cdot M / r}\end{array}$

3 0

3 years ago