Help please

How do think animals grow from babies to adults

Sladkaya [172]

Answer:

As animals get older, their bodies begin to change as well as their instincts and priorities. They get bigger and might adapt or develop.

7 0

3 years ago

What does the law of conservation of energy state?

VikaD [51]

The total amount of energy remains constant in an isolated system. It implies that energy can neither be created nor destroyed, but can be change from one form to another.

6 0

3 years ago

If the velocity of an object is doubled, its kinetic energy is ______

swat32

Increased by a factor of 4

4 0

3 years ago

La ecuación de la posición de una esferita está dada por: r(t)=(2.Cos(πt) i-3.Sen(πt) j) (m) ¿Cuál es la velocidad de la esferit

katovenus [111]

Answer:

v = (-4.44 i^ + 6.66 j^ ) m/s, a_average =( 0 i^ -2π j^) m/s²

Explanation:

The expression left corresponds to an oscillatory movement (MAS), the speed is defined by

v = dr / dt

the function of position

r = 2 cos πt i^ + 3 sin πt j^

let us note that it is a movement in two dimensions

let's perform the derivative

v = -2π sin πt i^ + 3π cos πt j^

we evaluate this expression for t = 0.25 s, remember that the angle is in radians

v = -2π sin (π 0.25) i^ + 3π cos (π 0.25) j^

v = (-4.44 i^ + 6.66 j^ ) m/s

To calculate the mean acceleration we use the expression

a = ( $v_{f} - v_{o}$ ) / Δt

indicates that the time is the first 3 s

we look for the initial velocity t = 0 s

v₀ = 0 i ^ + 3π j ^

we look for the fine velocity, t = 3 s

v_f = - 2π sin (π 3) + 3π cos (π 3) j ^

v_f = 0 i ^ - 3π j ^

we calculate the average acceleration

Δt = (3 -0) = 3 s

a_average = (0-0) / 3 i ^ + (-3π - 3π) / 3

a_average = (0 i ^ -2π j ^ ) m/s²

6 0

3 years ago

1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

Delvig [45]

1) $6.67\cdot 10^{-4} N$

The force of gravitation between the two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2}$ is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N$

2) $2.7\cdot 10^{-3} N$

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

$F \sim \frac{1}{r^2}$

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

$F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F$

So, the gravitational force increases by a factor 4. Therefore, the new force will be

$F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N$

3) 12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

$\tau=Fd$