162 x [ 6 ( 7 x 4²) ]<br><br> I got 108,864, Is this correct?

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

4 years ago

Under what circumstances will a binary star produce a nova?

Oksana_A [137]

When a star uses up all of it's energy and begins to die, it swells up to become a red giant star. This causes its surface gravity to decrease, thereby allowing some of its mass to escape into space. A binary star is a pair of stars that orbit each other because of their gravitational attraction to each other. When one member of the binary pair uses up all of its energy and begins to die, it loses mass due to the reduction in surface gravity. But instead of escaping into space, this mass is attracted to the companion star because of its gravitational pull. That increases the mass of the companion star. In a process that takes thousands of years, enough matter is transfered that causes the temperature and pressure to increase sufficiently to result in nuclear fusion reactions on the companion star. When these nuclear reactions become extremely violent, the released nuclear energy increases the brightness of this companion star dramatically, thereby creating a nova. Therefore, it is the dying of one of the stars in a binary system along with a sufficient transfer of star mass to sustain nuclear reactions that results in a nova.

5 0

3 years ago

Dry air will break down if the electric field exceeds about 3.0×106v/m. part a what amount of charge can be placed on a capacito

Genrish500 [490]

The solution for this problem is:The charge would be now equal to:(electric constant) multiplied by the (field strength) multiplied by the (area) so plugging in our values, will give us:8.85 * 10^-12 As / (V * m) * 3 * 10^6 V/m * 0.055 m^2 = 1.46 e-6 amperes would be the answer

7 0

3 years ago

A 4kg ball moving at 8m\sec

Llana [10]

Conservation of momentum: total momentum before = total momentum after

Momentum = mass x velocity

So before the collision:
4kg x 8m/s = 32
1kg x 0m/s = 0
32+0=32

Therefore after the collision
4kg x 4.8m/s = 19.2
1kg x βm/s = β
19.2 + β = 32

Therefore β = 12.8 m/s

8 0

3 years ago

An object oscillates back and forth on the end of a spring. Which of the following statements are true at some time during the c

Sindrei [870]

Answer:

a. The object can have zero acceleration and, simultaneously, nonzero velocity.

c. The object can have nonzero velocity and nonzero acceleration simultaneously.

d. The object can have zero velocity and, simultaneously, nonzero acceleration.

Explanation:

For an object in simple harmonic motion, the total mechanical energy (sum of elastic potential energy and kinetic energy) is constant:

$E=U+K=\frac{1}{2}kx^2 + \frac{1}{2}mv^2$ (1)

where

k is the spring constant

x is the displacement

m is the mass

v is the speed

We can also notice that the force on the spring is given by Hook's law:

$F=-kx$

And since according to Newton's law we have F = ma, this can be rewritten as

$ma=-kx\\a=-\frac{k}{m}x$

which means that the acceleration is proportional to the displacement.

So by looking again at eq.(1), we can now states that:

- when the displacement is zero, x=0, the acceleration is zero, a=0, and the velocity is maximum

- when the velocity is zero, v=0, the acceleration is maximum, which occurs when the displacement is maximum

- in all the other intermediate situations, both velocity and acceleration are non-zero

So the correct answers are

a. The object can have zero acceleration and, simultaneously, nonzero velocity.

c. The object can have nonzero velocity and nonzero acceleration simultaneously.

d. The object can have zero velocity and, simultaneously, nonzero acceleration.

8 0

3 years ago

162 x [ 6 ( 7 x 4²) ] I got 108,864, Is this correct?