N a scientific investigation, what is the name for a prediction that can be tested?

A satellite in outer space is moving at a constant velocity of 20.5 m/s in the +y direction when one of its on board thruster tu

Katyanochek1 [597]

Answer:

a) V_f = 25.514 m/s

b) Q =53.46 degrees CCW from + x-axis

Explanation:

Given:

- Initial speed V_i = 20.5 j m/s

- Acceleration a = 0.31 i m/s^2

- Time duration for acceleration t = 49.0 s

Find:

(a) What is the magnitude of the satellite's velocity when the thruster turns off?

(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.

Solution:

- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:

V_f = V_i + a*t

V_f = 20.5 j + 0.31 i *49

V_f = 20.5 j + 15.19 i

- The magnitude of the velocity vector is given by:

V_f = sqrt ( 20.5^2 + 15.19^2)

V_f = sqrt(650.9861)

V_f = 25.514 m/s

- The direction of the velocity vector can be computed by using x and y components of velocity found above:

tan(Q) = (V_y / V_x)

Q = arctan (20.5 / 15.19)

Q =53.46 degrees

- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.

4 0

3 years ago

if a torque of 55.0 N/m is required and the largest force that can be exerted by you is 135 N what is th e length of the lever a

Whitepunk [10]

Answer:

$r=0.41m$

Explanation:

Torque is defined as the cross product between the position vector ( the lever arm vector connecting the origin to the point of force application) and the force vector.

$\tau=r\times F$

Due to the definition of cross product, the magnitude of the torque is given by:

$\tau=rFsin\theta$

Where $\theta$ is the angle between the force and lever arm vectors. So, the length of the lever arm (r) is minimun when $sin\theta$ is equal to one, solving for r:

$r=\frac{\tau}{F}\\r=\frac{55\frac{N}{m}}{135N}\\r=0.41m$

7 0

3 years ago

Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)

sergeinik [125]

$v^2-{v_0}^2=2a(x-x_0)$

dónde $v$ es la velocidad de la esfera, $v_0$ es suya velocidad inicial, $a=-g$ la aceleración debida a la gravedad, $x$ la posición, y $x_0$ la posición inicial. Tomamos $x_0=0\,\mathrm m$ a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

$v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)$

$\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}$

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

4 0

3 years ago

37 Questions - All Questions Are Required

ANTONII [103]

Acceleration = vf-vi /t
10-22/3=2.6m/s^2

5 0

3 years ago

Read 2 more answers

What two factors directly affect the acceleration of an object?

marin [14]

The acceleration of an object depends on the mass of the object and the amount of the force applied to it. It states that the acceleration of an object increases with increased force and decreases with increased mass.

4 0

3 years ago