How is a manipulated variable different from a controlled variable?

NEED HELP FAST PLEASE

Fittoniya [83]

Answer:

if a star is moving towards earth it is blue shifting

4 0

3 years ago

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A bathysphere used for deep-sea exploration has a radius of 1.50 m and a mass of 1.19 104 kg. In order to dive, this submarine t

Hitman42 [59]

Answer:

m = 2,776.95 kg

Explanation:

given,

radius = 1.5 m

mass = 1.19 x 10⁴ Kg

constant speed, v = 1.4 m/s

Resistive force = 1105 N

density of sea water = 1.03 x 10³ kg/m³

Volume of vessel,

$v=\dfrac{4}{3}\pi r^3$

$v=\dfrac{4}{3}\pi\times 1.5^3$

v = 14.14 m³

Upthrust,

U = ρ g V

U = 1030 x 9.8 x 14.14

U = 142729.16 N

Since the vessel is moving at a constant speed, the resultant force on it should equal zero.

downward acting forces = upward acting forces

Mg = U + resistive force

M x 9.8 = (142729.16+1105)

M=14676.95 Kg

Mass of water,

m = M - mass of vessel

m = 14676.95 Kg- 11,900 kg

m = 2,776.95 kg

5 0

3 years ago

PLEASE HELP!!

leva [86]

Milk production i thank so.

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3 years ago

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A ball is rolled of the edge of a table with a

tatiyna

Answer:

Approximately $10\; \rm m \cdot s^{-1}$ at $5.6^\circ$ below the horizon.

Horizontal component of velocity: $10\; \rm m \cdot s^{-1}$ .
Vertical component of velocity: $0.981\; \rm m\cdot s^{-1}$ (downwards.)

(Assumption: air resistance on the ball is negligible; $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

Assume that the air resistance on the ball is negligible. The horizontal component of the velocity of the ball would stay the same at $10\; \rm m \cdot s^{-1}$ until the ball reaches the ground.

On the other hand, the vertical component of the ball would increase (downwards) at a rate of $g = 9.81\; \rm m\cdot s^{-2}$ (where $g$ is the acceleration due to gravity.) In $0.1\; \rm s$ , the vertical component of the velocity of this ball would have increased by $9.81\; \rm m \cdot s^{-2} \times 0.10\; \rm s = 0.981\; \rm m \cdot s^{-1}$ .

However, right after the ball rolled off the edge of the table, the vertical component of the velocity of this ball was $0\; \rm m\cdot s^{-1}$ . Hence, $0.10\; \rm s$ after the ball rolled off the table, the vertical component of the velocity of this ball would be $0\; \rm m \cdot s^{-1} + 0.981\; \rm m\cdot s^{-1} = 0.981\; \rm m \cdot s^{-1}$ .

Calculate the magnitude of the velocity of this ball. Let $v_{x}$ and $v_{y}$ and denote the horizontal and vertical component of the velocity of this ball, respectively. The magnitude of the velocity of this ball would be $\displaystyle \sqrt{{v_x}^{2} + {v_y}^{2}}$ .

At $0.10\; \rm s$ after the ball rolled off the table, $v_x = 10\; \rm m \cdot s^{-1}$ while $v_y = 0.981\; \rm m \cdot s^{-1}$ . Calculate the magnitude of the velocity of the ball at this moment:

$\begin{aligned} \| v \| &= \sqrt{{v_x}^{2} + {v_y}^{2}} \\ &= \sqrt{\left(10\; \rm m \cdot s^{-1}\right)^{2} + \left(0.981\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 10.0\; \rm m\cdot s^{-1}\end{aligned}$ .

Calculate the angle between the horizon and the velocity of the ball (a vector) at that moment. Let $\theta$ denote that angle.

$\displaystyle \tan \theta = \frac{\text{rise}}{\text{run}}$ .

For the vector representing the velocity of this ball:

$\displaystyle \tan \theta = \frac{0.981\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-1}} = 0.0981$ .

Calculate the size of this angle:

$\theta = \arctan 0.0981 \approx 5.62^\circ$ .

Notice that the vertical component of the velocity of this ball at that moment points downwards (towards the ground.) Hence, the corresponding velocity should point below the horizon.

5 0

3 years ago

An object moves along the x-axis with its position x given as a function of time t by x(t)= Ht^2 - Ft + G

Jet001 [13]

Answer:

v(t) = 2Ht - F

Explanation:

Since, the position of the object is given in terms of time (t) as follows:

x(t) = Ht² - Ft + G

where,

H, F, G are constants.

Therefore, the velocity of the object can also be found in terms of the time (t), by simply taking the derivative of the given position equation with respect to time. So, the velocity can be found as follows:

(d/dt) x(t) = (d/dt)(Ht² - Ft + G)

v(t) = (d/dt)(Ht²) - (d/dt)(Ft) + (d/dt)(G)

v(t) = H (d/dt)(t²) - F (d/dt)(t) + (d/dt)(G)

v(t) = H(2t) - F(1) + 0

3 0

3 years ago