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3 years ago

13

According to Newton’s first law of motion, what will an object in motion do when no external force acts on it?

2 answers:

Colt1911 [192]3 years ago

6 0

An object will continue in constant, uniform motion (in a straight
line at constant speed) until acted on by a net external force.

The speed may just as well be zero. That covers the separate case
that people always want to add, of "an object at rest".

KonstantinChe [14]3 years ago

4 0

By definition, we have to:

Newton's first law states that any object will remain in a state of rest or with a uniform rectilinear motion unless an external force acts on it.

Therefore, according to the first law of Newton, if the object is already in motion and has no force acting on it then, it will remain with a uniform rectilinear motion.

Answer:

The object will remain with a uniform rectilinear movement when the external force does not act on it.

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A Rankine cycle with one closed feedwater heater with its drain cascaded backward has a water mass flow rate through the steam g

mamaluj [8]

Answer:Draw a T-s diagram for the ideal Rankine Cycle

Explanation:

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3 years ago

The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of

adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = $ma_a (0.35)$

$1.8v^2(1.1)$ - 1200(9.8)(1.25) = 1200a(0.35)

$1.8v^2(1.1)$ - 14700 = 420 a ------- equation (1)

$F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a$ --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

${1.1 * 1200 \ a} - 14700 = 420 a$

1320 a - 14700 = 420 a

1320 a - 420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

5 0

3 years ago

What derived unit is used to measure the slope of the line in this graph?

Alexxx [7]

Answer:

g/cm³

Explanation:

From the question given above,

The y-axis is representing mass (g)

The x-axis is representing volume (cm³)

Unit of slope =?

Slope of a graph is simply defined as the change in y-axis divided by the change in x-axis. Mathematically it is expressed as:

Slope = change in y-axis (Δy)/change in x-axis (Δx)

Slope = Δy/Δx

Thus, with the above formula, we can obtain the unit used for measuring the slope as follow:

y-axis = mass (g)

x-axis = volume (cm³)

Slope =.?

Slope = Δy/Δx

Slope = mass (g) /volume (cm³)

Slope = g/cm³

Therefore, the derive unit used for measuring the slope is g/cm³

5 0

3 years ago

If a cat can exert 2000N Of Force to move a trailer 50m is 20 seconds how much power did the car use ?

inessss [21]

им putins брат, почему вы обманываете нашу систему образования, Это теперь запрещено в России.

8 0

3 years ago

A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it

natka813 [3]

Answer:

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Explanation:

The rotated angle is given by:

$\theta=\omega0*t+1/2*\alpha*t^2$

Since this is a quadratic equation it can be solved using:

$x=\frac{-b \± \sqrt{b^2-4*a*c} }{2*a}$

Rewriting our equation:

$1/2*\alpha*t^2+\omega0*t-\theta=0$

$t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Since $\sqrt{\omega0^2+2*\theta*\alpha} >\omega0$ we discard the negative solution.

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

8 0

3 years ago