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dimulka [17.4K]

2 years ago

8

The center of mass of the system is the point at which the total mass of the system could be concentrated without changing the _

___________.

1 answer:

Sveta_85 [38]2 years ago

8 0

The centre of mass of the system is the point at which the total

mass of system could be concentrated without changing the moment

of the system.

Centre of Mass is the point at which the whole mass of the system

is assumed to be concentrated.

The general formula for the COM is:

xₙ = Σmₐxₐ / Σmₐ where, a = 1,2,3.........n

Here the term Σ mₐ xₐ is called the first moment of the system and the

denominator expression is called total mass of the system.

Therefore, from this theory we can say that the moment of the system

remain unchanged while calculating the COM.

Hence, The centre of mass of the system is the point at which the total

mass of system could be concentrated without changing the moment

of the system.

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Suppose you are an engineer building a road across a mountain. From a prudential point of view, it would be easier and cheaper t

Alchen [17]

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7 0

2 years ago

An object of mass m travels along the parabola yequalsx squared with a constant speed of 5 units/sec. What is the force on the

Dmitriy789 [7]

Explanation:

The object is moving along the parabola y = x² and is at the point (√2, 2). Because the object is changing directions, it has a centripetal acceleration towards the center of the circle of curvature.

First, we need to find the radius of curvature. This is given by the equation:

R = [1 + (y')²]^(³/₂) / |y"|

y' = 2x and y" = 2:

R = [1 + (2x)²]^(³/₂) / |2|

R = (1 + 4x²)^(³/₂) / 2

At x = √2:

R = (1 + 4(√2)²)^(³/₂) / 2

R = (9)^(³/₂) / 2

R = 27 / 2

R = 13.5

So the centripetal force is:

F = m v² / r

F = m (5)² / 13.5

F = 1.85 m

7 0

3 years ago

In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.

HACTEHA [7]

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is $41.9^{\circ}$

Explanation:

A)

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

$s=u_y t + \frac{1}{2}at^2$ (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

$u_y = u sin \theta$ is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

$\theta=40.0^{\circ}$ is the angle of projection

So

$u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s$

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (downward)

Substituting the numbers, we get

$-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0$

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

$d=u_x t$

where

$u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s$ is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

$d=(9.19)(1.778)=16.34 m$

B)

In this second case,

$\theta=42.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.1t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s$

So, the range of the shot is

$d=u_x t = (8.85)(1.851)=16.38 m$

C)

In this third case,

$\theta=45^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.5t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s$

So, the range of the shot is

$d=u_x t = (8.49)(1.925)=16.34$ m

D)

In this 4th case,

$\theta=47.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.8t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s$

So, the range of the shot is

$d=u_x t = (8.11)(1.981)=16.07 m$

E)

From the previous parts, we see that the maximum range is obtained when the angle of releases is $\theta=42.5^{\circ}$ .

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

$s-u sin \theta t + \frac{1}{2}gt^2=0$

The solutions of this quadratic equation are

$t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}$

This is the time of flight: so, the horizontal range is

$d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta$

It can be found that the maximum of this function is obtained when the angle is

$\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})$

Therefore in this problem, the angle which leads to the maximum range is

$\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}$

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8 0

4 years ago

An air craft heads north at 320 km/hr relative to the wind. the wind velocity is 80km/hr from the north. find the relative veloc

Gnoma [55]

Answer:

Relative to the ground, the velocity of the aircraft is 240 km/hr

Explanation:

Relative velocity is different from normal velocity;

When 2 objects are moving in opposite directions towards each other, they will appear to be faster than they actually are;

This is known as the relative velocity;

The information tells us we have the aircraft moving 320 km/hr northwards relative to the wind;

The wind is in the opposite direction at 80 km/hr;

R = relative velocity of the aircraft

v = actual velocity of the aircraft

w = velocity of the wind

R = v + w

Note: if the wind was moving in the same direction, the formula would be R = v - w

320 = v + 80

v = 320 - 80

v = 240

The velocity relative to the ground is simply the actual velocity as the ground doesn't move;

So, relative to the ground, the velocity of the aircraft is simply 240 km/hr

7 0

3 years ago

Which term describes an educated guess about the outcome of an

bezimeni [28]

Answer:

Hypothesis

Explanation:

Refer to a trial solution to a problem as a hypothesis, often called an "educated guess" because it provides a suggested outcome based on the evidence.

5 0

3 years ago

Read 2 more answers