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2 years ago

2. What do you understand by balanced and unbalanced force

Physics

1 answer:

labwork [276]2 years ago

8 0

Answer:

forces that are equal in size and opposite in direction. Balanced forces do not result in any change in motion. unbalanced. forces: forces applied to an object in opposite directions that are not equal in size. Unbalanced forces result in a change in motion.

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4 years ago

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2) A skier stands at rest and begins to ski downhill with an acceleration of 3.0 m/s² {downhill). What is

Finger [1]

Answer:

337.5m

Explanation:

Kinematics

Under constant acceleration, the kinematic equation holds:

$s=\frac{1}{2}at^2+v_ot+s_o$ , where "s" is the position at time "t", "a" is the constant acceleration, " $v_o$ " is the initial velocity, and $s_o$ is the initial position.

Defining Displacement

Displacement is the difference in positions: $s-s_o$ or $\Delta s$
$s=\frac{1}{2}at^2+v_ot+s_o$

$s-s_o=\frac{1}{2}at^2+v_ot$

$\Delta s=\frac{1}{2}at^2+v_ot$

Using known information

Given that the initial velocity is zero ("skier stands at rest"), and zero times anything is zero, and zero plus anything remains unchanged, the equation simplifies further to the following:

$\Delta s=\frac{1}{2}at^2+v_ot$

$\Delta s=\frac{1}{2}at^2+(0)*t$

$\Delta s=\frac{1}{2}at^2+0$

$\Delta s=\frac{1}{2}at^2$

So, to find the displacement after 15 seconds, with a constant acceleration of 3.0 m/s², substitute the known values, and simplify:

$\Delta s=\frac{1}{2}at^2$

$\Delta s=\frac{1}{2}(3.0[\frac{m}{s^2}])(15.0[s])^2$

$\Delta s=337.5[m]$

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2 years ago

Usain Bolt (m = 86.2 kg)

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Answer: 138 N

Explanation:

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3 years ago

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Leni [432]

Answer: 12.0 m/s^2

Explanation:

Let $\alpha$ be the angular acceleration of the end of the rod

Taking torque about the link, we have:

$\tau = W \times OM\\ or\\ \tau = mg\times \left(\dfrac{L}{2}\right)\sin 55^\circ ....(i)$

Torque is also given in terms of moment of inertia of the rod and its angular acceleration i.e.

$\tau = I_{rod}\ \alpha......(ii)$

From equations (i) and (ii) we have:

$mg\times \left(\frac{L}{2}\right)\sin 55^\circ = \left(\frac{mL^2}{3}\right)\alpha\ \ \ \ \ \ \ (\because I_{rod} = \dfrac{mL^2}{3}) \\ \\ \alpha = \left(\frac{3g}{2L}\right)\sin 55^\circ\\ \\ \alpha = \left(\frac{3\times 9.8}{2\times 2.4}\right)\sin 55^\circ\\ \\ \boxed{\alpha = 5.02\ rad/s^2} }$

The acceleration of the end of the rod farthest from the link is given by:

$a = L\alpha\\ \\ a = (2.4\ m)(5.02\ rad/s^2)\\ \\ \boxed{a = 12.0\ m/s^2}$

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2 years ago

every object that has mass attracts every other object with a gravitational force. it has been proven that the size of the gravi

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Answer;

- This statement about matter and its behavior is best classified as a Law.

-It is the Law of Universal Gravitation.

Explanation;

-The Law of Universal Gravitation states that every point mass attracts every other point mass in the universe by a force pointing in a straight line between the centers-of-mass of both points, and this force is proportional to the masses of the objects and inversely proportional to their separation.This attractive force always points inward, from one point to the other.

-The Law applies to all objects with masses, big or small. Two big objects can be considered as point-like masses, if the distance between them is very large compared to their sizes or if they are spherically symmetric.

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3 years ago

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2. What do you understand by balanced and unbalanced force​

2. What do you understand by balanced and unbalanced force