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3 years ago

Which object has unbalanced forces acting on it

Physics

1 answer:

Basile [38]3 years ago

8 0

Answer:

If you kick a football and it moves from one place to another, it means that unbalanced forces are acting upon it. Ball moves from one place to another after kicking it. This is an example of unbalanced force.

Explanation:

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A simple pendulum is made from a 0.75-m-long string and a small ball attached to its free end. The ball is pulled to one side th

Alecsey [184]

Answer:

It takes 0.43 seconds before the pendulum attains the maxium speed.

Explanation:

L= 0.75m

g= 9.8 m/s²

T= 2π * √(L/g)

T=1.73 sec

T(vmax) = T/4

T(vmax) = 0.43 sec

8 0

4 years ago

I WILL MARK YOU TOP IF YOU HELP ME I NEED THIS ASAP!!!!

Pachacha [2.7K]

The basketballs and racquetball eventually stopped bouncing due to the first law. the first law states that an object at rest will stay at rest unless acted upon by an external force. this means that once the 2 balls lose friction, they will remain at rest until acted upon by an external force. once the 2 balls lose friction(energy), they come to a complete stop.

6 0

3 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

7 0

3 years ago

A tumbleweed is moving with a velocity of 5.4 m/s westward. A sudden gust of wind blows for 1.5 seconds and reduces its velocity

Klio2033 [76]

Answer:

$a=-2.4 m/s^{2}$

Explanation:

Let's recall the acceleration definition.

$a=\frac{\Delta v}{\Delta t}$ (1)

Δv is the velocity variation, Δv = v(f) - v(i). It is just the difference between final and initial velocities. So Δv = 1.8[m/s] - 5.4[m/s] = -3.6 [m/s]

And Δt is the time variation, in this case it is just 1.5 s.

Therefore, using the equation (1), the acceleration will be:

$a=\frac{-3.6}{1.5}=-2.4 m/s^{2}$

The minus sign means that the tumbleweed is slowing down.

I hope it helps you!

8 0

3 years ago

Temperature measures: A. how much kinetic energy molecules have. B. the potential energy of molecules. C. how heat flows. D. the

serg [7]

Temperature measures how much kinetic energy molecules have. The faster the kinetic energy, the higher the temperature.

8 0

3 years ago