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3 years ago

Why is rolling a ball down a hill considered an unbalanced force?

2 answers:

7 0

Friction is what makes the ball unbalanced. At one point the ball will stop rollin down the hill because of Friction the Friction made the ball balanced. Hopefully this helped you

SIZIF [17.4K]3 years ago

6 0

Answer:

An unbalanced force refers to the force that causes change in velocity that is causes acceleration of the body. An unbalanced force makes a body to change either its speed or direction or both.

Balanced forces are those which have equal magnitude but act in opposite direction. There is zero net acceleration.

A ball rolling down a hill, accelerates downwards because of gravitational force.

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2 years ago

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3 years ago

Read 2 more answers

An object begins with a speed of 20 meters per second and slows down to a speed of 10 meters per second over a time of 5 seconds

Andreyy89

Answer:

V = 6 m/s

Explanation:

Given that,

Initial speed of an object is 20 m/s

Final speed of an object is 10 m/s

Time, t = 5 s

We need to find the average speed of the object during these 5 seconds. Let it is equal to V. Here, time is same. The average speed is given by :

$V=\dfrac{x+y}{2}\\\\\text{Putting values}\\\\V=\dfrac{20+10}{5}\\\\V=6\ m/s$

So, the average speed of the object is 6 m/s.

3 0

3 years ago

Check the dimensional consistencies of s =vot+1/2at2

Leni [432]

Answer:

s is distance so it's dimensions become L.

Nd other side we have ut+1\2at^2.

as 1\2 is a constant it will have dimensions and apply the dimensions to other quantities.

on solving u will get L there also i,e ur LHS = RHS.

thus the equation is dimensionally consistent.

Explanation:

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3 years ago

a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the

tatuchka [14]

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
$W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N$
the force of the wind F, acting horizontally, with intensity
$F=12 N$
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
$T \cos \alpha -F=0$
$T \sin \alpha -W=$
By dividing the second equation by the first one, we get
$\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04$
From which we find
$\alpha = 63.8 ^{\circ}$
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
$T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N$

7 0

3 years ago