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2 years ago

Label these as balanced or unbalanced:

Physics

2 answers:

kvasek [131]2 years ago

4 0

Answer:

A book lying on a table - Balanced force
An airplane cruising in level flight - Balanced
A rock falling from a cliff - Unbalanced force
A bridge collapsing in an earthquake - Unbalanced force
A man sitting on a park bench - Balanced force
A space shuttle taking off - Unbalanced force
A car maintaining a constant speed on a straight road - Balanced force
An airplane landing - Unbalanced force

Explanation:

Usually, one or more forces act on a body at an instant of time. When these forces acting on a body and bring the body in the equilibrium position, the force is said to be balanced. The unbalanced force changes the equilibrium state of the body.

As in the case of an airplane cruising in a level flight, the weight of the plane will be equal to the lift force and the thrust is equal to the drag. So the plane is experiencing a balanced force.

erik [133]2 years ago

3 0

Answer:

1. A book lying on a table (Balanced force)

2.An airplane cruising in level flight (Balanced force)

3.A rock falling from a cliff (Unbalanced force)

4.A bridge collapsing in an earthquake (Unbalanced force)

5.A man sitting on a park bench (Balanced force)

6.A space shuttle taking off (Unbalanced force)

7.A car maintaining a constant speed on a straight road (Balanced force)

8.An airplane landing (Unbalanced force)

Explanation:

An object will be in equilibrium or balanced state if the sum off all the forces and toques on that object will be equal to zero.

in case of book lying on the table its weight is balanced by the normal force exerted by the table. So, balance force is acting on it. Same analogy can be applied to a man sitting on bench.

When air plane cruise in level flight, the weight of airplane is equal to the lift force and drag force is equal to the thrust force. Hence, balanced force acts on the airplane.

When car runs with the uniform speed on the road, engine force is equal to the frictional force. Hence, balanced force acts on the car.

In all other cases unbalance forces act which can be judged just by looking statements.

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What is the force acting on the object?(g=10m/s^2)

Vera_Pavlovna [14]

Answer:w=mxg

2x10 =20 N

Explanation:force acting downwards is mg mass into gravitional feild

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2 years ago

A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s aft

Anton [14]

Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

$m_1 v_1 = m_2 v_2$

$m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s$

Using the equation to find $v_1$ ,

$v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475$

Using the conservation of energy equation, we have,

$KE= \frac{1}{2}m*v^2$

$KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J$

$KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J$

$Total KE= 92077.75+13425530=92708.9J$

Now this energy over the cannonball

$KE=\frac{1}{2}m*v_2^2$

$92708.9=\frac{1}{2}15.5v_2^2$

$V_2 = 109.37m/s$

The gain in velocity is 0.37m/s

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2 years ago

Alan starts from his home and walks 1.3 km east to the library. He walks an additional 0.68 km east to a music store. From there

zepelin [54]

Answer: final Displacement = 0 km, total distance covered =7 km

Explanation:

Given the following :

From home to library = 1.3 km East

Library to music store = 0.68km East

Music store to friend's house = 1.1km North

Friend's house to grocery store = 0.42 km North

Displacement is the net change in distance traveled.

Eastward distance :

To = (1.3 + 0.68)km = 1.98km East

Fro = (0.68 + 1.3)km = 1.98 km East

Δ distance = (1.98 - 1.98) = 0

Northward direction:

To = (1.1 + 0.42)km = 1.52km north

Fro = (0.42 + 1.1)km = 1.52km North

Δ distance = (1.98 - 1.98) = 0

Hence final Displacement = 0

Total distance covered = 2 × (1.3 + 0.68 + 1.1 + 0.42) = 2 × 3.5

= 7km

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2 years ago

A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou

elena55 [62]

Answer:

$3.2\times 10^{-7}\ m$ or 0.32 μm.

Explanation:

Given:

The radiations are UV radiation.

The frequency of the radiations absorbed (f) = $9.38\times 10^{14}\ Hz$

The wavelength of the radiations absorbed (λ) = ?

We know that, the speed of ultraviolet radiations is same as speed of light.

So, speed of UV radiation (v) = $3\times 10^8\ m/s$

Now, we also know that, the speed of the electromagnetic radiation is related to its frequency and wavelength and is given as:

$v=f\lambda$

Now, expressing the above equation in terms of wavelength 'λ', we have:

$\lambda=\frac{v}{f}$

Now, plug in the given values and solve for 'λ'. This gives,

$\lambda=\frac{3\times 10^8\ m/s}{9.38\times 10^{14}\ Hz}\\\\\lambda=3.2\times 10^{-7}\ m\\\\\lambda=3.2\times 10^{-7}\times 10^{6}\ \mu m\ [1\ m=10^6\ \mu m]\\\\\lambda=3.2\times 10^{-1}=0.32\ \mu m$

Therefore, the wavelength of the radiations absorbed by the ozone is nearly $3.2\times 10^{-7}\ m$ or 0.32 μm.

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