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1 year ago

Example of a balanced force

1 answer:

4 0

An example of a balanced force would be a book sitting on a shelf untouched.

Isaac Newton’s First Law of Motion states that an object at motion stays in motion, and an object at rest stays at rest until acted on by an unbalanced force. A book sitting still is an example of a balanced force because nothing is acting on it; its potential energy is stored while it’s at rest. For this book to become an unbalanced force, an outside force would have to occur (i.e pushing the book or dropping it) that causes it to not be in a state of stillness.

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You are the design engineer in charge of testing the crashworthiness of new automobile models. Cars are tested by smashing them

BaLLatris [955]

Answer:

average force is 138900 N

Explanation:

given data

speed = 50 km/h = 13.89 m/s

mass = 1500 kg

time = 0.15 s

to find out

average force

solution

we know initial velocity of car is positive

we will apply here equation to find force that is

Force x change in time that is equal to mass x change in speed ................1

so put here all value of mass, time and velocity in equation 1

Force x 0.15 = 1500 x ( -13.89)

force = 138900 N

so average force is 138900 N

3 0

3 years ago

What question can you ask yourself if you are trying to determine if a mixture is heterogeneous or homogeneous? (This is not in

Zina [86]

Answer:

did it have more than one phase

Explanation:

this will be my question

7 0

4 years ago

Read 2 more answers

An object falling from a great height in Earth's atmosphere eventually reaches a speed called Terminal velocity. What cause a fa

Julli [10]

Answer:

Explanation:

If its at a height the Gratitude of it falling down with only Gravity if Any other Forces are acting on it so as Friction But Sideways.

5 0

3 years ago

A 6.00 kg ball is dropped from a height of 12.0 m above one end of a uniform bar that pivots at its center. The bar has mass 5.0

Andre45 [30]

Answer:

The height of the other ball go after the collision is 2.304 m.

Explanation:

Given that,

Mass of ball = 6.00 kg

Height = 12.0 m

Mass of bar =5.00 kg

Length = 4.00 m

Suppose we need to calculate how high will the other ball go after the collision

We need to calculate the velocity of ball

Using formula of velocity

$v=\sqrt{2gh}$

$v=\sqrt{2\times9.8\times12.0}$

$v=15.33\ m/s$

We need to calculate the angular momentum

Using formula of angular momentum

$l_{before}=mvr$

Put the value into the formula

$l_{before}=6.00\times15.33\times2.0$

$l_{before}=183.96\ kgm^2/s$

We need to calculate the angular momentum

Using formula of angular momentum

$l_{after}=I_{t}\omega$

$l_{after}=(\dfrac{ml^2}{12}+m_{1}r^2+m_{2}r^2)\omega$

Put the value into the formula

$l_{after}=(\dfrac{5\times4.00^2}{12}+6.00\times2.0^2+6.00\times2.0^2)\omega$

$183.96=54.66\omega$

$\omega=\dfrac{183.96}{54.66}$

$\omega=3.36\ rad/sec$

After collision the ball leaves with velocity

We need to calculate the velocity after collision

Using formula of the velocity

$v= r\omega$

$v=2.0\times3.36$

$v=6.72\ m/s$

We need to calculate the height

Using formula of height

$h=\dfrac{v^2}{2g}$

Put the value into the formula

$h=\dfrac{(6.72)^2}{2\times9.8}$

$h=2.304\ m$

Hence, The height of the other ball go after the collision is 2.304 m.

5 0

3 years ago

A wave on a string has a wavelength of 0.90 m at a frequency of 600 Hz. If a new wave at a frequency of 300 Hz is established in

kobusy [5.1K]

Answer:

λ₂ = 1.8 m

Explanation:

given,

wavelength of the string 1 = 0.90 m

frequency of the string 1 = 600 Hz

wavelength of string 2 = ?

frequency of the string 2 = 300 Hz

we now,

$f\ \alpha\ \dfrac{1}{\lambda}$

now,

$\dfrac{f_1}{f_2}=\dfrac{\lambda_2}{\lambda_1}$

$\dfrac{600}{300}=\dfrac{\lambda_2}{0.9}$

λ₂ = 2 x 0.9

λ₂ = 1.8 m

Hence, the wavelength of the second string is equal to λ₂ = 1.8 m

8 0

3 years ago