All categories

2 years ago

What is a Non-Example of Newton's 1st law?

Physics

1 answer:

guajiro [1.7K]2 years ago

5 0

Answer:

Say a 14 year old girl was at a construction site and she was asked to move something like a 10,000 pound brick( one brick). She would be acting on it as the unbalanced force but they would still not change their position.

so to say the girl would be doing everything she could to move that brick but the brick would still be in that same spot so the unbalanced force (the girl) would be acting on the thing that was at rest but it wouldn't move.

so the unbalanced force would not really be acting on the thing at rest; even though the unbalanced force was doing something to the brick.

( just think about it and you will eventually get it...just imagine in your head...)

Explanation:

You might be interested in

I will mark brainlist

tensa zangetsu [6.8K]

Answer:

False

Explanation:

A wave is a disturbance that transfers energy from one place to another without transferring matter.

6 0

3 years ago

Read 2 more answers

A car that increases it's speed from 20 km/h to 100 km/h undergoes

enot [183]

If a car increases its speed from 20km/h to 100 km/h undergoes positive acceleration. Hope this helped!

3 0

3 years ago

Why is the H-R diagram useful in plotting the life cycles of stars?

Roman55 [17]

The sun provides a handy benchmark for describing other stars. The mass of this solar system's sun gives us a unit for measuring other stars' masses.

4 0

4 years ago

You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 6 m at your feet, then a

Zigmanuir [339]

Answer:

4.5 m/s

Explanation:

The rock must barely clear the shelf below, this means that the horizontal distance covered must be

$d_x = 5 m$

while the vertical distance covered must be

$d_y = 6 m$

The rock is thrown horizontally with velocity $v_x$ , so we can rewrite the horizontal distance as

$d_x = v_x t$

where t is the time of flight. Re-arranging the equation,

$t=\frac{d_x}{v_x}$ (1)

The vertical distance covered instead is

$d_y = \frac{1}{2}gt^2$

where we omit the term $ut$ since the initial vertical velocity is zero. From this equation,

$t=\sqrt{\frac{2d_y}{g}}$ (2)

Equating (1) and (2), we can solve the equation to find $v_x$ :

$\frac{d_x}{v_x}=\sqrt{\frac{2d_y}{g}}\\\frac{d_x^2}{v_x^2}=\frac{2d_y}{g}\\v_x = d_x \sqrt{\frac{g}{2d_y}}=5\sqrt{\frac{9.8}{2(6)}}=4.5 m/s$

6 0

3 years ago

If the car's speed decreases at a constant rate from 77 mi/h to 50 mi/h in 3.0 s, what is the magnitude of its acceleration, ass

Lostsunrise [7]

The acceleration of the car,

$a = \frac{v-u}{t}$

Here, v is final velocity, u is initial velocity and t is time taken by the car.

Given $u =77 \ mil/h$ , $v = 50 mi/h$ and $t = 3.0 s = 3.0 \times \frac{1 \ h}{3600} = 8.3 \times 10^{-4} h$

Therefore, from above equation

$a = \frac{50 \ mi/h -77 \ mi/h}{8.33 \times 10^{-4} h} = - \frac{27 \ mi/h }{8.33 \times 10^{-4} h} = - 3.2 \times 10^{4} \ mi/h^2$ .

Here, negative sign shows deceleration of a car.

Thus the the magnitude of car acceleration is $3.2 \times 10^{4} \ mi/h^2$ .

3 0

3 years ago