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2 years ago

Is knocking dominoes newton's 1st law ?

Physics

1 answer:

myrzilka [38]2 years ago

3 0

No because Newton’s first law is inertia

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consider the photoelectric effect experiment. in one experiment yellow light shines on a piece of potassium metal. a current is

kompoz [17]

Explanation:

If the intensity of the yellow light increased, meaning more photons will strike the Potassium metal per unit area. This will cause more ejection of electrons from the metal and hence, the strength of current will also increase as we know that

I = Q/t, as the charge increase , the current will also increase.

4 0

3 years ago

2. A 55 kg woman has a momentum of 200 kg m/s. What is her velocity?

NeTakaya

Answer:

$\boxed {\tt 3.63636364 \ m/s}$

Explanation:

Velocity can be found using the following formula:

$v=\frac{p}{m}$

where p is the momentum and m is the mass.

The woman has a mass of 55 kilograms and a momentum of 200 kilogram meters per second.

$p= 200 \ kgm/s\\m=55 \ kg$

Substitute the values into the formula.

$v=\frac{200 \ kg m/s}{55 \ kg}$

Divide. Note that the kilograms, or kg, will cancel each other out.

$v=\frac{200 \ m/s}{55}$

$v= 3.63636364 \ m/s$

The woman's velocity is 3.63636364 meters per second.

6 0

3 years ago

Two bikers are riding a circular

OverLord2011 [107]

ANSWER.no cause they have to be going a the same speed and I doubt that are gonna go at the same speed.

8 0

3 years ago

Here is more points drip students - drip king

levacccp [35]

Answer
Thanks
Explanation

4 0

3 years ago

Read 2 more answers

How do i solve the ones in red? A student fires a cannonball horizontally with a speed of 24.0m/s from a height of 51.0m. Neglec

Rus_ich [418]

Horizontal speed = 24.0 m/s

height of the cliff = 51.0 m

For the initial vertical speed will are considering the vertical component. Therefore,

Since the student fires the canonical ball at the maximum height of 51 m, the initial vertical velocity will be zero. This means

$\text{ Initial vertical velocity = 0 m/s}$

let's find how long the ball remained in the air.

$\begin{gathered} 0=51-\frac{1}{2}(9.8)t^2 \\ 4.9t^2=51 \\ t^2=\frac{51}{4.9} \\ t^2=10.4081632653 \\ t=\sqrt[]{10.4081632653} \\ t=3.22 \\ t=3.22\text{ s} \end{gathered}$

Finally, let's find the how far from the base of the building the ball landed(horizontal distance)

$\begin{gathered} S_x=u_xt+\frac{1}{2}a_xt^2 \\ S_x=24\times3.23 \\ S_x=\text{horizontal distance=77.28 m} \\ \text{note} \\ a_x=0m/s^2 \end{gathered}$

6 0

1 year ago