Please help I just need help for question 4. And 5. (They’re at the top)

Physics

1 answer:

olya-2409 [2.1K]4 years ago

7 0

Explanation:

4.he invented the atomic model with all different electrons proton and neutrons

5. this was important as this was developed and is still used in the modern day

You might be interested in

During which strokes does the piston move downward in a four-stroke internal combustion engine?

yuradex [85]

I believe that it is the third stroke

3 0

4 years ago

Read 2 more answers

If a ping pong ball and a golf ball are both moving in the same direction with the same amount of kinetic energy, the speed of t

Liono4ka [1.6K]

If the kinetic energy of each ball is equal to that of the other,
then

(1/2) (mass of ppb) (speed of ppb)² = (1/2) (mass of gb) (speed of gb)²

Multiply each side by 2:

(mass of ppb) (speed of ppb)² = (mass of gb) (speed of gb)²

Divide each side by (mass of gb) and by (speed of ppb)² :

(mass of ppb)/(mass of gb) = (speed of gb)²/(speed of ppb)²

Take square root of each side:

√ (ratio of their masses) = ( 1 / ratio of their speeds)²

By trying to do this perfectly rigorously and elegantly, I'm also
using up a lot of space and guaranteeing that nobody will be
able to follow what I have written. Let's just come in from the
cold, and say it the clear, easy way:

If their kinetic energies are equal, then the product of each
mass and its speed² must be the same number.

If one ball has less mass than the other one, then the speed²
of the lighter one must be greater than the speed² of the heavier
one, in order to keep the products equal.

The pingpong ball is moving faster than the golf ball.

The directions of their motions are irrelevant.

5 0

3 years ago

If it takes a ball dropped from rest 2.261 s to fall to the ground, from what height H was it released? Express your answer in m

algol [13]

Answer:

Height, H = 25.04 meters

Explanation:

Initially the ball is at rest, u = 0

Time taken to fall to the ground, t = 2.261 s

Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :

$H=ut+\dfrac{1}{2}at^2$

Here, a = g

$H=\dfrac{1}{2}gt^2$

$H=\dfrac{1}{2}\times 9.8\times (2.261)^2$

H = 25.04 meters

So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.

7 0

3 years ago

A box of mass m1 = 20.0 kg is released from rest at a warehouse loading dock and slides down a 3.0 m-high frictionless chute to

Harrizon [31]

´To develop this problem we will use the concepts related to the conservation of momentum and the application of energy conservation equations to find the velocity of the mass after the collision, like this:

Velocity of the mass $m_1$ just before the collision