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2 years ago

How do you decrease the reaction force felt on the egg if its change in momentum, and therefore total impulse, stays

Physics

1 answer:

Ivenika [448]2 years ago

3 0

Answer:

Hey, how you doing.

I think (key word think, i haven't done this test) that you would make the egg smaler and make it go faster, so that the atomic bonds are stronger.

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Two identical freight cars roll without friction (one at 1 m/s, the other at 2 m/s) toward each other on a level track. They col

Kazeer [188]

Answer:

They collide, couple together, and roll away in the direction that <u>the 2m/s car was rolling in.</u>

Explanation:

We should start off with stating that the conservation of momentum is used here.

Momentum = mass * speed

Since, mass of both freight cars is the same, the speed determines which has more momentum.

Thus, the momentum of the 2 m/s freight car is twice that of the 1 m/s freight car.

The final speed is calculated as below:

mass * (velocity of first freight car) + mass * (velocity of second freight car) = (mass of both freight cars) * final velocity

(m * V1) + (m * V2) = (2m * V)

Let's substitute the velocities 1m/s for the first car, and - 2m/s for the second. (since the second is opposite in direction)

We get:

$m*1 + m*(-2) = 2m*V$

solving this we get:

V = - 0.5 m/s

Thus we can see that both cars will roll away in the direction that the 2 m/s car was going in. (because of the negative sign in the answer)

7 0

3 years ago

What is being described in the following definition: An opening in the floor, platform, or pavement that measures 12 inches or m

azamat

Answer is Floor Opening

3 0

2 years ago

In the photoelectric effect, a photon with an energy of 5.3 × 10–19 J strikes an electron in a metal. Of this energy, 3.6 × 10–1

valentina_108 [34]

Answer:

The velocity of the photo electron is $6.11\times 10^5\ m/s$ .

Explanation:

Given that,

Supplied energy, $E_s=5.3\times 10^{-19}\ J$

Minimum energy of the electron to escape from the metal, $E_e=3.6\times 10^{-19}\ J$

We need to find the velocity of the photo electron. The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So,

$5.3\times 10^{-19}\ J=3.6\times 10^{-19}\ J+K\\\\K=5.3\times 10^{-19}-3.6\times 10^{-19}\\\\K=1.7\times 10^{-19}\ J$

The formula of kinetic energy is given by :

$K=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2K}{m}} \\\\v=\sqrt{\dfrac{2\times 1.7\times 10^{-19}}{9.1\times 10^{-31}}} \\\\v=6.11\times 10^5\ m/s$

So, the velocity of the photo electron is $6.11\times 10^5\ m/s$ .

4 0

3 years ago

A 10 kg package is delivered to your house.

kolbaska11 [484]

In ur explanation make sure to use the terms

7 0

2 years ago

A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at

stellarik [79]

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

$\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft$

Also, at any instant t

$\Rightarrow l^2=x^2+y^2$

differentiate w.r.t.

$\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s$

5 0

3 years ago