Can some one help me with this so i can bring my grade up

Pavel [41]

The answer is D. If you aren't consistent with your drop positions, then your data may be invalid. To be frank: it basically screws over the experiment.

5 0

3 years ago

How many photons with 10 ev are required to produce 20 joules of energy?

oee [108]

One electron Volt (eV) is equal to 1.6 x 10^-19 Joules. Therefore, 10 eV is equal to 1.6 x 10^-18 Joules. In order to produce 20 Joules of energy from 10 eV photons, we would require 20 x 1/(1.6 x 10^-18) = 1.25 x 10^19 particles. This demonstrates that in the world of particle physics, the Joule is a massive energy unit relative to the commonly used electron Volt.

6 0

3 years ago

A batter hits a pop fly, and the baseball (with a mass of 148 g) reaches an altitude of 265 ft. If we assume that the ball was 3

den301095 [7]

Answer:

The increase in potential energy of the ball is 115.82 J

Explanation:

Conceptual analysis

Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:

U = m × g × h

U: Potential Energy in Joules (J)

m: mass in kg

g: acceleration due to gravity in m/s²

h: height in m

Equivalences

1 kg = 1000 g

1 ft = 0.3048 m

1 N = 1 (kg×m)/s²

1 J = N × m

Known data

$h_2 = 265ft * \frac{0.3048m}{ft} = 80.77m$

$h_1 = 3ft * \frac{0.3048m}{ft} = 0.914m$

$m = 148g*\frac{1kg}{1000g} = 0.148kg$

$g = 9.8 \frac{m}{s^2}$

Problem development

ΔU: Potential energy change

ΔU = U₂ - U₁

U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁

U₂ - U₁ = mₓg(h₂ - h₁)

$U_2 - U_1 = 0.148kg * 9.8 \frac{m}{s^2}*(80.77m - 0.914m) = 115.82 N * m = 115.82J$

The increase in potential energy of the ball is 115.82 J

5 0

2 years ago

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

The kinetic energy of a 1000-kg roller coaster car that is moving with a speed of 20.0 m/s.

eimsori [14]

E=(mV^2)/2
m=1000kg, V=20m/s
then, E=(1000kg*(20m/s)^2)/2
E=(1000*400)/2 J = 200000J

8 0

3 years ago