A graph titled energy versus velocity is shown with energy on the vertical axis and velocity on the horizontal axis. The graph is a curved line opened up from bottom left to top right.

Which of the following best describes the relationship between the two variables?
A. Velocity is inversely related to energy.
B. Energy is inversely related to the square of velocity.
C. Velocity is directly related to energy.
D. Energy is directly related to the square of velocity.

Answer:

Explanation:

The answer is **(c) 9.39 m/s** for the velocity of the bullet-block system after it's hit, and **(g) 209.39 m/s** for the bullet's velocity before it hit the box.

The velocity of the bullet-block system after it's hit can be found using the conservation of energy. The potential energy of the box before it was hit is mgh, where m is the mass of the box, g is the acceleration due to gravity, and h is the height that the box was displaced. After the bullet hits the box, the potential energy of the box is zero, but the kinetic energy of the bullet-block system is mv^2/2, where m is the total mass of the bullet-block system and v is the velocity of the bullet-block system. Setting these two expressions equal to each other, we get:

```

mgh = mv^2/2

```

Solving for v, we get:

```

v = sqrt(2mgh)

```

In this case, we have:

* m = 0.05 kg + 1.3 kg = 1.35 kg

* g = 9.8 m/s^2

* h = 4.5 m

So, the velocity of the bullet-block system after it's hit is:

```

v = sqrt(2 * 1.35 kg * 9.8 m/s^2 * 4.5 m) = 9.39 m/s

```

The velocity of the bullet before it hit the box can be found using the conservation of momentum. The momentum of the bullet before it hit the box is mv, where m is the mass of the bullet and v is the velocity of the bullet. After the bullet hits the box, the momentum of the bullet-block system is (m + M)v, where M is the mass of the box and v is the velocity of the bullet-block system. Setting these two expressions equal to each other, we get:

```

mv = (m + M)v

```

Solving for v, we get:

```

v = mv/(m + M)

```

In this case, we have:

* m = 0.05 kg

* M = 1.3 kg

* v = 9.39 m/s

So, the velocity of the bullet before it hit the box is:

```

v = 0.05 kg * 9.39 m/s / (0.05 kg + 1.3 kg) = 209.39 m/s

```

The velocity of the bullet-block system after the collision is approximately a) 6.76 m/s, and the bullet's velocity before it hit the box is approximately e) 196.76 m/s.

To solve this problem, we can apply the principle of conservation of momentum and the principle of conservation of mechanical energy.

First, let's calculate the velocity of the bullet-block system after the collision. We can use the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

Let m1 be the mass of the bullet (0.05 kg) and m2 be the mass of the box (1.3 kg). Let v1 be the velocity of the bullet before the collision (which we need to find) and v2 be the velocity of the bullet-block system after the collision.

Using the conservation of momentum:

m1 * v1 = (m1 + m2) * v2

0.05 kg * v1 = (0.05 kg + 1.3 kg) * v2

0.05 kg * v1 = 1.35 kg * v2

Now, let's calculate the velocity of the bullet-block system (v2). Since the system goes up by a height of 4.5 m, we can use the principle of conservation of mechanical energy.

m1 * v1^2 = (m1 + m2) * v2^2 + m2 * g * h

0.05 kg * v1^2 = 1.35 kg * v2^2 + 1.3 kg * 9.8 m/s^2 * 4.5 m

Now, we can substitute the value of v2 from the momentum equation into the energy equation and solve for v1.

By solving these equations, we find that v1 is approximately 196.76 m/s.

Therefore, the bullet's velocity before it hit the box is approximately 196.76 m/s. (e) 196.76 m/s

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Answer:

3400 K

Explanation:

use wiens displacement law

λ_max = 853 nm = 853 × 10^−9 m

T = b / λ_max

substiute

T = (2.898 × 10^−3 m·K) / (853 × 10^−9 m)

T = 2.898 × 10^−3 / 853 = 3.4 × 10^3 K

The resistance of a platinum resistance thermometer is 200 Ω at 0°C and 255.8 Ω at the melting point. Using the resistance-temperature relationship and calculations, the estimated melting point is approximately 19.93°C.

The resistance of a platinum resistance thermometer is 200 Ω when placed in a 0°C ice bath and 255.8 Ω when immersed in a crucible containing a melting substance. To determine the melting point of the substance, we need to calculate the temperature at which the resistance reaches 255.8 Ω.

First, we find the temperature coefficient of resistance (α) using the formula α = (R - R₀) / (R₀ * T), where R is the resistance at the melting point, R₀ is the resistance at 0°C, and T is the temperature at the melting point.

Substituting the given values, we have α = (255.8 - 200) / (200 * T₀), where T₀ is the known room temperature of 20°C.

Calculating α, we find α ≈ 0.014.

Next, we use the resistance-temperature relationship equation R = R₀(1 + αT) to solve for the melting point temperature (T). Substituting the known values, we have 255.8 = 200(1 + 0.014 * T).

Simplifying the equation, we find 1.279 = 1 + 0.014T.

Solving for T, we get T ≈ 19.93°C.

Therefore, based on the given data, the estimated melting point of the substance is approximately 19.93°C.

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The displacement needed to get the ball into the hole on the first stroke is approximately 9.8731 m at an angle of 12.01° east of north

To calculate the displacement needed to get the ball into the hole on the first stroke, we need to consider the vector components of each stroke and combine them to find the resultant displacement.

First, we break down the second stroke of 3.0 m southeast into its north and east components. Since it's at a 45° angle, both components will have the same magnitude of 3.0 m.

North component = 3.0 m * cos(45°) = 3.0 m * 0.7071 ≈ 2.1213 m

East component = 3.0 m * sin(45°) = 3.0 m * 0.7071 ≈ 2.1213 m

Next, we break down the third stroke of 2.0 m southwest into its north and west components. Since it's at a 45° angle, both components will have the same magnitude of 2.0 m.

North component = 2.0 m * cos(45°) = 2.0 m * 0.7071 ≈ 1.4142 m

West component = 2.0 m * sin(45°) = 2.0 m * 0.7071 ≈ 1.4142 m

Now, we can sum up the north and east components from the first, second, and third strokes:

North displacement = 6.0 m + 2.1213 m + 1.4142 m = 9.5355 m

East displacement = 0 m + 2.1213 m - 0 m = 2.1213 m

Finally, we can calculate the magnitude and direction of the resultant displacement using the Pythagorean theorem and trigonometry:

Resultant displacement = √(North displacement^2 + East displacement^2) ≈ √(9.5355 m^2 + 2.1213 m^2) ≈ 9.8731 m

The direction can be found using the tangent function:

Direction = arctan(East displacement / North displacement) ≈ arctan(2.1213 m / 9.5355 m) ≈ 12.01°

Therefore, the displacement needed to get the ball into the hole on the first stroke is approximately 9.8731 m at an angle of 12.01° east of north.

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Answer:

B. Selective Pressures