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3 years ago

According to the Declaration of Independence, all people are created in what manner?

Physics

2 answers:

FinnZ [79.3K]3 years ago

7 0

Answer:

All men are created equal.

Explanation:

We hold these truths to be self-evident, that all men are created equal, that they are endowed by their Creator with certain unalienable Rights, that among these are Life, Liberty and the pursuit of Happiness.

-Declaration of Independence

djyliett [7]3 years ago

5 0

The second paragraph of the United States Declaration of Independence starts as follows: "We hold these truths to be self-evident, that all men are created equal, that they are endowed by their Creator with certain unalienable Rights, that among these are Life, Liberty and the Pursuit of Happiness.

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If you add two batteries to a circuit, the resistance will

aksik [14]

<span>If you add two batteries to a circuit, the resistance will stay the same.

Unless you're working with toasters, electric stoves, or high-power
floodlights, the circuit voltage has no effect on the total resistance.</span>

7 0

4 years ago

train travels 1/4of its total distance over half of its time with a constant velocity of 5m/s. calculate the average velocity

Katarina [22]

Let D be the total distance (say in meters) traveled by the train and T the time (say in seconds) it takes to do so. (Assume the train moves in a straight line in only one direction.) Then the average velocity of the train as it covers this distance is

v (ave) = D/T

We're told the train can traverse a distance of D/4 in a matter of T/2 seconds if it moves at a speed of 5 m/s. This means

D/4 = (5 m/s) (T/2)

⇒ 5 m/s = 1/2 D/T

⇒ v (ave) = D/T = 10 m/s

3 0

3 years ago

A 100 g ball collides elastically with a 300 g ball that is at rest. If the 100 g ball was traveling

sammy [17]

Answer:

The magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

Explanation:

We can find the velocity of the two balls after the collision by conservation of linear momentum and energy:

$P_{1} = P_{2}$

$m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

Where:

m₁: is the mass of the ball 1 = 100 g = 0.1 kg

m₂: is the mass of the ball 2 = 300 g = 0.3 kg

$v_{1_{i}}$ : is the initial velocity of the ball 1 = 6.20 m/s

$v_{2_{i}}$ : is the initial velocity of the ball 2 = 0 (it is at rest)

$v_{1_{f}}$ : is the final velocity of the ball 1 =?

$v_{2_{f}}$ : is the initial velocity of the ball 2 =?

$m_{1}v_{1_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

$v_{1_{f}} = v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}}$ (1)

Now, by conservation of kinetic energy (since they collide elastically):

$\frac{1}{2}m_{1}v_{1_{i}}^{2} = \frac{1}{2}m_{1}v_{1_{f}}^{2} + \frac{1}{2}m_{2}v_{2_{f}}^{2}$

$m_{1}v_{1_{i}}^{2} = m_{1}v_{1_{f}}^{2} + m_{2}v_{2_{f}}^{2}$ (2)

By entering equation (1) into (2) we have:

$m_{1}v_{1_{i}}^{2} = m_{1}(v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}})^{2} + m_{2}v_{2_{f}}^{2}$

$0.1 kg*(6.20 m/s)^{2} = 0.1 kg*(6.2 m/s - \frac{0.3 kg*v_{2_{f}}}{0.1 kg})^{2} + 0.3 kg(v_{2_{f}})^{2}$

By solving the above equation for $v_{2_{f}}$ :

$v_{2_{f}} = 3.1 m/s$

Now, $v_{1_{f}}$ can be calculated with equation (1):

$v_{1_{f}} = 6.20 m/s - \frac{0.3 kg*3.1 m/s}{0.1 kg} = -3.1 m/s$

The minus sign of $v_{1_{f}}$ means that the ball 1 (100g) is moving in the negative x-direction after the collision.

Therefore, the magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

I hope it helps you!

5 0

3 years ago

3. Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If

Daniel [21]

Answer:

Object 2 has the larger drag coefficient

Explanation:

The drag force, D, is given by the equation:

$D = 0.5 c \rho A v^2$

Object 1 has twice the diameter of object 2.

If $d_2 = d$

$d_1 = 2d$

Area of object 2, $A_2 = \frac{\pi d^2 }{4}$

Area of object 1:

$A_1 = \frac{\pi (2d)^2 }{4}\\A_1 = \pi d^2$

Since all other parameters are still the same except the drag coefficient:

For object 1:

$D = 0.5 c_1 \rho A_1 v^2\\D = 0.5 c_1 \rho (\pi d^2) v^2$

For object 2:

$D = 0.5 c_2 \rho A_2 v^2\\D = 0.5 c_2 \rho (\pi d^2/4) v^2$

Since the drag force for the two objects are the same:

$0.5 c_1 \rho (\pi d^2) v^2 = 0.5 c_2 \rho (\pi d^2/4) v^2\\4c_1 = c_2$

Obviously from the equation above, c₂ is larger than c₁, this means that object 2 has the larger drag coefficient

8 0

3 years ago

A tuning fork vibrates at a frequency of 512 hertz

Naily [24]

Answer: The correct answer is " longitudinal wave with air molecules

vibrating parallel to the direction of travel".

Explanation:

In longitudinal wave, the particles vibrate parallel to the direction to the propagation of the wave. For example, sound wave is a longitudinal wave.

It needs a medium for its propagation. It can travel in solid, liquid and gas. It travel faster in solid in comparison to the liquid.

In the given problem, a tuning fork vibrates at a frequency of 512 hertz when struck with a rubber hammer. The sound produced by the tuning fork will travel through air. Here, a longitudinal wave with air molecules vibrating parallel to the direction of travel.

Therefore, the correct option is (1).

4 0

3 years ago

Read 2 more answers