A bottle rocket takes off with a = 34.5 m/s2. It

Hydrochloric acid and sodium hydroxide can be combined in equal ratios to form sodium chloride and water. Which best explains th

Oksanka [162]

It's an acid-base type of redox reaction, in which neutralization occurs between the acid component [H+] of HCl and the base component [OH-] of NaOH. Equal parts of acid + base form H2O and the metal and halide combine to form neutral salt (NaCl).

7 0

3 years ago

In a study conducted by a University of Illinois researcher, the football team at Unity High School in Tolono, IL was equipped f

Angelina_Jolie [31]

The duration (i.e time) of the impact, given that the head impact was 1770 N is 0.005 s

<h3>How do I determine the duration (i.e time)?</h3>

Impulse is defined as the change in momentum of an object. It is expressed as:

Impulse = change in momentum

Impulse = final momentum – Initial momentum

Impule = m(v - u)

Impulse = force × time

Impulse = Ft

Thus,

Ft = m(v - u)

Where

F is the force
t is the time
m is the mass
u is the initial velocity
v is the final velocity

Now, we can obtain the duration (i.e time) of the impact. Details below:

Force (F) = 1770 N
Initial velocity = 2.09 m/s
Mass (m) = 4.12 Kg
Final velocity = 0 m/s
Duration (t) =?

Ft = m(v + u) since the collision came to a stop

1770 × t = 4.12 × (0 + 2.09)

1770 × t = 4.12 × 2.09

Divide both sides by 1770

t = (4.12 × 2.09) / 1770

t = 0.005 s

Thus, the duration is 0.005 s

Learn more about time:

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6 0

1 year ago

The lung volume that represents the maximum amount of exchangeable air is the _____

Savatey [412]

The lung volume that represents the maximum amount of exchangeable air is the: <u>vital capacity</u>

<u></u>

The vital capacity is the volume of air that can be taken into the lungs by inhaling and expelling air by inspiration and expiration.

<h3>What are the lungs? </h3>

The lungs are one of the largest organs of the body and are in charge of the respiratory function. They have a spongy appearance and are located in the thorax, protected by the ribs, one on each side of the heart.

Learn more about the lungs at: brainly.com/question/27010145

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5 0

1 year ago

explain how potential and kinetic energy are at play when we talk about Newton's second law of motion

KIM [24]

Potential and kinetic energy are at play when we talk about Newton's second law of motion through the various positions in relation to the bodies involved.

<h3>What is Newton's second law of motion?</h3>

This law states that force is equal to the rate of change of momentum and is denoted as F = mv where m is mass and v is velocity.

Potential energy is the energy is possessed by a body by virtue of its position while kinetic energy is possessed by a body by virtue of its motion. Both forms of energy are influenced by forces and are equal to the total momentum.

Read more about Newton's second law of motion here brainly.com/question/2009830

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7 0

1 year ago

A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th

Scorpion4ik [409]

1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k = 7 N/m is the spring constant

m = 1 kg is the mass attached to the spring

Substituting these numbers into the formula, we find

$f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz$

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

$T=\frac{1}{0.42 Hz}=2.38 s$

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: $x=0$ , so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

$E=K=\frac{1}{2}mv^2$

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$

Since the total energy must be conserved, we have:

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m$

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}kA^2$ is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m$

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}mv_{max}^2$ is the mechanical energy of the system when x=0, which is when the system has maximum velocity, $v_{max}$

Equalizing the two expressions, we can solve to find $v_{max}$ , the maximum velocity:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m$

4 0

3 years ago

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