A fountain of water and steam that erupts from the ground

When measuring the diameter of a cylinder the following measurements are obtained.

Arlecino [84]

Answer:

a. 1.80 %

b. 0.60 %

c. 1.20 %

d. 0.60 %

e. 0.60 %

Explanation:

In order to calculate percent uncertainties in each case, we forst need to calculate the average value:

Average Diameter = (16.4 mm + 16.8 mm + 16.9 mm + 16.6 mm +16.8 mm)/5

Average Diameter = 16.7 mm

Now, the formula for percent uncertainty is:

% Uncertainty = (Uncertainty/Average) * 100 %

where,

Uncertainty = |Value - Average|

For Each Case:

a. 16.4 mm:

Uncertainty = |16.4 mm - 16.7 mm| = 0.3 mm

Therefore,

% Uncertainty = (0.3 mm/16.7 mm) * 100%

% Uncertainty = 1.80 %

b. 16.8 mm:

Uncertainty = |16.8 mm - 16.7 mm| = 0.1 mm

Therefore,

% Uncertainty = (0.1 mm/16.7 mm) * 100%

% Uncertainty = 0.60 %

c. 16.4 mm:

Uncertainty = |16.9 mm - 16.7 mm| = 0.2 mm

Therefore,

% Uncertainty = (0.2 mm/16.7 mm) * 100%

% Uncertainty = 1.20 %

d. 16.6 mm:

Uncertainty = |16.6 mm - 16.7 mm| = 0.1 mm

Therefore,

% Uncertainty = (0.1 mm/16.7 mm) * 100%

% Uncertainty = 0.60 %

e. 16.8 mm:

Uncertainty = |16.8 mm - 16.7 mm| = 0.3 mm

Therefore,

% Uncertainty = (0.1 mm/16.7 mm) * 100%

% Uncertainty = 0.60 %

7 0

3 years ago

Define work What is the unit for work? For Physics, thanks.

Minchanka [31]

Answer:

Work, in physics, measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement

Explanation:

5 0

3 years ago

What is carried by a wave?

Nata [24]

Waves carry energy from one place to another. Because waves carry energy, some waves are used for communication, eg radio and television waves and mobile telephone signals. ... Some types of waves need to be transmitted through matter, either a solid, liquid or a gas. For example, water waves have to travel in water.

3 0

3 years ago

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GIVING BRAINLIEST PLEASE HELP!!

Marrrta [24]

The answer is D hope it helps

8 0

3 years 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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