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dusya [7]
3 years ago
12

Yasmin determines the velocity of a car to be 15m/s. The accepted value for the velocity of the car is 15.6m/s. What is yasmins

percent error ?
Physics
2 answers:
zaharov [31]3 years ago
7 0

Percent error is calculated as follows:

% = ( |15-15.6| / 15.6 ) *  100%

% = (0.6/15.6) * 100%

% = 0.0385 * 100%

% = 3.85%

Hope this helped!

densk [106]3 years ago
6 0

For this case we have that the percentage of error is given by the following equation:

percent\ error= \frac {|v_ {a} -v_ {e}|} {v_ {e}} * 100

Where:

v_ {a}: Approximate (measured) value: 15

v_ {e}:Exact value: 15.6

Substituting:

percent\ error= \frac {| 15-15.6 |} {15.6} * 100\\percent\ error = \frac {0.6} {15.6} * 100\\% error = 3.85

Thus, the percentage of error is 3.85%

Answer:

3.85%

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vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses
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Answer:

The effective spring constant of the firing mechanism is 1808N/m.

Explanation:

First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

v_0_x=\frac{x}{t}\\ \\v_0\cos\theta=\frac{x}{t}\\\\v_0=\frac{x}{t\cos\theta}

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v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

E_0=E_f\\\\U_e=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\\implies k=\frac{mv^2}{x^2}

Then, plugging in the given values, we obtain:

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Finally, the effective spring constant of the firing mechanism is 1808N/m.

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