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1 year ago

Use the information from the graph to answer the question.

Physics

1 answer:

galina1969 [7]1 year ago

5 0

The displacement of the object as determined from the velocity-time graph is 562.5 m.

<h3>What is a velocity-time graph?</h3>

A velocity-time graph is a graph of the velocity of an object plotted in the vertical or y-axis of the graph against the time taken on the horizontal or x-axis.

The displacement of an object can be obtained from its velocity-time graph by calculating the total area under the graph.

The total area under the graph = area of triangle + area of rectangle

Area of triangle = b*h/2 =

Area of triangle = 25 * (35 - 10)/2 = 312.5 m

Area of rectangle = l * b

Area of rectangle = 10 * 25 = 250 m

Total area = (312.5 + 250) m

Total area = 562.5 m

Therefore, the displacement of the object is 562.5 m

In conclusion, the total area of a velocity-time graph gives the displacement.

Learn more about velocity-time graph at: brainly.com/question/28064297

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A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of

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(C) is correct option.

Explanation:

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We need to calculate the time

Using equation of motion

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3 years ago

Consider a uniform horizontal wooden board that acts as a pedestrian bridge. The bridge has a mass of 300 kg and a length of 10

gayaneshka [121]

Answer:

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Explanation:

In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.

$\Sigma \tau=0$

Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.

Then, you obtain the following formula:

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d: distance to the pivot of the torque, in this case, distance to the right support.

You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:

$-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N$

Where d1, d2 and d3 are distance to the right support.

You solve the equation for F and replace the values of the other parameters:

$F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N$

The force applied by the left support is 2123.33 N

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