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yan [13]
3 years ago
12

What differentiates baseline actives from health enhancing activities?

Physics
1 answer:
german3 years ago
7 0
A. The difference lies in the intensity of the activity
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What kind of symmetry do you have
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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:

F = 2123.33N

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:

-\tau_l+\tau_p+\tau_{cm}=0          (1)

τl: torque produced by the left support

τp: torque produced by the person

τcm: torque produced by the center of mass of the wooden

The torque is given by:

\tau=Fd           (2)

F: force applied

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

8 0
3 years ago
Two spheres, 1.00 kg each, whose centers are 2.00 m apart, would have what gravitational force between them? A. 3.14 X 10-17 N
Angelina_Jolie [31]

Answer: B

Explanation: the teacher just told us the answer

4 0
2 years ago
A boat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, space, start fraction, m, divided by, s,
zepelin [54]

Answer:

d=42m(-\hat{i})

Explanation:

For this problem, we need to apply the formulas of constant accelerated motion.

To obtain the boat displacement we need to calculate the displacement because of the river flow and the displacement done because of the boat motor.

for the river:

d_r=v*t\\d_r=5m/s*6s\\d_r=30m(\hat{i})

for the boat:

x=\frac{1}{2}*a*t^2\\\\x=\frac{1}{2}*4.0m/s*(6s)^2\\\\\\x=72m(-\hat{i})

So the final displacement is given by:

d=dr+x\\d=30m-72m\\d=42m(-\hat{i})

8 0
3 years ago
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