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ki77a [65]
3 years ago
15

1000 lb boulder B is resting on a 500 lb platform A when truck C accidentally accelerates to the right (truck in reverse). Which

of the following statements are true (select two answers)?
a. The tension in the cord connected to the truck is 200 lb.
b. The tension in the cord connected to the truck is 1200 lb.
c. The tension in the cord connected to the truck is greater than 1200 lb.
d. The normal force between A and B is 1000 lb.
e. The normal force between A and B is 1200 lb.
f. None of the above are true.
Engineering
1 answer:
Setler79 [48]3 years ago
4 0

Answer:

<em>c. The tension in the cord connected to the truck is greater than 1200 lb</em>

<em>e. The normal force between A and B is 1200 lb.</em>

Explanation:

The correct question should be

A 1000 lb boulder B is resting on a 200 lb platform A when truck C accidentally accelerates to the right (truck in reverse). Which of the following statements are true (select two answers)?

A free body diagram is shown below.

The normal force between the the boulder and the platform will be the sum of the force, i.e 1000 lb + 200 lb = 1200 lb

For the combination of the bodies to accelerate upwards, then the tension must be greater than the normal force, i.e T > 1200 lb

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3 years ago
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3 years ago
A student takes 60 voltages readings across a resistor and finds a mean voltage of 2.501V with a sample standard deviation of 0.
Lena [83]

Answer:

There are 2 expected readings greater than 2.70 V

Solution:

As per the question:

Total no. of readings, n = 60 V

Mean of the voltage, \mu = 2.501 V

standard deviation, \sigma = 0.113 V

Now, to find the no. of readings greater than 2.70 V, we find:

The probability of the readings less than 2.70 V, P(X\leq 2.70):

z = \frac{x - \mu}{\sigma} = \frac{2.70 - 2.501}{0.113} = 1.761

Now, from the Probability table of standard normal distribution:

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Now, for the expected no. of readings greater than 2.70 V:

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No. of readings expected to be greater than 2.70 V = P(X\geq 2.70)\times Total\ no.\ of\ readings

No. of readings expected to be greater than 2.70 V = 0.0392\times 60 = 2.352 ≈ 2

7 0
3 years ago
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