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FromTheMoon [43]
4 years ago
10

How long did the trip from camp wood to the pacific ocean and back again take?

Physics
1 answer:
maria [59]4 years ago
6 0

The trip from Camp Wood to the Pacific Ocean and back again took 1.5 years to complete.<span>

</span>

<span>The </span>Lewis<span> and Clark </span>Expedition<span> from May 1804 to September 1806, also known as the Corps of Discovery </span>Expedition, was the first American expedition<span> to cross what is now the western portion of the United States.</span>
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I would say B but I’m not 100%
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3 years ago
two children having a disagreement pull on a 50N sled in opposite directions. one pulls with a force of 300N east, the other wit
prohojiy [21]
<span>Since forces are vector quantities, we must indicate direction using positive and negative values. East will be assigned positive and west will be negative. Friction will act as a negative force since it impedes action. To calculate the net force we sum the vector quantities, as follows. Net force equals 50n which is derived by the following calculation: 300n-220n-30n.</span>
7 0
3 years ago
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An athlete is working out in the weight room. He steadily holds 50 kilograms above his head for 10 seconds. Which statement is t
postnew [5]

Answer:

A. The athlete isn’t doing any work because he doesn’t move the weight.

Explanation:

We must remember the definition of work, which says that work is equal to the product of mass by the distance displaced. In this case, the athlete only does work when he lifts the weight from the ground to the point where he holds the weight suspended.

So when he's holding the weight, he doesn't do any work.

3 0
3 years ago
A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal
dsp73

Answer:

v\approx 8.570\,\frac{m}{s}

Explanation:

The equation of equlibrium for the box is:

\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a

The formula for the acceleration, given in \frac{m}{s^{2}}, is:

a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}

Velocity can be derived from the following definition of acceleration:

a = v\cdot \frac{dv}{dx}

v\, dv = a\, dx

\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx

\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg}  \int\limits^{17\,m}_{0\,m}\, dx  - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx

\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}

v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}]  }

The speed after the box has travelled 17 meters is:

v\approx 8.570\,\frac{m}{s}

3 0
3 years ago
a fan acquires a speed of 180 rpm in 4s, starting from rest. calculate the speed of the fan at the end of the 5th second startin
KengaRu [80]

Answer:

225 rpm

Explanation:

The angular acceleration of the fan is given by:

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

where

\omega_f is the final angular speed

\omega_i is the initial angular speed

\Delta t is the time interval

For the fan in this problem,

\omega_i = 0\\\omega_f = 180 rpm\\\Delta t=4 s

Substituting,

\alpha = \frac{180-0}{4}=45 rpm/s

Now we can find the angular speed of the fan at the end of the 5th second, so after t = 5 s. It is given by:

\omega' = \omega_i + \alpha t

where

\omega_i = 0\\\alpha = 45 rpm/s\\t = 5 s

Substituting,

\omega' = 0 + (45)(5)=225 rpm

7 0
3 years ago
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