3 years ago

7) The acceleration due to gravity near the surface of Mars is about one-third of the value

Physics

1 answer:

Doss [256]3 years ago

8 0

Answer:

C) three times slow than on earth

You might be interested in

In the mixtures lab, you waited to see if the liquid would form lumpy or fluffy masses, which would indicate it was a _____.

Over [174]

B) colloid because i took the test and that the answer

5 0

3 years ago

Read 2 more answers

Michelle recently started selling her invention: A bed that looks like it floats in mid-air. The bed is actually suspended by ma

Tamiku [17]

Answer:

Explanation:

designer

illusionist

engineer

entrepreneur

salesperson

human

inventor

8 0

3 years ago

Where can classic examples of shield volcanoes be found?

Darina [25.2K]

The largest is Mauna Loa on the Big Island of Hawaii; all the volcanoes in the Hawaiian Islands are shield volcanoes. There are also shield volcanoes, for example, in Washington, Oregon, and the Galapagos Islands

4 0

3 years ago

The lower the frequency the _____ the pitch sound.

stepan [7]

The lower the frequency the lower the pitch sound.

8 0

3 years ago

Read 2 more answers

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