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3 years ago

Sketch a position-time graph for a bear starting

Physics

1 answer:

Dmitrij [34]3 years ago

3 0

Explanation:

hopefully that makes sense. the position doesn't change over the 5 seconds, meaning it's stopped but time still continues. then when the slope is negative this shows the bear's position becoming negative (backing up, changing direction).

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Question 2 The gravitational force between two objects with identical masses that are 10 m apart, is 2.67 x10-10 N. To the neare

xxMikexx [17]

Answer 888990,0 kg

Explanation:

3 0

3 years ago

Once juno reaches jupiter, what is the minimum amount of time it takes for the transmitted signals to travel from the spacecraft

Komok [63]

In very very very round figures . . .

-- Jupiter is about 5.2 times as far from the sun as the earth is.

-- So when Jupiter and the EARTH are aligned in both orbits, Jupiter is about

(4.2) x (150 million kilometers) = 630 million kilometers

Time = (distance) / (speed)

The speed of light and radio is 300,000 km/second

Time = (630 million / 300 thousand)

<em>Time = 2,100 seconds</em>

That's 35 minutes.

4 0

3 years ago

Read 2 more answers

The number of employees for a certain company has been decreasing each year by 4%. If the company currently has 650 employees an

nika2105 [10]

The answer would probably be 649.96

8 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

Page 40-44 earth science regents<br> just post the picture of the pages please

Anna007 [38]

I don’t know what book you’re talking about so I can’t help but have a look online, you may be able to find it if you search up the book name and look around a few websites

8 0

3 years ago