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Alik [6]
3 years ago
13

Is the distance we walk equal to the magnitude of our displacement?

Physics
1 answer:
Yuliya22 [10]3 years ago
8 0

Answer:

If you take the first example of the walk around the desk, it should be apparent that sometimes the distance is the same as the magnitude of the displacement. ... Since the displacement is measured along the shortest path between two points, its magnitude is always less than or equal to the distance.

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Explain why some places can use only certain kinds of energy to generate electricity.
alukav5142 [94]

Answer:

please mark me as a brainleast and follow me please my soster

6 0
2 years ago
Answer either one or both questions! Must explain and show work to receive brainliest!
MariettaO [177]
<span>F x L = W x X whereW=weight is total load = 80, L is length from fulcrum which is the unknown and what we are solving for. x= length we know. and F equals 50 force we know. So (W*X)/F=LL equals 64</span>
3 0
3 years ago
Fiora starts riding her bike at 20 mi/h. After a while, she slows down to 14 mi/h, and maintains that speed for the rest of the
goblinko [34]

Answer:

3.5 hours

Explanation:

Speed = distance/time

Let the distance that Fiora biked at 20 mi/h through be x miles and the time it took her to bike through that distance be t hours at 20 mi/h

Then, the rest of the distance that she biked at 14 mi/h is (112 - x) miles

And the time she spent biking at 14 mi/h the rest of the distance = (6.5 - t) hours

Her first biking speed = 20 mph = 20 miles/hour

Speed = distance/time

20 = x/t

x = 20 t (eqn 1)

Her second biking speed = 14 mph = 14 miles/hour

14 = (112 - x)/(6.5 - t)

112 - x = 14 (6.5 - t)

112 - x = 91 - 14t (eqn 2)

Substitute for x in (eqn 2)

112 - 20t = 91 - 14t

20t - 14t = 112 - 91

6t = 21

t = 3.5 hours

x = 20t = 20 × 3.5 = 70 miles.

(112 - x) = 112 - 70 = 42 miles

(6.5 - t) = 6.5 - 3.5 = 3 hours

Meaning that she travelled at 20 mi/h for 3.5 hours.

4 0
3 years ago
Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the
Evgen [1.6K]

Answer:

g(h) = g ( 1 - 2(h/R) )

<em>*At first order on h/R*</em>

Explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

In order to do this, we must expand the newton's law of universal gravitation around r=R

Remember that this law is:

F = G \frac{m_1m_2}{r^2}

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

F = m_2a

Therefore, we can see that

a(r) = G \frac{m_1}{r^2}

With a the acceleration due to the earth's mass.

Now, the taylor series is going to be (at first order in h/R):

a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R}

a(R) is actually the constant acceleration at sea level

and

a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}

Therefore:

a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:

a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R} + \frac{h^2}{2!} \frac{d^2a(r)}{dr^2}_{r=R}

6 0
3 years ago
Read 2 more answers
What are the like terms in the expression? 4m-9+3n+2
wel
-9 and +2
Hope I helped :)

5 0
3 years ago
Read 2 more answers
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