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Masteriza [31]
3 years ago
13

If a pebble is being transported in a stream by rolling, how does the velocity of it compare to the velocity of the stream?

Physics
1 answer:
AleksandrR [38]3 years ago
3 0
Streams carry sediment, like pebbles, in their flows. The pebbles can be in a variety of locations in the flow, depending on it's size, the balance between the upwards velocity on the pebble (drag and lift forces), and it's settling velocity.
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Calculate the work required to lift a 2000 kg vehicle on a lift, 2.0 m at constant speed, assuming friction averages 500 N..a. 3
hoa [83]

Answer:

c. 40200 J

Explanation:

Assume gravitational constant g = 9.8m/s2. The weight of the 2000kg vehicle is

W = mg = 2000*9.8 = 19600 N

In addition to the friction averaging at 500N, the total force is

F = 20000 + 500 = 20100 N

The work required to generate this force over a distance of 2m would be

F*s = 20500 * 2 = 40200 J

So c.40200 J is the correct answer

7 0
3 years ago
Which of the following motions has a straight line?
skad [1K]
The answer would probably be B.
8 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
4 years ago
Read 2 more answers
Under what atmospheric condition is fog most often formed in the san joaquin valley?
GalinKa [24]
It seems that you have missed the necessary options for us to answer this question so I had to look for it. Anyway,here is the answer. The atmospheric condition in which <span>fog is most often formed in the san Joaquin valley is stable stability. Hope this helps.</span>
4 0
3 years ago
A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 108.7 g of water at 60.2°C with 108
Oduvanchick [21]

Answer : The heat capacity of the calorimeter is, 6.72J/g^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of water = 4.184J/g^oC

c_2 = specific heat of calorimeter = ?

m_1 = mass of water = 108.7 g

m_2 = mass of calorimeter = 108.7 g

T_f = final temperature of mixture = 35.0^oC

T_1 = initial temperature of water = 60.2^oC

T_2 = initial temperature of calorimeter = 19.3^oC

Now put all the given values in the above formula, we get

(108.7g)\times (4.184J/g^oC)\times (35.0-60.2)^oC=-(108.7g)\times c_2\times (35.0-19.3)^oC

c_2=6.72J/g^oC

Therefore, the heat capacity of the calorimeter is, 6.72J/g^oC

3 0
3 years ago
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