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bonufazy [111]
2 years ago
5

Place a small object on the number line below at the position marked zero. Draw a circle around the object. Mark the center of t

his circle with the symbol for “initial position”. Move the object 5.0cm to the right and stop. Label this circle with the correct symbol for “final position.”  (A) What was the initial position of the object?     (B) What is the final position of the object?     (C) What is the distance traveled by the object?     (D) What is the displacement of the object?     (E) Of the three underlined quantities, which are numerically equal?
​

Physics
1 answer:
Viefleur [7K]2 years ago
6 0
It’s is D bc it is and idk you should do it
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In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water d
Alexxandr [17]

Answer:

a) a = g / 3

b) x (3.0) = 14.7 m

c) m (3.0) = 29.4 g

Explanation:

Given:-

- The following differential equation for (x) the distance a rain drop has fallen has the form:

                             x*g = x * \frac{dv}{dt} + v^2

- Where,                v = Speed of the raindrop

- Proposed solution to given ODE:

                             v = a*t

Where,                  a = acceleration of raindrop

Find:-

(a) Using the proposed solution for v find the acceleration a.

(b) Find the distance the raindrop has fallen in t = 3.00 s.

(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.

Solution:-

- We know that acceleration (a) is the first derivative of velocity (v):

                             a = dv / dt   ... Eq 1

- Similarly, we know that velocity (v) is the first derivative of displacement (x):

                            v = dx / dt  , v = a*t ... proposed solution (Eq 2)

                             v .dt = dx = a*t . dt

- integrate both sides:

                             ∫a*t . dt = ∫dt

                             x = 0.5*a*t^2  ... Eq 3

- Substitute Eq1 , 2 , 3 into the given ODE:

                            0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2

                                                = 1.5 a^2 t^2

                            a = g / 3

- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:

                           x (t) = 0.5*a*t^2

                           x (t = 3.0) = 0.5*9.81*3^2 / 3

                           x (3.0) = 14.7 m  

- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:

                           m (t) = k*x (t)

                           m (3.0) = 2.00*x(3.0)

                           m (3.0) = 2.00*14.7

                           m (3.0) = 29.4 g

6 0
2 years ago
The water level in a tank is 20 m above the ground. a hose is connected to the bottom of the tank, and the nozzle at the end of
Damm [24]

Answer:

P_(pump) = 98,000 Pa

Explanation:

We are given;

h2 = 30m

h1 = 20m

Density; ρ = 1000 kg/m³

First of all, we know that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,

Thus, it can be expressed as;

P_(tank)+ P_(pump) = P_(nozzle)

Now, the pressure would be given by;

P = ρgh

So,

ρgh_1 + P_(pump) = ρgh_2

Thus,

P_(pump) = ρg(h_2 - h_1)

Plugging in the relevant values to obtain;

P_(pump) = 1000•9.8(30 - 20)

P_(pump) = 98,000 Pa

5 0
2 years ago
Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a
Naily [24]

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

m_1v_1 + m_2v_2 = 0

final they both comes to rest so here we can say that final momentum must be zero

now we have

34 v + 54 (3 m/s) = 0

v = -4.76 m/s

8 0
2 years ago
Similar to what you see in your textbook, you can generally omit the multiplication symbol as you answer questions online, excep
lidiya [134]

Answer:

ma= ma

m⋅a = m⋅a

And equivalently:

am=ma

a⋅m = m⋅a

Explanation:

Question

Assuming this question "Similar to what you see in your textbook, you can generally omit the multiplication symbol as you answer questions online, except when the symbol is  needed to make your meaning clear. For example, 1*10^5 is not the same as 110^5 . When you need to be explicit, type * (Shift + 8) to insert the multiplication operator. You will see a multiplication dot (⋅) appear in the answer box. Do not use the symbol x. For example, for the expression ma,

typing m⋅a would be correct, but mxa would be incorrect".

Solution to the problem

For this case we want to write a expression for ma, and based on the previous info we can write:

ma= ma

m⋅a = m⋅a

And equivalently:

am=ma

a⋅m = m⋅a

But is not correct do this:

mxa=mxa

axm = mxa

8 0
2 years ago
Read 2 more answers
How are chargeable cells different from ordinary dry cells​
topjm [15]
Ordinary cells can convert chemical energy to electrical energy only, but rechargeable cells can also store electrical energy into chemical energy and vice versa. You will study more about it in your higher classes. secondary cells can be recharged and used again but dry cells cannot be recharged.
6 0
2 years ago
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