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2 years ago

6

An object with a mass of 0.5 kilometre start from rest and achieves a maximum speed of 20 metre per second in 0.01 second, what

average unbalanced force accelerated this object

1 answer:

Katarina [22]2 years ago

3 0

Answer:

Hiii how are you <u>doing?</u><u>?</u><u>I </u><u>don't</u><u> </u><u>understand</u><u> </u><u>that</u>

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The potential energy of a pair of hydrogen atoms separated by a large distance x is given by U(x)=−C6/x6, where C6 is a positive

kirza4 [7]

Answer:

The correct answer will be " $-\frac{6C_{6}}{x^{7}}$ ". The further explanation is given below.

Explanation:

The potential energy will be,

⇒ $U(x)= -\frac{C_{6}}{x^6}$

The expression of force will be,

⇒ $F=-\frac{dU(x)}{dx}$

⇒ $=-(C_{6}(-6)x^{-7})$

⇒ $=-\frac{6C_{6}}{x^{7}}$

Force seems to be appealing because the expression has been negative. It therefore means that the force or substance is acting laterally in on itself.

6 0

3 years ago

Lemon trees receiving the most water produced the most lemons. What is the independent variable, the dependent variable, and the

Veseljchak [2.6K]

The lemon trees will always be the same so they are the constants. The independent variable is the water amount because it is changing. The dependent variable is the amount of lemons because it is the variable being tested. (If we change the amount of water how many lemons will be produced?)

3 0

2 years ago

6. Most organizations have not banned the use of performing enhancing drugs.

Setler79 [48]

The answer is false

4 0

2 years ago

QuestionDetails:

sleet_krkn [62]

To solve the two parts of this problem, we will begin by considering the expressions given for gravitational potential energy and finally kinetic energy (to find velocity). From the potential energy we will obtain its derivative that is equivalent to the Force of gravitational attraction. We will start considering that all the points on the ring are same distance:

$r = \sqrt{x^2+R^2}$

Then the potential energy is

$U = \frac{-GMm}{\sqrt{x^2+R^2}}$

PART A) The force is excepted to be along x-axis.

Therefore we take a derivative of U with respect to x.

$F = -\frac{dU}{dx}$

$F = -\frac{d}{dx}(GMm(\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}}))$

$F = \frac{GMmx}{(x^2+R^2)^{3/2}}$

This expression is the resultant magnitude of the Force F.

PART B) The magnitude of loss in potential energy as the particle falls to the center

$U = GMm(\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}})$

According to conservation of energy,

$\frac{1}{2}mv^2 = GMm (\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}})$

$\therefore v = \sqrt{2GM(\frac{1}{R}-\frac{1}{x^2+R^2})}$

7 0

2 years ago

A projectile is fired horizontally off the top of a cliff with an initial velocity of 30 m/s. It hits the ground 2.0 seconds lat

Mandarinka [93]

Answer:

The horizontal distance traveled by the projectile is 60 m

Explanation:

Given;

initial horizontal velocity of the projectile, Vₓ = 30 m/s

time of the motion of the projectile, t = 2 s

The horizontal distance traveled by the projectile is given by the range of the projection;

X = Vₓt

X = 30 x 2

X = 60 m

Therefore, the horizontal distance traveled by the projectile is 60 m

Therefoe

6 0

2 years ago