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Mandarinka [93]

4 years ago

13

A spring is stretched from equilibrium to 0.90 m, and the work done on the spring equals 18 J. What is the value of the spring c

onstant?

1 answer:

Nataliya [291]4 years ago

7 0

W = (K * x^2)/2
36 = K*0.9^2
K = 36/0.81

I know you can do it from here, mate ;)

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A constant amount of charge passes through a conductor. How is current affected if the same amount of charge passes in less time

baherus [9]

Answer:

B. The current increases.

Explanation:

As we know that rate of flow of charge through the conductor is known as electric current

So we have

$i = \frac{q}{t}$

here we know that charge Q flowing through the conductor is constant while the time in which it passes through it is decreased

so we can say that the ratio of charge and time will increase

so here we have

$i = increased$

So correct answer will be

B. The current increases.

4 0

4 years ago

A baseball was hit and reached its maximum height in 3.00s. Find (a) its initial velocity; (b) the height it reaches.[29.4 m/s;

kicyunya [14]

Answer:

u= 29.43 m/s

h=44.14 m

Explanation:

Given that

t= 3 s

We know that acceleration due to gravity ,g = 9.81 m/s² (Downward)

Initial velocity = u

Final velocity ,v= 0 (At maximum height)

We know v = u +a t

v=final velocity

u=initial velocity

a=Acceleration

Now by putting the values in the above equation

0 = u - 9.81 x 3

u= 29.43 m/s

The maximum height h is given as

v² = u ² - 2 g h

0² = 29.43 ² - 2 x 9.81 x h

$h=\dfrac{29.43^2}{2\times 9.81}\ m$

h=44.14 m

4 0

4 years ago

A simple generator has a square armature 9.0 cm on a side. The armature has 95 turns of 0.59-mm-diameter copper wire and rotates

Marrrta [24]

Answer:

f = 3.102 Hz

Explanation:

In this case you have that the required voltage is the maximum induced emf produced by the rotating generator.

In order to calculate the frequency of rotation of the generator that allows one to obtain 12.0V you use the following formula:

$emf_{max}=NBA\omega$ (1)

N: turns of the armature = 95

B: magnitude of the magnetic field = 0.800T

A: area of the square armature = (9.0cm)^2 = (0.09m)^2 = 8.1*10^-3 m^2

emf_max = 12.0V

w: angular frequency

you solve the equation (1) for w:

$\omega=\frac{emf_{max}}{NBA}=\frac{12.0V}{(95)(0.800T)(8.1*10^{-3}m^2)}\\\\\omega=19.49\frac{rad}{s}$

Then, the frequency is:

$f=\frac{\omega}{2\pi}=\frac{19.49rad/s}{2\pi}=3.102Hz$

7 0

3 years ago

An object moving due to gravity can be described by the motion equation y=y0+v0t−12gt2, where t is time, y is the height at tha

LenaWriter [7]

Answer:

$t=6.4534 s$

Explanation:

This is an exercise where you need to use the concepts of <em>free fall objects</em>

Our <u>knowable variables</u> are initial high, initial velocity and the acceleration due to gravity:

$y_{0}=75m$

$v_{oy} =20m/s$

$g=9.8 m/s^{2}$

At the end of the motion, the <u><em>rock hits the ground</em></u> making the final high y=0m

$y=y_{o}+v_{oy}*t-\frac{1}{2}gt^{2}$

If we <em>evaluate the equation</em>:

$0=75m+(20m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$

This is a classic form of <u><em>Quadratic Formula</em></u>, we can solve it using:

$t=\frac{-b ± \sqrt{b^{2}-4ac } }{2a}$

$a=-4.9\\b=20\\c=75$

$t=\frac{-(20) + \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=-2.37s$

$t=\frac{-(20) - \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=6.4534s$

Since the <u><em>time can not be negative</em></u>, the <em>reasonable answer</em> is

$t=6.4534s$

8 0

3 years ago

An oxygen atom picks up two additional, free-floating electrons. Is the charge of the newly formed oxygen ion positive, negative

kogti [31]

Answer:

The atom is now a negative charge

Explanation:

Adding electrons makes things negative.

6 0

3 years ago