Work and energy both have the unit?

An ideal horizontal spring-mass system has a mass of 1.0 kg and a spring with constant 78 N/m. It oscillates with a period of 0.

steposvetlana [31]

Answer:

T = 0.71 seconds

Explanation:

Given data:

mass m = 1Kg, spring constant K = 78 N/m, time period of oscillation T = 0.71 seconds.

We have to calculate time period when this same spring-mass system oscillates vertically.

As we know

$T = 2\pi \sqrt{\frac{m}{K} }$

This relation of time period is true under every orientation of the spring-mass system, whether horizontal, vertical, angled or inclined. Therefore, time period of the same spring-mass system oscillating vertically too remains the same.

Therefore, T = 0.71 seconds

6 0

3 years ago

Marked as brainiest if correct

Naily [24]

Answer:

I think the the answer is creating and layout and template style ( C )

Explanation:

I did it before and i was checking my notes and i wrote that down , Hope this Helps :)

7 0

3 years ago

Calculate the kinetic energy of Earth due to its spinning about its axis. Krot = J Compare your answer with the kinetic energy o

Studentka2010 [4]

Answer:

Explanation:

Answer: Let ke = 1/2 IW^2 = 1/2 kMr^2 W^2 be Earth's rotational KE. W = 2pi/24 radians per hour rotation speed and k = 2/5 for a solid sphere M is Earth mass, r = 6.4E6 m.

Then ke = 1/2 2/5 6E24 (6.4E6)^2 (2pi/(24*3600))^2 = ? Joules. You can do the math, note W is converted to radians per second for unit consistency.

Let KE = 1/2 KMR^2 w^2 be Earth's orbital KE. w = 2pi/(365*24) radians per hour K = 1 for a point mass. Note I used 365 days, a more precise number is 365.25 days per year, which is why we have Leap Years.

Find KE/ke = 1/2 KMR^2 w^2//1/2 kMr^2 W^2 = (K/k)(w/W)^2 (R/r)^2 = (5/2) (365)^2 (1.5E11/6.4E6)^2 = 7.81E9 ANS

8 0

4 years ago

A jet of water squirts out horizontally from a hole on the side of the tank as shown below. 0.579 m 1.45 m h If the hole has a d

klemol [59]

Answer:

The height of the water above the hole in the tank is 58 mm

Explanation:

In order to solve this problem we need to draw a sketch of the dimensions that include the input variables of the problem.

Where:

x = 0.579[m]

y = 1.45 [m]

Using the following kinematic equation we can find the time that takes the water to hit the ground, and then with this time, we can find the velocity of the water in the x-component.

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

It is necessary to clarify the value of each of the respective variables below

y = - 1.45 [m] "It is negative because this point is below the water outlet"

yo = 0

vo = 0 "The velocity is zero because the component of the speed on the Y-axis does not exist"

therefore:

$-1.45=0.5*(-9.81)*t^{2} \\t = \sqrt{\frac{1.45}{0.5*9.81} } \\t = 0.543[s]$

The next step is to determine the velocity in component x, knowing the time.

$v=\frac{x}{t} \\v=\frac{0.579}{.543} \\v = 1.06[m/s]$

Now using torricelli's law we can find the elevation.

$v=\sqrt{2*g*h} \\h=\frac{v^{2} }{2*g} \\h=\frac{1.06^{2} }{2*9.81} \\h= 0.057[m] = 57.95[mm]$

3 0

3 years ago

If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If i

Blababa [14]

Explanation:

An object is attached to the spring and then released. It begins to oscillate. If it is displaced a distance 0.125 m from its equilibrium position and released with zero initial speed. The amplitude of a wave is the maximum displacement of the particle. So, its amplitude is 0.125 m.

After 0.800 seconds, its displacement is found to be a distance 0.125 m on the opposite side. The time period will be, $t=2\times 0.8=1.6\ s$

We know that the relation between the time period and the time period is given by :

$f=\dfrac{1}{t}$

$f=\dfrac{1}{1.6}$

f = 0.625 Hz

So, the frequency of the object is 0.625 Hz. Hence, this is the required solution.

6 0

3 years ago

Read 2 more answers