I need the right answer ASAP NO LINKS!!!

A student in Denver (altitude = 1 mile = 1609 m above sea level) brings a physics book of mass 1.3 kg to the top of a ten story

Nesterboy [21]

Answer:

Explanation:

Potential energy is the energy of a body due to is virtue of rest.

Potential energy is given as mgh

g is a constant and it is 9.81m/s²

And also the mass of the body is given as 1.3kg

Now the height of the body is

He took a book to a storey building of height 26m

He still holds the book 151 cm (1.51m) above the house.

The house is on an altitude of 1609m from the sea level.

Total Ug with out the sea level is

Ug=mgh

Ug=1.3 × 9.81 ×(26+1.51)

Ug=350.84J

Then, the potential energy due to the sea level is given as

Ug=mgh

Where g = 1/6371 m/s²

Therefore

Ug=mgh

Ug=1.3 × 1/6371 ×1609

Ug=0.328J

Total energy = 0.328+350.84

Ug=351.17J

8 0

3 years ago

What is the elevation of point B?<br> What about F?

kolbaska11 [484]

Explanation:

Since I can only do this by observation, the elevation of F is approximately 850km and the elevation of B is 925km.

7 0

3 years ago

A thin rod of length 0.64 m and mass 120 g is suspended freely from one end. It is pulled to one side and then allowed to swing

valina [46]

Answer:

1. Kinetic Energy = 0.0161 Joules

2. Height = 0.0137m

Explanation:

Given

Length of Rod, l = 0.64m

Mass, m = 120g = 0.12kg

Angular speed, w = 1.40 rad/s

a.

Calculating the Rod's kinetic energy

This is calculated by

Kinetic Energy = ½Iw²

Where I = rotational inertia of the rod about an axis.

This is calculated as follows;

I = Icm + mh²

I = ImL² + m(L/2)²

I = 1/12 * 0.12 * 0.64² + 0.12 * (0.64/2)²

I = 0.016384 kgm²

By substituton

KE = ½Iw² becomes

KE = ½ * 0.016384 * 1.40²

KE = 0.01605632J

KE = 0.0161 Joules

2. Using the total conservation of momentum;

K + U = Kf + V

Where K = Initial Kinetic Energy of the rod at lowest point.

U = Initial gravitational potential energy of the rod at lowest point

Kf = Final Kinetic Energy of the rod at maximum height = 0 J

V = Final gravitational potential energy of the rod at maximum height

So, K + U = Kf + V become

K + U = 0 + V

K + U = V

K = V - U = mgh

substitute 0.01605632J for K

0.01605632J = mgh

h = 0.01605632J/mg

h = 0.01605632J/(0.12 * 9.8)

h = 0.013653333333333

h = 0.0137m

4 0

4 years ago

What is the value of the resulting force and its direction acting on the following block: * 6N, left 1N, right 4N, left 1N, left

kherson [118]

1 N to the right as shown on photo

5 0

3 years ago

A normal blood pressure reading is less than 120/80 where both numbers are gauge pressures measured in millimeters of mercury (m

KIM [24]

The pressure value is given by the equation,

$P= \rho gh$

Where,

$\rho$ represents the density of the liquid

g= gravity

h= Heigth

A) For the measurement of the guage pressure we have the data data,

$\rho_{mercury} = 13.6*10^3kg/m^3$

$h=0.0930m$

$g=9.8m/s^2$

Replacing we get,

$P_g=\rho_{mercury}gh$

$[tex]P_g = (13.6*10^3)(9.8)(0.0930)$ P_g = 12395Pa[/tex]

In order to find the Absolute pressure, we perform a sum between the atmospheric pressure and that of the Gauge,

B) The atmospheric pressure at sea level is 101325Pa, assuming ideal conditions, we will take this pressure for our calculation, so

$P_T = P_a+P_g\\P_T = 101325Pa+12395Pa\\P_T = 113720Pa\\P_T = 113.7kPa$

6 0

3 years ago