Why do we use the name electromagnetism to describe the force of an

A solid cylinder with a mass of 2.72 kg and a radius of 0.083 m starts from rest at a height of 4.20 m and rolls down a 88.7 ◦ s

Bingel [31]

Explanation:

According to the law of conservation of energy ,

Potential energy = kinetic energy

$mgh = \frac{1}{2} \times mv^{2} + \frac{1}{2} \times I \times \omega^{2}$

I = $\frac{mr^{2}}{2}$

$\omega = \frac{v}{r}$

$mgh = [\frac{1}{2} \times mv^{2}] + [\frac{1}{2} \times (\frac{mr^{2}}{2}) \frac{v^{2}}{r^{2}}]$

mgh = $[\frac{1}{2} \times mv^{2}] + [\frac{1}{4} \times mv^{2}]$

$g \times h = \frac{3}{4} \times v^{2}$

$9.8 \times 4.2 = \frac{3}{4} \times v^{2}$

v = 7.4 m/s

thus, we can conclude that the translational speed of the cylinder when it leaves the incline is 7.4 m/s.

5 0

3 years ago

A 7-hp (shaft) pump is used to raise water to an elevation of 19 m. if the mechanical efficiency of the pump is 82 percent, dete

Hoochie [10]

Convert the units of power,W = 7 hp = 7 * 745.69 = 5219.83 WCalculate the power input to the pump using the efficiency of the pump equationn=Wpump/Wshaft
Substitute 0.82 for n and 5219.83 for Wshaft0.82=Wpump/5219.83 Wpump=0.82*5219.83=4280.26 WCalculate the mass flow ratein=Wpump/(gz_2 )Where g is the acceleration due to gravity, and z_2 is the elevation of water. Substitute 4280.26 for Wpump, 9.81 m/s^2 for g, and 19m for z_2in = 4280.26 / 9.81 * 19 = 22.9640 m^3/sCalculate the volume flow rate of waterV=m/ρWhere ρ is the density of water. Substitute 22.9640 m^3/s for in and 1000 m^3/kg for ρ, we get V = 22.9640 / 1000 = 0.0230 kg/sTherefore, the volume flow rate of water is 0.0230 kg/s

7 0

3 years ago

Birdman is flying horizontally at a

den301095 [7]

Answer:

68 m

Explanation:

Given that the horizontal velocity of the birdman = 17 m/s and

the height, h= 78 m.

The gravitational force is acting in the downward direction, so it will not change the horizontal speed.

The horizontal speed will remains be constant and will be equal to the initial horizontal speed of the turd.

Initially, the turd was also flying horizontally with the birdman, so the initial velocity of the turd is the same as the horizontal velocity of the birdman, i.e In the horizontal direction, $u_0=17$ m/s.

In the vertical direction, u = 0,

The distance to be traveled, in the direction of application of force, is equals to the height of the turd, i.e

s= 78 m

Let t be the time taken to cover a distance of 78 m.

Now, applying the equation of motion in the vertical direction,

$s=ut+\frac 12 at^2$

where u is the initial velocity and a is the acceleration due to gravity in the direction of displacement,s.

Here, $a=g=9.81 m/s^2$ , so

$78=0\times t +\frac 12 (9.81)t^2$

$\Rightarrow t^2=(78\times2)/9.81$

$\Rightarrow t = 4$ seconds.

Hence, the time taken to reach the ground is 4 seconds.

As the horizontal speed, $u_0=17$ m/s, is constant throughout the journey, so

the horizontal distance covered by turd

$= u\times t$

$= 17 \times 4 = 68$ m.

So, the distance of landing from the start of the field is 68 m as the birdman releases a turd directly above the start of the field.

Hence, the robot must hold the bucket at a distance of 68 m from the start of the field.

5 0

3 years ago

Scientific evidence documents the pattern of evolution. The evidence exists in a variety of categories. What are these categorie

Sauron [17]

Answer:

Explanation:

Scientific evidence takes note of the pattern of evolution. The evidence exists in a variety of categories, including direct observation of evolutionary change, the fossil record, homology, and biogeography.

Examples of the categories with their respective examples are:

Direct observation of evolutionary change: Development of drug resistant bacteria

Fossil record: Discovery of transitional forms of horses, Discovery of shells of extinct species

Homology: Similarities in mammalian forelimbs, Same genetic code in fireflies and tobacco plants, Vestigial pelvis in right whales

Biogeography: Similarity of endemic island species to nearby mainland species, The high concentration of marsupial species in Australia

3 0

3 years ago

a 90 kg architect is standing 2 meters from the center of a scaffold help up by a rope on both sides. the scaffold is 6m long an

Mademuasel [1]

We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.

1) Equilibrium of forces:
$T_1 - W_p - W_s + T_2 =0$
where
$W_p = (90kg)(9.81 m/s^2)=883 N$ is the weight of the person
$W_s = (200kg)(9.81 m/s^2)=1962 N$ is the weight of the scaffold
Re-arranging, we can write the equation as
$T_1 = 2845 N-T_2$ (1)

2) Equilibrium of torques:
$T_1 \cdot 3 m - W_p \cdot 2 m - T_2 \cdot 3m =0$
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using $W_p = 883 N$ and replacing T1 with (1), we find
$2845 N \cdot 3 m - T_2 \cdot 3 m - 833 N \cdot 2 m - T_2 \cdot 3 m=0$
from which we find
$T_2 = 1128 N$

And then, substituting T2 into (1), we find
$T_1 = 1717 N$

8 0

3 years ago