Which electromagnetic waves have the shortest wavelengths and highest frequencies?

Determine the weight in newtons of a woman whose weight in pounds is 157. Also, find her mass in slugs and in kilograms. Determi

DerKrebs [107]

Answer:

Weight of the woman in Newton

$x = 698.6 \ N$

Mass of the woman in slug

$Mass = 4.86 \ slug$

Mass of the woman in kg

$Mass = 16 \ kg$

My weight in Newton

$W = 784 \ N$

Explanation:

From the question we are told that

The weight of the woman in pounds is $W = 157 \ lb$

Converting to Newton

1 N = 0.22472 lb

x N = 157

=> $x = \frac{157 * 1}{0.22472}$

=> $x = 698.6 \ N$

Obtaining the mass in slug

$Mass = \frac{W}{g}$

Here $g = 32.2 ft/s^2$

So

$Mass = \frac{157 }{32.2}$

$Mass = 4.86 \ slug$

Obtaining the mass in kilogram

$Mass = \frac{W}{g}$

Here $g = 9.8 \ m/s$

So

$Mass = \frac{157 }{9.8}$

$Mass = 16 \ kg$

Generally weight is mathematically represented as

$W = m * g$

Given that my mass is 80 kg then my weight is

$W = 80 *9.8$

$W = 784 \ N$

6 0

3 years ago

What are noble gases?

Bas_tet [7]

Answer:

noble gases are basically a group of gases that are similar in their chemical compounds, theres six of them : helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and the radioactive radon (Rn).

~batmans wife dun dun dun.....

6 0

3 years ago

A can, containing only air, has its lid tightly screwed on and is left in strong sunlight

allsm [11]

Answer:

Because everyone knows that when you increase temperature activity withn molecules increases they will collide more making the can probably explode

3 0

3 years ago

A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coeffici

Ivenika [448]

Answer:

Coefficient of friction will be 0.296

Explanation:

We have given initial speed of the stone u = 8 m /sec

It comes to rest so final speed v = 0 m /sec

Distance traveled before coming to rest s = 11 m

According to third equation of motion

$v^2=u^2+2as$

So $0^2=8^2+2\times a\times 11$

$a=\frac{-64}{22}=-2.90m/sec^2$

Acceleration due to gravity $g=9.8m/sec^2$

We know that acceleration is given by

$a=\mu g$

So $2.90=9.8\times \mu \\$

$\mu =\frac{2.9}{9.8}=0.296$

So coefficient of friction will be 0.296

3 0

3 years ago

A simple hydraulic lift is made by fitting a piston attached to a handle into a 3.0-cm diameter cylinder. The cylinder is connec

stiv31 [10]

Answer:

Approximately $3.1 \times 10^4 \; \rm N$ (assuming that the acceleration due to gravity is $g = 9.81\; \rm kg \cdot N^{-1}$ .)

Explanation:

Let $A_1$ denote the first piston's contact area with the fluid. Let $A_2$ denote the second piston's contact area with the fluid.

Similarly, let $F_1$ and $F_2$ denote the size of the force on the two pistons. Since the person is placing all her weight on the first piston:

$F_1 = W = m \cdot g = 50\; \rm kg \times 9.81 \; \rm kg \cdot N^{-1} =495\; \rm N$ .

Since both pistons fit into cylinders, the two contact surfaces must be circles. Keep in mind that the area of a square is equal to $\pi$ times its radius, squared:

$\displaystyle A_1 = \pi \times \left(\frac{1}{2} \times 3.0\right)^2 = 2.25\, \pi\;\rm cm^{2}$ .
$\displaystyle A_2 = \pi \times \left(\frac{1}{2} \times 24\right)^2 = 144\, \pi\;\rm cm^{2}$ .

By Pascal's Law, the pressure on the two pistons should be the same. Pressure is the size of normal force per unit area:

$\displaystyle P = \frac{F}{A}$ .

For the pressures on the two pistons to match:

$\displaystyle \frac{F_1}{A_1} = \frac{F_2}{A_2}$ .

$F_1$ , $A_1$ , and $A_2$ have all been found. The question is asking for $F_2$ . Rearrange this equation to obtain:

$\displaystyle F_2 = \frac{F_1}{A_1} \cdot A_2 = F_1 \cdot \frac{A_2}{A_1}$ .

Evaluate this expression to obtain the value of $F_2$ , which represents the force on the piston with the larger diameter:

$\begin{aligned}F_2 &= F_1 \cdot \frac{A_2}{A_1} \\ &= 495\; \rm N \times \frac{2.25\, \pi\; \rm cm^2}{144\, \pi \; \rm cm^2} \approx 3.1 \times 10^4\; \rm N\end{aligned}$ .

6 0

3 years ago