Hey guys, I don't know what this is. Any help?

Which is the best estimate for the mass of a typical lineman on a high school football team

fgiga [73]

We have that the best estimate for the mass of a typical lineman on a high school football team is

$100,000 g=100kg$

From the options

1. 100,000 g

2. 100,000 kg

3. 100,000 mg

Generally

A Typical lineman cannot be as small as

$100000mg=0.1kg=100g$

A Typical linesman cannot be as massive as

$100,000kg$ as that is unrealistic

With the Average Mass of A Typical linesman can be

$100,000 g=100kg$

In conclusion

The best estimate for the mass of a typical lineman on a high school football team is

$100,000 g=100kg$

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7 0

3 years ago

A scientist conducting a field investigation records measurements of very low pressure and high relative humidity at the top of

Ivan

Answer:

Please find the answer in the explanation

Explanation:

Given that a scientist conducting a field investigation records measurements of very low pressure and high relative humidity at the top of a mountain.

Since a weather map indicates that a warm front is approaching the mountain, according to the conventional current, the warm front is approaching because the weather must have been in higher relative humidity in cool air.

The warm front is approaching to replace it so that the cool air can conventionally replace the warmth air too.

The condition the scientists will most likely observe at the top of the mountain will be high relative humidity.

5 0

4 years ago

A very long, solid insulating cylinder has radius R; bored along its entire length is a cylindrical hole with radius a. The axis

lawyer [7]

Answer:

The value of the electric field is $E_{net} = \dfrac{r \textbf{b}}{2\epsilon_{0}}$

Explanation:

We know that the electric field inside a solid cylinder at a distance $\textbf{r}$ from the centre is given by

$E = \dfrac{\rho \textbf{r}}{2 \epsilon_{0}}$

Let's consider the cross-section of the cylinder as shown in the figure. Let `O' be the centre of the long solid insulating cylinder having radius 'R'. Also consider that $O'$ be the cetre of the hole of radius 'a' situated at a distance 'b' from 'O'. Given, the volume charge density of the material is 'r'. So, the volume charge density inside the hole will be '-r'. Let's consider 'P' be any arbitrary point inside the hole situated at a distance 's' from $O'$ .

So, the electric field ' $E_{O}$ ' due to the long cylinder at point 'P' is given by

$E_{O} = \dfrac{r \textbf{c}}{2 \epsilon_{0}}$

and the electric field ' $E_{O'}$ 'due to the hole at point 'P' is given by

$E_{O'} = \dfrac{\rho \textbf{s}}{2 \epsilon_{0}}$

So the net electric field ( $E_{net}$ ) inside the hole is given by

$E_{net} = E_{O} - E_{O'} = \dfrac{r}{2\epsilon_{0}}(\textbf{c - s}) = \dfrac{r \textbf{b}}{2\epsilon_{0}}$

5 0

4 years ago

Someone please help me

bagirrra123 [75]

Oxygen. Plants need water sunlight and CO2 to make their food. They get the carbon and the energy from the CO2 and the sun, and they need water as well. Then they release oxygen

3 0

3 years ago

Master of physics needed

Delicious77 [7]

Hey JayDilla, I get 1/3. Here's how:
Kinetic energy due to linear motion is:
$E_{linear}= \frac{1}{2}mv^2$
where
$v=r \omega$
giving
$E_{linear}= \frac{1}{2}mr^2 \omega ^2$

The rotational part requires the moment of inertia of a solid cylinder
$I_{cyl} = \frac{1}{2}mr^2$
Then the rotational kinetic energy is
$E_{rot}= \frac{1}{2}I \omega ^2= \frac{1}{4}mr^2 \omega ^2$
Adding the two types of energy and factoring out common terms gives
$\frac{1}{2}mr^2 \omega ^2(1+ \frac{1}{2})$
Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part. Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.

8 0

4 years ago