This picture represents A: Reflection B: Refraction C: Interference D: Diffraction

An airplane flies 1000 miles in 2 hours. What is its average speed in miles per hour?

OLEGan [10]

Answer:

500km per hour

Explanation:

if in 2 hours the airplane flies 1000 km then 1000 divided by 2 is 500km per hour.

5 0

3 years ago

) What is the electric potential due to the nucleus of hydrogen at a distance of 7.50× 10-11 m? Assume the potential is equal to

ohaa [14]

For this, we need the formula:
V = k q / r
where k is the Coulombs law constant = 9 x 10^9 N
q is the charge of the hydrogen nucleus (proton) = 1.6 x 10^-19 C
r is the distance
Simply plug in the values and solve for V

5 0

3 years ago

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A box of mass 10 kg is pulled from the hold of a ship with an acceleration of 1 m/s^2 by a vertical rope attached to it .find th

wlad13 [49]

F=ma
Mass times acceleration
We have g (10ms^_2) and a (1 given)
So total would be
10 kg times (10+1) =
110 N

5 0

3 years ago

A 100 kg bag of sand has a weight on 100 N. When dropped its acceleration is what?

Anettt [7]

Answer:

The acceleration of the bag of sand is $1\ m/s^2$ .

Explanation:

We have,

Mass of a bag of sand is 100 kg

Weight of the bag of sand is 100 N

It is required to find the acceleration of the bag when it is dropped. The weight of an object is given by :

$F =ma$

When it is dropped, a = g

$g=\dfrac{F}{m}\\\\g=\dfrac{100}{100}\\\\g=1\ m/s^2$

So, the acceleration of the bag of sand is $1\ m/s^2$ .

4 0

4 years ago

Find its moment of inertia about an axis perpendicular to its plane and passing through the midpoint of the line connecting its

antoniya [11.8K]

A) Moment of inertia about an axis passing through the point where the two segments meet : $I_A=\frac{1}{12} M L^2$

B) Moment of inertia passing through the point where the midpoint of the line connects to its two ends: $I x=\frac{1}{3} M L^2$

What is Moment of inertia?

The term "moment of inertia" refers to a physical quantity that quantifies a body's resistance to having its speed of rotation along an axis changed by the application of a torque (turning force). The axis might be internal or exterior, fixed or not.

A) The moment of inertia about an axis passing through the point where the two segments meet is $I_A=\frac{1}{12} M L^2$ given that the rod is bent at the center and distance from all the points to the axis remains the same, the moment of inertia about the center will remain the same.

B) Determine the moment of inertia about an axis passing through the point midpoint of the line which connects the two ends

First step: determine the distance between the ends ( d )

After applying Pythagoras theorem $\mathrm{d}=\frac{\sqrt{2}}{2} L$

Next step : determine distance between the two axis $(\mathrm{x})$

After applying Pythagoras theorem

$\mathrm{x}=\frac{\sqrt{2}}{4} L$$$

Final step : Calculate the value of $\mathrm{I}_{\mathrm{x}}$

applying Parallel Axis Theorem

$I_x=I_8+M x^2$

$\begin{aligned}& =\frac{1}{12} M L^2+\frac{1}{4} M L^2 \\& \therefore \quad I x=\frac{1}{3} M L^2 \\&\end{aligned}$

Hence we can conclude that Moment of inertia about an axis passing through the point where the two segments meet: $I_A=\frac{1}{12} M L^2$ , Moment of inertia passing through the point where the midpoint of the line connects its two ends: $I x=\frac{1}{3} M L^2$

To learn more about moment of inertia visit:brainly.com/question/15246709

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5 0

1 year ago