Ajdaifsgodtistizhxtsgoachoach

Which one of the following is dimensional expressions speed?

Evgesh-ka [11]

D)LT^-1 speed=distance(L)/time(T)——>L/T

5 0

3 years ago

You place a 0.17 kg can of soup and a 0.31 kg jar of pickles on the kitchen counter, separated by a distance of 0.42 m. What is

tangare [24]

$1.984 \times 10^{-11} \mathrm{N} \text { is the force of gravity exerted on the jar of pickles. }$

<u>Explanation</u>:

According to Newton's third law that each force has an equal and opposite reaction force in this case both of the jars will exert the same force an each other

. The force is given by

$\mathrm{F}=\frac{G \times M_{1} \times M_{2}}{r^{2}}$

Where, F = force, $G=\text { gravitational constant }=\left(6.67 \times 10^{-11}\right)$ , $mass \left(\mathrm{M}_{1}\right)=0.17 \mathrm{kg}$ , $mass \left M_{2}= 0.31 \mathrm{kg}$ and Distance(r) = 0.42 m.

Substitute the values in the formula.

$\mathrm{F}=\frac{6.67 \times 10^{-11} \times 0.17 \times 0.31}{0.42^{2}}$

$\mathrm{F}=\frac{3.51 \times 10^{-12}}{0.176}$

$\mathrm{F}=1.984 \times 10^{-11} \mathrm{N}$

$\text { The force of gravity exerted on the jar of pickles is } 1.984 \times 10^{-11} \mathrm{N} \text { . }$

3 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

Can you guys help to understand this graph I'm so confused why I getting wrong?

dedylja [7]

The one fact that needs to be mentioned but isn't given anywhere on or around the graph is: The distance, on the vertical axis, is the distance FROM home. So any point on the graph where the distance is zero ... the point is in the x-axis ... is a point AT home.

Segment D ...
Walking AWAY from home; distance increases as time increases.

Segment B ...
Not walking; distance doesn't change as time increases.

Segment C ...
Walking away from home, but slower than before; distance increases as time increases, but not as fast. Slope is less than segment-D.

Segment A ...
Going home; distance is DEcreasing as time increases. Walking pretty fast ... the slope of the line is steep.

4 0

3 years ago

Read 2 more answers

A book that weighs 19 Newtons sits on a table. With what force

iVinArrow [24]

Answer:

We know there's two forces acting on a book while it sits on a table:the force of gravity pulling it down, and the normal force of the table acting upward on the book. The book isn't accelerating while it sits there. That's because the weight of the book is being counteracted by the normal force of the table.

Explanation:

There are two forces acting upon the book. One force - the Earth's gravitational pull - exerts a downward force. The other force - the push of the table on the book (sometimes referred to as a normal force) - pushes upward on the book.

5 0

3 years ago