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denis-greek [22]
4 years ago
8

5. From the definition of work, explain why only the vertical height of the stairs is measured. Hint: Think of the data table co

lumn headed “Increase of GPE."
Physics
1 answer:
natulia [17]4 years ago
8 0
The motivation behind why the vertical stature of the stairs is the main thing measured is that it uncovers to us how much gravity is up against the individual and their weight, so we require this data to decide how much vitality and power we have to get up the stairs.
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In any vector space au=bu implies a=b ? Trou or False​
mixas84 [53]

Answer:

False

Explanation:

(I guess if it were written "properly" it would be ax=bx implies a=b).

Given the axioms we were given, it would seem that the statement should be true, no?

A related statement -- also listed as false -- is that "in any vector space, ax=ay implies that x=y." Again, given the axioms we have.

5 0
3 years ago
Read 2 more answers
What is the application of physics<br>​
arsen [322]

Answer:

1) We can estimate the age of the earth

2) we can calculate the speed of anything

3) we can also calculate gravity, e.t.c.

Explanation:

I could give you more just ask

4 0
4 years ago
Select the statements that describe a vector. Check all that apply
lozanna [386]
Vectors have both size and direction.
Vectors are indicated with arrows.
Vectors can have positive or negative values.
5 0
3 years ago
Read 2 more answers
An object on a vertical spring oscillates up and down in simple harmonic motion with an angular frequency of 14.2 rad/s. Calcula
zubka84 [21]

Answer:

0.04865 m

Explanation:

k = Spring Constant

m = Mass

d = Distance

g = Acceleration due to gravity = 9.81 m/s²

Angular frequency is given by

\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{k}{m}=\omega^2\\\Rightarrow \dfrac{k}{m}=14.2^2

At equilibrium we have

kd=mg\\\Rightarrow d=\dfrac{mg}{k}\\\Rightarrow d=\dfrac{g}{\omega^2}\\\Rightarrow d=\dfrac{9.81}{14.2^2}\\\Rightarrow d=0.04865\ m

The distance by which the spring stretches from its unstrained length is 0.04865 m

7 0
3 years ago
A constant electric field of magnitude E = 148 V/m points in the positive x-direction. How much work (in J) does it take to move
DochEvi [55]

Answer:

W=-2.1405\times 10^9\,J

Explanation:

Given:

electric field, E=148\,V.m^{-1}

charge, Q=-13\,\mu C=-13\times 10^{-6}\,C

initial position coordinates, p1 =(-18,-131)

final position coordinates,   p2 =(107,76)

We find the distance through which the charge has been moved:

d=\sqrt{(x1-x2)^2+(y1-y2)^2}

Where we have (x1,y1) & (x2,y2) as the initial and final coordinates of the points.

d=\sqrt{(107-(-81))^2+(76-(-131))^2}

d= 279.63\,m

Now we need the angle through which displacement is made with respect to the direction of electric field.

tan\,\theta= \frac{y2-y1}{x2-x1}

\theta= tan^{-1}[\frac{76-(-131)}{107-(-81)} ]

\theta= 47.75^{\circ}

Now from the relation between the change in potential difference:

\Delta V= E.d.cos\,\theta

\Delta V= 148\times 279.63\times cos\,47.75^{\circ}

\Delta V= 27826.06 V

∵The change in voltage is defined as the work done per unit charge.

∴\Delta V=\frac{W}{Q}

W=\frac{\Delta V}{Q}

Putting the respective values

W=\frac{27826.06 }{-13\times 10^{-6}}

W=-2.1405\times 10^9\,J

3 0
3 years ago
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