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2 years ago

An object is thrown straight up into the air at 10 m/s. how fast is the object traveling after 2 seconds?

Physics

1 answer:

ziro4ka [17]2 years ago

7 0

First write down all the known variables:

vi = 10m/s

t = 1

a = -9.8m/s^2

vf = ?

Now choose the kinematic equation to relate these variables:

vf = vi + at

vf = (10) + (-9.8)(1)

vf = 0.2 m/s

The speed of the object travelling after a second would be 0.2m/s in reality, however, it is closest to the choice A, which would suffice as the answer for your question.

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If the moment acting on the cross section is M=630N⋅m, determine the maximum bending stress in the beam. Express your answer to

-BARSIC- [3]

Answer:

2.17 Mpa

Explanation:

The location of neutral axis from the top will be

$\bar y=\frac {(240\times 25)\times \frac {25}{2}+2\times (20\times 150)\times (25+(\frac {150}{2}))}{(240\times 25)+2\times (20\times 150)}=56.25 mm$

Moment of inertia from neutral axis will be given by $\frac {bd^{3}}{12}+ ay^{2}$

Therefore, moment of inertia will be

$\frac {240\times 25^{3}}{12}+(240\times 25)\times (56.25-25/2)^{2}+2\times [\frac {20\times 150^{3}}{12}+(20\times 150)\times ((25+150/2)-56.25)^{2}]=34.5313\times 10^{6} mm^{4}}$

Bending stress at top= $\frac {630\times 10^{3}\times (175-56.25)}{34.5313\times 10^{6}}=2.1665127\approx 2.17 Mpa$

Bending stress at bottom= $\frac {630\times 10^{3}\times 56.25}{34.5313\times 10^{6}}=1.026242858\approx 1.03$ Mpa

Comparing the two stresses, the maximum stress occurs at the bottom and is 2.17 Mpa

8 0

2 years ago

The function H(t) = -16t^2 + vt + s shows the height H (t), in feet, of a projectile launched vertically from s feet above the g

Reptile [31]

H(t) = -16t^2 + vt + s
Part A:
Using the given data:

H(t)= -16*t² + 60*t + 82;

Part B:
Put H(t)=0
0= -16*t² + 60*t + 82;
Use the quadratic formula to find t.
See the attachment...'t' is replaced with 'x'.

3 0

2 years ago

The field is strongest at which point?

Licemer1 [7]

Answer:The poles

Explanation:

The field is strongest at the poles

4 0

2 years ago

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

Nostrana [21]

Answer:

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

$E_1 = E_2$

Let the position where net field is zero will lie at distance "r" from q1

$\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}$

now we will have

$\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}$

now square root both sides

$\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}$

now we have

$\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1$

so we have

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

8 0

2 years ago

Help meh in this question plzzz

iragen [17]

The Moment of Inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

$I = I_{D} - I_{H}$ (1)

Where:

$I_{D}$ - Moment of inertia of the Disk.

$I_{H}$ - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

$I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right)$ (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole ( $m$ ):