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tensa zangetsu [6.8K]

2 years ago

10

What happens during the fourth stage of technological design that is not necessary during the sixth or seventh stages of scienti

fic investigation?

2 answers:

riadik2000 [5.3K]2 years ago

8 0

<span>A design is remodeled after analysis and tested again.</span>

julsineya [31]2 years ago

6 0

Edg_nuity:

C. identify a problem

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If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it oscillates. Suppose

Volgvan

Answer:

a) A=0.125 m

b) T = 1.72 s

c) f= 0.58 Hz

Explanation:

a) As we are told that the maximum displacement from the equilibrium position was 0.125 m (from which it was released at zero initial speed), this is the amplitude of the resultant SHM, so, A=0.125 m

b) In order to find the period, we must get the total time needed to complete a full cycle (which means that the block must pass twice through the equilibrium point). We are told that at t=0.860 sec, the block has reached to the other end of the trajectory, and it has passed through the equilibrium point only once.

This means that the period must be exactly the double of this time:

T = 2*0. 860 sec = 1.72 sec.

c) In a SHM, the frequency is defined just as the inverse of the period (like in a uniform circular movement), so we can get the frequency f as follows:

f = 1/T = 1/ 1.72 s= 0.58 Hz

8 0

2 years ago

Light cannot escape the intense gravitational pull of a _____.

natima [27]

Black hole
hope it helped

3 0

2 years ago

Read 2 more answers

Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a −80-nC charge at x = −4 m on

disa [49]

Answer:

V = 48 Volts

Explanation:

Since we know that electric potential is a scalar quantity

So here total potential of a point is sum of potential due to each charge

It is given as

$V = V_1 + V_2 + V_3$

here we have potential due to 50 nC placed at y = 6 m

$V_1 = \frac{kQ}{r}$

$V_1 = \frac{(9\times 10^9)(50 \times 10^{-9})}{\sqrt{6^2 + 8^2}}$

$V_1 = 45 Volts$

Now potential due to -80 nC charge placed at x = -4

$V_2 = \frac{kQ}{r}$

$V_2 = \frac{(9\times 10^9)(-80 \times 10^{-9})}{12}$

$V_2 = -60 Volts$

Now potential due to 70 nC placed at y = -6 m

$V_3 = \frac{kQ}{r}$

$V_3 = \frac{(9\times 10^9)(70 \times 10^{-9})}{\sqrt{6^2 + 8^2}}$

$V_3 = 63 Volts$

Now total potential at this point is given as

$V = 45 - 60 + 63 = 48 Volts$

3 0

2 years ago

A boy whirls a stone in a horizontal circle of radius 1.50m and at height 2.00m above level ground. The string breaks, and the s

Eddi Din [679]

Answer:

$a_{cp}=162.2m/s^2$

Explanation:

The time the stone takes to fall can be calculated considering only the vertical component with the formula:

$y=y_0+v_{0y}t+\frac{a_yt^2}{2}$

Taking the inital height as 0m and downward direction positive, since it departs from (vertical) rest we have:

$y=\frac{gt^2}{2}$

Which gives us a time:

$t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2(2m)}{(9.8m/s^2)}}=0.64s$

Horizontally, on that time the stone travelled a distance x=10m, which means its horizontal speed was:

$v_x=\frac{x}{t}=\frac{10m}{0.64s}=15.6m/s$

Since <u>this speed is the tangential velocity</u> while whirling, the centripetal acceleration of the stone was:

$a_{cp}=\frac{v_x^2}{r}=\frac{(15.6m/s)^2}{(1.5m)}=162.2m/s^2$

7 0

3 years ago

A 2.00 meter long horizontal plank whose mass is 3.00 kg is placed so that the very end (the right end) rests on a retaining wal

avanturin [10]

Answer:

$R_g=\frac{61}{7} =8.7143\ N$

$R_w=45.1857\ N$

Explanation:

Given that:

mass of plank, $m=3\ kg$

length of plank, $l=2\ m$

From the image we can visualize the given situation.

Consider the given plank to be mass-less and having a uniformly distributed mass of 1.5 kg per meter.

<u>Now in the balanced condition:</u>

$\sum F=0$

$R_g+R_w=(3+2.5)\times 9.8$

$R_g+R_w=53.9\ N$ .......................(1)

$\sum M_w=0$

$1.75\times R_g=3\times 1+(2.5\times 9.8)\times 0.5$

$R_g=\frac{61}{7} =8.7143\ N$ ...........................(2)

is the force acted by the tailgate on the plank.

<u>Substitute the value from (2) into (1):</u>

$R_w=45.1857\ N$ is the force acted by the wall upon the plank.

6 0

3 years ago