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9 months ago

14

At the circus, the Human Cannonball is

1 answer:

svlad2 [7]9 months ago

6 0

The correct answer is 1.4285714.

In physics, velocity is characterised as a vector measurement of the motion's direction and speed. To be more precise, the rate of change in an object's position relative to a frame of reference and time is another way to describe velocity. The definition of velocity simply states the rate of motion of an object in a specific direction. It determines how quickly or slowly something is going.

Velocity = distance/ time

Thus time = distance/velocity

Here velocity = 350m/s

diatnce = 500 m

time = 500/350

time = 1.42857142857

t= 200m /350m/s = 1.4285714

To learn more about velocity refer the link:

brainly.com/question/18084516

#SPJ9

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Yakvenalex [24]

Answer:

B

Explanation:

6 0

2 years ago

A short-wave radio antenna is supported by two guy wires, 155 ft and 175 ft long. Each wire is attached to the top of the antenn

Vladimir79 [104]

Answer:

163.8 ft

Explanation:

In triangle ABD

$AB$ = 155 ft

$Cos63 = \frac{BD}{AB} = \frac{BD}{155}\\BD = 155 Cos63 \\BD = 70.4 ft$

$Sin63 = \frac{AD}{AB} = \frac{AD}{155} \\AD = 166 Sin63\\AD = 148 ft$

Using Pythagorean theorem in triangle ADC

$AC^{2} = AD^{2} + DC^{2} \\175^{2} = 148^{2} + DC^{2} \\DC = 93.4 ft$

$d$ = distance between the anchor points

distance between the anchor points is given as

$d = BD + CD = 70.4 + 93.4\\d = 163.8 ft$

5 0

3 years ago

Read 2 more answers

Ruthford's theory was identified by

8090 [49]

Answer:

your mom

Explanation:

6 0

3 years ago

Read 2 more answers

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

<u>Distance of the center from each corners</u> $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

<u>For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:</u>

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

<u>Force by each of the charges at the remaining corners:</u>

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

<u> Now, net electric field in the vertical direction:</u>

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

<u>Now, net electric field in the horizontal direction:</u>

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

2 years ago

A charge Q is transferred from an initially uncharged plastic ball to an identical ball 10.0 cm away. The force of attraction is

victus00 [196]

Answer:

a charge Q is transferred from an initially uncharged

Explanation:

Hope this helps!

5 0

1 year ago