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jonny [76]
3 years ago
14

When it goes through a chemical change, a substance changes what

Physics
1 answer:
Anestetic [448]3 years ago
8 0

Answer:

form

I'm pretty sure...

let me know if I'm right, if I am give brainliest

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A positive point charge (q = +8.65 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 2.97 m. A p
ella [17]

Answer:

r_B = 1.88 m

Explanation:

As we know that work done by electric force is given as

W_e = -q\Delta V

so here we know that charge is moving from

r_A = 2.97 m

to another position

so we will have

W_{AB} = \frac{kq_1q_2}{r_A} - \frac{kq_1q_2}{r_B}

-6.95\times 10^{-9} = (9\times 10^9)(8.65\times 10^{-8})(4.56\times 10^{-11})(\frac{1}{2.97} - \frac{1}{r_B})

-0.196 = (\frac{1}{2.97} - \frac{1}{r_B})

r_B = 1.88 m

7 0
3 years ago
A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit
vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

\omega =\sqrt{\dfrac{K}{m}}

The maximum speed v given as

v=\omega A

A=Amplitude

v=\sqrt{\dfrac{K}{m}}\times A

0.32=\sqrt{\dfrac{13.5}{0.75}}\times A

A=0.075 m

A= 0.75 cm

The speed at distance x

v=\omega \sqrt{A^2-x^2}

v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}

v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}

v= 0.21 m/s

5 0
2 years ago
A force acts on a 9.90 kg mobile object that moves from an initial position of to a final position of in 5.40 s. Find (a) the wo
horrorfan [7]

Given that,

Mass of object = 9.90 kg

Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

We need to calculate the displacement

Using formula of displacement

s=r_{2}-r_{1}

Where, r_{1} = initial position

r_{2} = final position

Put the value into the formula

s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)

s= -6.80i+6.20j-0.1k

(a). We need to calculate the work done on the object

Using formula of work done

W=F\cdot s

Put the value into the formula

W=(2.00i + 9.00j + 5.30k)\cdot (-6.80i+6.20j-0.1k)

W=-13.6+55.8-0.53

W=41.67\ J

(b). We need to calculate the average power due to the force during that interval

Using formula of power

P=\dfrac{W}{t}

Where, P = power

W = work

t = time

Put the value into the formula

P=\dfrac{41.67}{5.40}

P=7.71\ Watt

(c). We need to calculate the angle between vectors

Using formula of angle

\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})

Put the value into the formula

\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})

\theta=79.7^{\circ}

Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.

(b). The average power due to the force during that interval is 7.71 Watt.

(c).  The angle between vectors is 79.7°

7 0
3 years ago
What is the height of the image? Round the answer
Maru [420]
It’s a concave mirror
8 0
2 years ago
A sample of chlorine has two naturally occurring isotopes. The isotope Cl-35 (mass 35.0 amu) makes up 75.8% of the sample, and t
irinina [24]

Answer:

M_{av}=35.521amu

Explanation:

As in any sample you will have 75.8% of Cl-35 iosotopes and 24.3% of Cl-37 iosotopes you can get the average atomic mass as:

M_{av}=(35amu*75.8+37amu*24.3)/100=35.521amu

4 0
2 years ago
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