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Anton [14]
3 years ago
8

classify the following elements as halogens alkali metals alkaline earth metals, transition elements or inner transitional eleme

nts calcium,gold,iron,magnesium,plutonium,potassium,sodium and uranium
Physics
1 answer:
Sliva [168]3 years ago
6 0
<span>Calcium - alkali earth metals is the answer</span>
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An electron of mass 9.11 1031 kg has an initial speed of 3.00 105 m/s. It travels in a straight line, and its speed increases to
elena55 [62]

Explanation:

It is given that,

Mass of an electron, m=9.11\times 10^{-31}\ kg

Initial speed of the electron, u=3\times 10^5\ m/s

Final speed of the electron, v=7\times 10^5\ m/s

Distance, d = 5 cm = 0.05 m

(a) The acceleration of the electron is calculated using the third equation of motion as :

a=\dfrac{v^2-u^2}{2d}

a=\dfrac{(7\times 10^5)^2-(3\times 10^5)^2}{2\times 0.05}

a=4\times 10^{12}\ m/s^2

Force exerted on the electron is given by :

F=m\times a

F=9.11\times 10^{-31}\times 4\times 10^{12}

F=3.64\times 10^{-18}\ N

(b) Let W is the weight of the electron. It can be calculated as :

W=mg

W=9.11\times 10^{-31}\times 9.8

W=8.92\times 10^{-30}\ N

Comparison,

\dfrac{F}{W}=\dfrac{3.64\times 10^{-18}}{8.92\times 10^{-30}}

\dfrac{F}{W}=4.08\times 10^{11}

Hence, this is the required solution.

8 0
3 years ago
A tennis ball is dropped from 1.13 m above the
Alex73 [517]

Answer:

-4.71 m/s

Explanation:

Given:

y₀ = 1.13 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.13 m)

v = -4.71 m/s

7 0
3 years ago
A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-
alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, q_1=-3\ nC

It is placed at a distance of 9 cm at x axis

Charge, q_2=+4\ nC

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0

Here,

r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}

So,

\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}

Squaring both sides,

3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0
3 years ago
A wave is a _____ that transmits energy
galben [10]

Answer is disturbance

3 0
3 years ago
Read 2 more answers
What is an output force?
Vinil7 [7]
The answer would be D

5 0
3 years ago
Read 2 more answers
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