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ludmilkaskok [199]

3 years ago

10

Students perform a probability experiment by throwing wads of a paper at a trash can 55.8 cm in diameter.What is the area of the

trash can?

1 answer:

Neko [114]3 years ago

5 0

The area of the trash can would be 12 inches

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Three charges 1.5*10-6, 3*10-6, -3*10-6 are placed at three vertices of an equilateral triangle of side 30cm. Find the net force

gulaghasi [49]

Answer:

F = 0N

Explanation:

The force between two charges is given by

$F=k\frac{q_1q_2}{r^2}$

where r is the distance between the charges and K is the Coulomb's constant

(k=8-89*10^9Nm^2/C^2)

The force in the first charge is only the sum of the forces due to the other charges. Hence we have

$F_T=F_1+F_2=k\frac{q_2q_1}{r^2}+k\frac{q_3q_1}{r^2}$

$F_T=(8.89*10^9\frac{Nm^2}{C^2})\frac{(3*10^{-6}C)(1.5*10^{-6}C)}{(0.3m)^2}+(8.89*10^9\frac{Nm^2}{C^2})\frac{(-3*10^{-6}C)(1.5*10^{-6}C)}{(0.3m)^2}\\\\F_T=0.445N-0.445N=0N$

Ft=0N

Hope this helps!!

5 0

3 years ago

Read 2 more answers

Neutrons can change the charge of an atom<br><br> True<br> False

IceJOKER [234]

Answer:

False. They can’t change the charge but they can change the mass

8 0

3 years ago

What is the distance write the equation

Sedbober [7]

Answer:

135 m

Explanation:

Given:

vi = 12 m/s

vf = 18 m/s

t = 9s

Find: x

x = ½ (vf + vi) t

x = ½ (18 m/s + 12 m/s) (9 s)

x = 135 m

5 0

3 years ago

The energy required to start a reaction is called the activation energy. true false

Gemiola [76]

The energy required to start a reaction IS called the activation energy. TRUE

3 0

3 years ago

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g

Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:

$k=\sqrt{\frac{I}{M} }$

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration $k=\sqrt{\frac{I}{M} }$

So, moment of rotational inertia (I) of a cylinder about it axis = $\frac{MR^2}{2}$

$k=\sqrt{\frac{\frac{MR^2}{2}}{M} }$

$k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }$

$k=\sqrt{{\frac{R^2}{2}}$

$k={\frac{R}{\sqrt{2}}$

$k={\frac{1.20m}{\sqrt{2}}$

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) = $\frac{2}{3}MR^2$

$k = \sqrt{\frac{\frac{2}{3}MR^2}{M} }$

$k = \sqrt{\frac{2}{3} R^2}$

$k = \sqrt{\frac{2}{3} }*R$

$k = \sqrt{\frac{2}{3}} *1.20$

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of radius

(I) = $\frac{2}{5}MR^2$

$k = \sqrt{\frac{\frac{2}{5}MR^2}{M} }$

$k = \sqrt{\frac{2}{5} R^2}$

$k = \sqrt{\frac{2}{5} }*R$

$k = \sqrt{\frac{2}{5}} *1.20$

k = 0.7560

k ≅ 0.76 m

6 0

4 years ago