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3 years ago

How is coulomb's law similar to newton's law of gravity?

1 answer:

4 0

<span>Newton's law of gravitation is attractive, whereas Coulomb's law is attractive or repulsive. Both are proportional to the inverse square of distance.</span>

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11. A sensitive thermometer is one which

Marta_Voda [28]

Answer:

A. ls sensitive to heat

this is correct answer

I hope it's helpful for you.......

4 0

3 years ago

The drawing shows a loudspeaker A and point C, where a listeer is positioned. A second loudspeaker B is located somewhere to the

Liono4ka [1.6K]

Answer: 4.17m

Explanation:

The observer at C will hear a sound on no sound upon whether the interference is constructive or destructive.

If the listeners hears sounds it is caled constructive interference but if he hears no sound its called destructive interference.

d2 - d1 = (n *lamba)/ 2

Where n=1,3,5

lamda=v/f =349/62.8

lamda=5.56m

d2= d1 + nlamda/2

d2= 1 + 5.56/2

d2= 3.78m

X'= 1 cos 60= 0.5m

Y= 1 sin60= 0.866m

X"^2 + Y^2 =d2^2

X" =√(y^2 - d2^2)

X"=√(3.78^2 - 0.886^2)

X"= 3.67m

So therefore the closest that speaker A can be to speaker B so the listener does not hear any sound is X' + X"= 0.5 + 3.67

4.17m

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

3 years ago

A flashlight uses four batteries to power it. Which type of current flows inside the flashlight?

Troyanec [42]

A flashlight has a flow of a direct current.

5 0

2 years ago

Read 2 more answers

A sample of 2.50 kg of water is held at a temperature of 100°C. How much energy must be added to completely turn the liquid wate

kap26 [50]

Answer:

Explanation:

Since 100C is the boiling temperature for water, for this problem we don't need to calculate the energy needed to get to the boiling point, just the heat or energy needed to vaporize the water to steam at 100C.

The formula for this is q=m(delta)

q is Joules of heat needed to vaporize the water to steam at 100C

m is mass in grams

Delta is in Joules per gram and can be looked up for water at this temperature. Here, it is approximately 2260J/g. This online lecture should help ease understanding: https://cabrillo.instructure.com/courses/10267/modules/items/256219

Therefore...

q=2.5g (2260J/g)= 5650J = 5.65kJ

I do not do Physics tutoring but am happy to answer questions here.

7 0

1 year ago