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4 years ago

An element's atomic number is 55. How many protons would an atom of this element have?

1 answer:

6 0

The atomic number of an element is the number of protons the element has, so the element with the atomic number 55 has 55 protons.

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10.

myrzilka [38]

Answer:

<em>The new period of oscillation is D) 3.0 T</em>

Explanation:

<u>Simple Pendulum</u>

A simple pendulum is a mechanical arrangement that describes periodic motion. The simple pendulum is made of a small bob of mass 'm' suspended by a thin inextensible string.

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

Where L is its length and g is the local acceleration of gravity.

If the length of the pendulum was increased to 9 times (L'=9L), the new period of oscillation will be:

$T'=2\pi \sqrt{\frac{L'}{g}}$

$T'=2\pi \sqrt{\frac{9L}{g}}$

Taking out the square root of 9 (3):

$T'=3*2\pi \sqrt{\frac{L}{g}}$

Substituting the original T:

$T'=3*T$

The new period of oscillation is D) 3.0 T

4 0

3 years ago

HELP CHECK ANSWERS ONLY 3 QUESTIONS:

maksim [4K]

Answer:

answer 1 is A, Answer 2 is A, Answer 3 is c

8 0

3 years ago

A 36-N crate is suspended from the left end of a plank. The plank weighs 21 N, but it is not uniform, so its center of gravity d

riadik2000 [5.3K]

Answer:

0.51 m

Explanation:

from the question we are given the following

weight of the crate = 36 N

weight of the plank = 21 N

distance of of the balance from the left end = 0.3 m

from the diagram attached, CG is the center of gravity ( the point near or within a body through which its weight can be assumed to act ), taking the CG as where the weight of the plank acts and it being at a distance L from the support

we can find the distance of the center of gravity just as we would find the moment about the support balance

therefore

36 x 0.3 = 21 x L

10.8 = 21L

L = 0.51 m

The center of gravity is 0.51 m to the right of the support.

4 0

3 years ago

The electric field on the surface of an irregularly shaped conductor varies from 74.0 kN/C to 14.0 kN/C.

mrs_skeptik [129]

Answer:

(a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Explanation:

Given that,

Electric field $E_{1}=74.0\ kN/C$

Electric field $E_{2}=14.0\ kN/C$

When the radius of curvature is greatest, the electric field at the surface will be smaller.

Where the radius of curvature is greatest

(a). We need to calculate the local surface charge density at the point on the surface

Using formula of charge density

$\sigma=\epsilon_{0}E_{2}$

Put the value into the formula

$\sigma=8.85\times10^{-12}\times14\times10^{3}$

$\sigma=1.239\times10^{-7}\ C/m^2$

$\sigma=123.9\times10^{-9}\ C/m^2$

$\sigma=123.9\ nC/m^2$

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). We need to calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is smallest

Using formula of charge density

$\sigma=\epsilon_{0}E_{1}$

Put the value into the formula

$\sigma=8.85\times10^{-12}\times74\times10^{3}$

$\sigma=6.549\times10^{-7}\ C/m^2$

$\sigma=654.9\times10^{-9}\ C/m^2$

$\sigma=654.9\ nC/m^2$

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Hence, (a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

7 0

3 years ago

Which of the following is a correct principle of relative-age dating?

elena55 [62]

The anwser choice is A

7 0

3 years ago

Read 2 more answers