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Igoryamba
3 years ago
10

A spherical ball of lead (density 11.3 g/cm 3) is placed in a tub of mercury (density 13.6 g/cm 3). Which answer best describes

the result
Physics
1 answer:
aniked [119]3 years ago
5 0

The lead ball will float with about 17% of its volume above the surface of the mercury.

We know that density is defined as mass per unit volume of a substance. The density of a substance is an intrinsic property which can be used to identify a substance.

Given that Lead is less dense that mercury, we know that lead will float on mercury. Since the density of mercury is 13.6 g/cm3 and that of lead is 11.3 g/cm3, lead ball will float with about 17% of its volume above the surface of the mercury.

Learn more: brainly.com/question/12108425

Missing parts;

A spherical ball of lead (density 11.3 g/cm3) is placed in a tub of mercury (density 13.6 g/cm3). Which answer best describes the result?

A.The lead ball will float with about 83% of its volume above the surface of the mercury.

B.The lead ball will float with about 17% of its volume above the surface of the mercury.

C.The lead ball will float with its top exactly even with the surface of the mercury.

D.The lead will sink to the bottom of the mercury.

E.none of the above

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3 years ago
At a certain distance from a point charge, the potential and electric field magnitude due to that charge are 4.98 V and 12.0 V/m
Finger [1]

Answer:

1. d = 0.415 m.

2. Q = 2.285 x 10^{-10} C.

Explanation:

The electric field and potential can be found by the following equations:

E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\\V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}

Applying these equations to the given variables yields

E = 12 = \frac{1}{4\pi\epsilon_0}\frac{Q}{d^2}\\V = 4.98 = \frac{1}{4\pi\epsilon_0}\frac{Q}{d}

Divide the first line to the second line:

\frac{12}{4.98} = \frac{ \frac{1}{4\pi\epsilon_0}\frac{Q}{d^2}}{\frac{1}{4\pi\epsilon_0}\frac{Q}{d}}\\\frac{12}{4.98} = \frac{1}{d}\\d = 0.415~m

Using this distance in either of the equations give the magnitude of the charge.

12 = \frac{1}{4\pi\epsilon_0}\frac{Q}{(0.415)^2}\\12 = \frac{1}{4\pi (8.8\times 10^{-12})}\frac{Q}{(0.415)^2}\\Q = 2.285 \times 10^{-10}~C

8 0
3 years ago
Can anyone help me solve number 5
Shtirlitz [24]
What subject is this?
6 0
3 years ago
A stiff wire 50.0 cm long is bent at a right angle in the middle. One section lies along the z axis and the other is along the l
zloy xaker [14]

Answer:

Magnitude of the force is 2.135N and the direction is 41.8° below negative y-axis

Explanation:

The stiff wire 50.0cm long bent at a right angle in the middle

One section lies along the z axis and the other is along the line y=2x in the xy plane

\frac{y}{x} = 2

tan θ = 2

Therefore,

slope m = tan θ = y / x

\theta=\tan^-^1(2)=63.4^0

Then length of each section is 25.0cm

so, length vector of the wire is

\hat I= (-25.0)\hat k +(25.0) \cos 63.4^0 \hati +(25.0) \sin63.4^0 \hatJ\\\\\hat I = (11.2) \hat i + (22.4) \hat j - (23.0) \hat k

And magnetic field is B = (0.318T)i

Therefore,

\bar F = \hat I (\bar l \times \bar B)

\bar F = (20.0)[(0.112m)i +(0.224m)j-(0.250m)k \times 90.318T)i]

= (20.0)(i(0)+j(-0.250)(0.318T)+k[0-(0.224m)(0.318T)]\\\\=(20.0)(-0.250)(0.318)j-(20.0)(0.224)(0.318T)\\\\=-(1.59N)j-(1.425N)k

Magnitude of the force is

F = \sqrt{(-1.59N)^2+(-1.425N)^2\\} \\F = 2.135N

Direction is

\alpha = \tan^-^1(\frac{-1.425N}{-1.59N} )\\\\= 41.8^0

Magnitude of the force is 2.135N and the direction is 41.8° below negative y-axis

5 0
3 years ago
Read 2 more answers
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