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Nesterboy [21]
2 years ago
5

A 10.0 kg ball weighs 98.0 N in air and weighs 65.0 N when submerged in water. The volume of the ball is:_________.A) 0.00245 m3

. B) 0.00337 m3. C) 0.00457 m3. D) 0.00766 m3 E) 0.00980 m3
Physics
1 answer:
Kamila [148]2 years ago
7 0

Answer: B) 0.00337 m3.

Explanation:

Given data:

Mass of the ball = 10kg

Weight of the ball in air = 98N

Weight of the ball in water = 65N

Solution:

To get the Volume of the ball when submerged in water, we divide the weight of the ball in water with the difference in apparent weight by 9.8m/s^2.

= 98 - 65 / 9.8

= 33 / 9.8

= 3.37kg

The volume of the ball is 3.37kg

The density of water is 1kg per Liter.

So 3.37 kg of water would have a volume of 3.37 Liters.

Therefore the ball would have a volume of 3.37 Liters (or 0.00337 cubic meters).

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Rina8888 [55]

Answer:

The magnetic induction of the magnetic field is  0.0005293 mT

Explanation:

Data given

I = 7 A = the total current in the wire

r = 23 cm = the radius of the wire = 0.23 meter

r' = 2cm = the measurement point, which should be inside the wire = 0.02 meter

Let's consider the current density is constant in the wire, ⇒  the current enclosed is a function of the enclosed area

I(enclosed) = Jπ r ²

we can  consider the current density  as the total current over the whole area:

I(enclosed) = I / (πr ²)  * πr' ²

I(enclosed) = (I* r'²)/ (r ²)  

with I =  total current in the wire = 7A

With r = the radius of wire = 0.23 meter

with r' = the distance of point from the center of wire  0.02 meter

We plug this into ampere's law:

∮ *B *dl =μ 0  * (I* r'²)/ (r ²)  

with B = Magnetic flux density (in Tesla) or magnetic induction

with dl = an infinitesimal element (a differential) of the curve C

with µ0 = the magnectic constant =  4π*10^−7 H/m

We can simplify this, by using an Amperian loop can write this as:

B *( 2 π r') =  μ 0  * (I* r'²)/ (r ²)  

Because the circumference of a circle is  2 π r , when we integrate over length at a distance  r ′  from the center of wire whose crossection is a circle we get  2 π r ′

When we isolate B, we get:

B = µo *(Ir'/2 π r ²)

B =  4π*10^−7 * ((7*0.02)/2*π*0.23²)

B =5.293 *10 ^-7 T  = 0.0005293 mT

The magnetic induction of the magnetic field is  0.0005293 mT

6 0
3 years ago
An isolated system consists of a 1.5 kg mass moving in the presence of the following potential energy function: U open parenthes
pickupchik [31]

Answer: Period T = 4.44secs

Explanation: see attachment

7 0
3 years ago
Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy
erma4kov [3.2K]

Answer:

Explanation:

⁵⁷Co₂₇  + e⁻¹  =  ²⁷Fe₂₆

mass defect = 56.936296 + .00055 - 56.935399

= .001447 u

equivalent energy

= 931.5 x .001447 MeV

= 1.3479 MeV .

= 1.35 MeV

energy of gamma ray photons = .14  + .017

= .157 MeV .

Rest of the energy goes to neutrino .

energy going to neutrino .

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5 0
3 years ago
The electric field 2.5 mm from a uniform sheet of charge is σ=800, NC. How much charge is contained in a 5.0x5.0 cm section of t
slamgirl [31]

Answer:

The charge is 2.75\times10^{-13}\ C

Explanation:

Given that,

Distance = 2.5 mm

Electric field = 800 NC

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We need to calculate the linear charge density

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E=\dfrac{2k\lambda}{r}

\lambda=\dfrac{Er}{2k}

Put the value into the formula

\lambda=\dfrac{800\times2.5\times10^{-3}}{2\times9\times10^{9}}

\lambda=1.1\times10^{-10}\ C/m

We need to calculate the charge

Using formula of charge

Q=\lambda\timesL

Put the value into the formula

Q=1.1\times10^{-10}\times(5.0\times5.0\times10^{-4})

Q=2.75\times10^{-13}\ C

Hence, The charge is 2.75\times10^{-13}\ C

5 0
3 years ago
Is xenon a pure substance​
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\large\huge\green{\sf{Yes}}

6 0
2 years ago
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