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Oliga [24]
2 years ago
11

What is the density of a block of wood with a mass of 12g and a volume of 6 cm3?

Physics
1 answer:
scZoUnD [109]2 years ago
5 0
<h2>Answer: 2 g/cm³</h2>

Explanation:

Density = mass ÷ volume

             = 12 g ÷ 6 cm³

             =  2 g/cm³

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Which feature of the sun releases large amounts of magnetic activity that can cause the communication systems of the Earth to ma
mihalych1998 [28]
I would say sunquakes
4 0
3 years ago
Read 2 more answers
Two large conducting parallel plates A and B are separated by 2.4 m. A uniform field of 1500 V/m, in the positive x-direction, i
Lapatulllka [165]

Answer:

a. 1.027 x 10^7 m/s b. 3600 V c. 0 V and d. 1.08 MeV

Explanation:

a. KE =1/2 (MV^2) where the M is mass of electron

b. E = V/d

c. V= 0 V (momentarily the pd changes to zero)

d KE= 300*3600 v = 1.08 MeV

6 0
3 years ago
If the distance of a galaxy is 2,000 Mpc, how many years back into the past are we looking when we observe this galaxy
ruslelena [56]

The age of the galaxy when we look back is 13.97 billion years.

The given parameters:

  • <em>distance of the galaxy, x = 2,000 Mpc</em>

According Hubble's law the age of the universe is calculated as follows;

v = H₀x

where;

H₀ = 70 km/s/Mpc

T = \frac{x}{V} \\\\T = \frac{x}{xH_0} \\\\T = \frac{1}{H_0} \\\\T = \frac{1}{70 \ km/s/Mpc} \\\\T = \frac{1 \ sec}{70 \times 3.24 \times 10^{-20} } \\\\T = 4.41 \times 10^{17} \ sec\\\\T = \frac{4.41 \times 10^{17} \ sec\  \times \ years}{3600 \ s \ \times\  24\ h\  \times \ 365.25 \ days} \\\\T = 1.397  \times 10^{10} \ years\\\\T = 13.97 \ billion \ years

Thus, the age of the galaxy when we look back is 13.97 billion years.

Learn more about Hubble's law here: brainly.com/question/19819028

8 0
2 years ago
If a small number of tree frogs migrate to a mat of vegetation that is already home to an established population of tree frogs a
Bingel [31]

Answer:

D

Explanation:

Gene flow is the transfer of genetic variation from one population to another

5 0
3 years ago
The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is
astraxan [27]

Complete question:

Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.

Answer:

The magnitude of this field is 826 N/C

Explanation:

Given;

The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m

PEsinθ = T

where;

E is the magnitude of the electric field

P is the dipole moment

First, we determine the magnitude dipole moment;

Magnitude of dipole moment = q*r

P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m

Finally, we determine the magnitude of this field;

E = \frac{T}{P*sin(\theta)}=  \frac{7.3 X 10^{-9}}{1.476X10^{-11}*sin(36.8)}\\\\E = 825.6 N/C

E = 826 N/C (in three significant figures)

Therefore, the magnitude of this field is 826 N/C

6 0
3 years ago
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