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igor_vitrenko [27]
3 years ago
14

What are the units for mass and weight?

Physics
2 answers:
Arlecino [84]3 years ago
8 0

Answer:

pounds or lbs

Explanation:

weight is measured in pounds.

Luda [366]3 years ago
6 0

The kilogram is the SI unit of mass and it is the almost universally used standard mass unit. The associated SI unit of force and weight is the Newton, with 1 kilogram weighing 9.8 Newtons under standard conditions on the Earth's surface.

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Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.
Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge q_{1}= 3.8\times10^{-6}\ C

Second charge q_{2}=3.2\times10^{-6}\ C

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

E_{1}=E_{2}

\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}

\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}

\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}

x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}

Put the value into the formula

x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}

x=1.69\ m

Hence, The distance is 1.69 m.

5 0
3 years ago
4. Using the bone density of 2.0 kg/m3, calculate the mass of an adult femur bone that has a volume of 0.00027 m3.
kolbaska11 [484]

Answer:

\boxed{\sf Mass \ of \ an \ adult \ femur \ bone = 0.00054 \ kg}

Given:

Bone density = 2.0 kg/m³

Volume of bone (V) = 0.00027 m³

To Find:

Mass of an adult femur bone (m).

Explanation:

\sf \implies Density = \frac{Mass (m)}{Volume (V)} \\ \\ \sf \implies \frac{Mass}{Volume} = Density \\ \\ \sf \implies Mass = Density \times Volume \\ \\ \sf \implies Mass = 2.0 \ kg/ \cancel{m^{3}} \times 0.00027 \ \cancel{m^{3}} \\ \\ \sf \implies Mass = 0.00054 \ kg

6 0
3 years ago
A dipole of moment 0.5 e·nm is placed in a uniform electric field with a magnitude of 8 times 104 N/C. What is the magnitude of
Over [174]

Answer:

The net torque is zero

Explanation:

Let's assume that the dipole is compose of two equal but oposite charges e, and it cam be represented by a rod with one end having a charge e and the other end with a charge of -e. Notice that the dipole is parallel to the electric field thus the force felt by both of the charges will be parallel to the electric field. This means that there will be no components of the forces that are perpendicular to the rod which is a requirement for it to have a torque.

8 0
3 years ago
An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a ver
Goshia [24]

Explanation:

It is given that,

An electron is released from rest in a weak electric field of, E=2.3\times 10^{-10}\ N/C

Vertical distance covered, s=1\ \mu m=10^{-6}\ m

We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

v^2-u^2=2as

v^2=2as.............(1)

Electric force is F_e and force of gravity is F_g. As both forces are acting in downward direction. So, total force is:

F=mg+qE

F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}

F=4.57\times 10^{-29}\ N

Acceleration of the electron, a=\dfrac{F}{m}

a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}

a=50.21\ m/s^2

Put the value of a in equation (1) as :

v=\sqrt{2as}

v=\sqrt{2\times 50.21\times 10^{-6}}

v = 0.010 m/s

So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.

6 0
3 years ago
Need help with this plz
Lisa [10]

Answer:

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Explanation:

7 0
3 years ago
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