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ohaa [14]
3 years ago
9

What is the name of the unit you use to measure force?

Physics
1 answer:
Anettt [7]3 years ago
8 0

Answer:

that would be newtons.

Explanation:

I hope this helps

You might be interested in
Which labels best complete the diagram?
nlexa [21]

Answer: D

Explanation: Look at the figure, we can conclude that the correct answer is

X: Low potential energy

Y: High potential energy

Z: Flow of electrons

Because electrons flow where there is difference in potential energy. And electrons move from a region of high potential energy to a region of low potential energy.

Since the arrow is pointing to X, that means

Z is the flow of electrons, X is of low potential energy and Y is of high potential energy.

6 0
3 years ago
A particle moves in a circular path of radius 0.10 m with a constant angular speed of 5 rev/s. The acceleration of the particle
Darya [45]

Answer:

2.5 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s²

For circular motion, the expression for acceleration is given as,

a = ω²r ................ Equation 1

Where a = acceleration of the particle, ω = angular speed of the particle, r = radius of the circular path.

Given: ω = 5 rev/s = 31.42 rad/s, r = 0.10 m.

Substitute into equation 1

a = 5²(0.10)

a = 25(0.10)

a = 2.5 m/s²

Hence the acceleration of the particle = 2.5 m/s²

Hence, none of the option  is correct

7 0
3 years ago
two point charges of magnitude 4.0 μc and -4.0 μc are situated along the x-axis at x1 = 2.0 m and x2 = -2.0 m, respectively. wha
user100 [1]

The electric potential at the origin of the xy coordinate system is negative infinity

<h3>What is the electric field due to the 4.0 μC charge?</h3>

The electric field due to the 4.0 μC charge is E = kq/r² where

  • k = electric constant = 9.0 × 10 Nm²/C²,
  • q = 4.0 μC = 4.0 × 10 C and
  • r = distance of charge from origin = x₁ - 0 = 2.0 m - 0 m = 2.0 m

<h3>What is the electric field due to the -4.0 μC charge?</h3>

The electric field due to the -4.0 μC charge is E = kq'/r² where

  • k = electric constant = 9.0 × 10 Nm²/C²,
  • q' = -4.0 μC = -4.0 × 10 C and
  • r = distance of charge from origin = 0 - x₂ = 0 - (-2.0 m) = 0 m + 2.0 m = 2.0 m

Since both electric fields are equal in magnitude and directed along the negative x-axis, the net electric field at the origin is

E" = E + E'

= -2E

= -2kq/r²

<h3>What is the electric potential at the origin?</h3>

So, the electric potential at the origin is V = -∫₂⁰E".dr

= -∫₂⁰-2kq/r².dr

Since E and dr = dx are parallel and r = x, we have

= -∫₂⁰-2kqdxcos0/x²

= 2kq∫₂⁰dx/x²

= 2kq[-1/x]₂⁰

= -2kq[1/x]₂⁰

= -2kq[1/0 - 1/2]

= -2kq[∞ - 1/2]

= -2kq[∞]

= -∞

So, the electric potential at the origin of the xy coordinate system is negative infinity

Learn more about electric potential here:

brainly.com/question/26978411

#SPJ11

3 0
2 years ago
A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00
Nutka1998 [239]

Answer:

6.99535\times 10^{-6}\ V/m

Explanation:

P = Power Output = 1000 W

r = Radius = 35000000 m

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

c = Speed of light = 3\times 10^8\ m/s

Intensity of Electric radiation is given by

I=\dfrac{P}{A}\\\Rightarrow I=\dfrac{P}{4\pi r^2}\\\Rightarrow I=\dfrac{1000}{4\pi\times 35000000^2}\ W/m^2

Intensity of Electric radiation is given by

I=\dfrac{1}{2}c\epsilon_0E_0\\\Rightarrow E_0=\sqrt{\dfrac{2I}{c\epsilon_0}}\\\Rightarrow E_0=\sqrt{\dfrac{2\times \dfrac{1000}{4\pi\times 35000000^2}}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E_0=6.99535\times 10^{-6}\ V/m

The amplitude of the electric field vector is 6.99535\times 10^{-6}\ V/m

6 0
4 years ago
A metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. Assume the test specimen is 12.8-m
ser-zykov [4K]

To solve this problem, we will start by defining each of the variables given and proceed to find the modulus of elasticity of the object. We will calculate the deformation per unit of elastic volume and finally we will calculate the net energy of the system. Let's start defining the variables

Yield Strength of the metal specimen

S_{el} = 414Mpa

Yield Strain of the Specimen

\epsilon_{el} = 0.002

Diameter of the test-specimen

d_0 = 12.8mm

Gage length of the Specimen

L_0 = 50mm

Modulus of elasticity

E = \frac{S_{el}}{\epsilon_{el}}

E = \frac{414Mpa}{0.002}

E = 207Gpa

Strain energy per unit volume at the elastic limit is

U'_{el} = \frac{1}{2} S_{el} \cdot \epsilon_{el}

U'_{el} = \frac{1}{2} (414)(0.002)

U'_{el} = 414kN\cdot m/m^3

Considering that the net strain energy of the sample is

U_{el} = U_{el}' \cdot (\text{Volume of sample})

U_{el} =  U_{el}'(\frac{\pi d_0^2}{4})(L_0)

U_{el} = (414)(\frac{\pi*0.0128^2}{4}) (50*10^{-3})

U_{el} = 2.663N\cdot m

Therefore the net strain energy of the sample is 2.663N\codt m

6 0
3 years ago
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