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blondinia [14]
4 years ago
13

A positive charge is placed in an electric field that points west. What direction is the force on the positive particle?

Physics
2 answers:
MatroZZZ [7]4 years ago
4 0
The direction of the electric field is defined as the direction of the force on a
positive charge immersed in the field. The force on the particle points west.
algol [13]4 years ago
4 0
The direction of positive charge is along electric field
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Answer:what are the answer options?

Explanation

3 0
3 years ago
A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone
Lemur [1.5K]

Answer:

Fy = 14.3 [N]

Explanation:

To be able to solve this problem we must know that the force is a vector and has magnitude and direction, therefore it can be decomposed into the force in the X & y components:

When we have the components on the horizontal and vertical axes we must use the Pythagorean theorem.

F = \sqrt{F_{x}^{2} +F_{y}^{2}  }

where:

F = 15 [N]

Fx = horizontal component = 4.5 [N]

Fy = vertical component [N]

15=\sqrt{4.5^{2}+F_{y}^{2}}\\ 15^{2}= (\sqrt{4.5^{2}+F_{y}^{2}})^{2} \\225 = 4.5^{2}+F_{y} ^{2}\\  F_{y}^{2} =225 -4.5^{2}\\ F_{y}^{2}=204.75\\F_{y}=\sqrt{204.75}\\  F_{y}=14.3 [N]

7 0
3 years ago
If a liquid has a density of 1.67 g/cm^3, what is the volume of 45g of the liquid?
Aloiza [94]

Answer:

V = 26.95 cm³

Explanation:

Density is given by the formula :

ρ = m÷V

Density = mass ÷ Volume

Given both density and mass we rearrange, substitute and solve for Volume :

Rearranging the equation to make Volume the subject :

ρ = m÷V

ρV = m

V = m÷ ρ

Now substitute :

V = 45 ÷ 1.67

V = 26.9461077844

Take 2 decimal places as the density is 2 decimal places :

V = 26.95

Units will be cm³ as it is volume

Hope this helped and have a good day

8 0
2 years ago
Please Help! When a metal is heated, the free electrons gain _____________.
lakkis [162]
It’s B potential energy
7 0
4 years ago
Read 2 more answers
The ammonia molecule (NH3) has a dipole moment of 5.0×10?30C?m. Ammonia molecules in the gas phase are placed in a uniform elect
Neko [114]

Question (continuation)

(a) What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to E S from parallel to perpendicular?

(b) At what absolute temperature T is the average translational kinetic energy 3/2kT of a molecule equal to the change in potential energy calculated in part (a)?

Answer:

a. 9.0 * 10^-24 Joules

b. 0.44K

Explanation:

Given

Let p = dipole moment = 5.0 * 10^-30 Cm

Let E = Magnitude = 1.8 * 10^6 N/m

a.

The charge in electric potential = Final Charge - Initial Charge

Initial Charge = Potential Energy

Initial Energy = -pE cosФ where Ф = 0

So, initial Energy = - 5.0 * 10^-30 * 1.8 * 10^6

Initial Energy = -9 * 10^-24 Joules

Final Energy = 0

Charge = 0 - (-9.0 * 10^-24)

Charge = 9.0 * 10^-24 Joules

b.

Absolute Temperature

Change in Kinetic Energy = Change in Potential Energy = 9.0 * 10^-24

Change in Kinetic Energy = 3/2kT where k is Steven-Boltzmann constant = 1.38 * 10^-23

So,

9.0 * 10^-24 = 3/2 * 1.38 * 10^-23 * T

T = (9.0 * 10^-24 * 2)/(3 * 1.38 * 10^-23)

T = (18 * 10^-24)/(4.14 * 10^-23)

T = 0.44K

6 0
3 years ago
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