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Ksju [112]
3 years ago
5

The force that keeps two surfaces at rest from sliding over each other is

Physics
1 answer:
natka813 [3]3 years ago
8 0
Friction is the correct answer.
You might be interested in
PLEASE HELP ASAP Which one of the following is a step used for balancing chemical equations?
scZoUnD [109]

Answer:

C. Count the atoms in each substance in the reactants and products.

Explanation:

A chemical reaction can be defined as a chemical process which typically involves the transformation or rearrangement of the atomic, ionic or molecular structure of an element through the breakdown and formation of chemical bonds to produce a new compound or substance.

In order for a chemical equation to be balanced, the condition which must be met is that the number of atoms in the reactants equals the number of atoms in the products.

This ultimately implies that, the mass and charge of the chemical equation are both balanced properly.

In Chemistry, all chemical equation must follow or be in accordance with the Law of Conservation of Mass, which states that mass can neither be created nor destroyed by either a physical transformation or a chemical reaction but transformed from one form to another in an isolated (closed) system.

One of the step used for balancing chemical equations is to count the atoms in each substance in the reactants and products.

For example;

NH3 + O2 -----> NO + H2O

The number of atoms in each chemical element are;

For the reactant side:

Nitrogen, N = 1

Hydrogen, H = 3

Oxygen, O = 2

For the product side;

Nitrogen, N = 1

Hydrogen, H = 2

Oxygen, O = 2

When we balance the chemical equation, we would have;

NH3 + 3O2 -----> 4NO + 2H2O

3 0
3 years ago
A highway curves to the left with radius of curvature of 36 m and is banked at 28 ◦ so that cars can take this curve at higher s
iVinArrow [24]

Answer: 30.34m/s

Explanation:

The sum of forces in the y direction 0 = N cos 28 - μN sin28 - mg

Sum of forces in the x direction

mv²/r = N sin 28 + μN cos 28

mv²/r = N(sin 28 + μcos 28)

Thus,

mv²/r = mg [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]

v²/r = g [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]

v²/36 = 9.8 [(0.4695 + 0.87*0.8829) - (0.8829 - 0.87*0.4695)]

v²/36 = 9.8 [(0.4695 + 0.7681) / (0.8829 - 0.4085)]

v²/36 = 9.8 (1.2376/0.4744)

v²/36 = 9.8 * 2.6088

v²/36 = 25.57

v² = 920.52

v = 30.34m/s

5 0
4 years ago
4 A magnet can exert a force of attraction or a force of repulsion on another magnet.
GenaCL600 [577]

Answer:

Push -repulsion

Pull - attraction

Explanation:

When two magnets are brought together, a push happens when a force of repulsion is experienced where the magnets move away from each other. This means their polarity is the same and this will cause the magnet to push away from each other.

When two magnets are brought together , a pull happens when a force of attraction is experienced where the magnets move close to each other. This means their polarity is different and thus causes the magnets to pull closer to each other.

7 0
3 years ago
A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
8 0
3 years ago
Three forces act on an object. Two of the forces are at an angle of 100◦to each other and have magnitude 25N and 12N. The third
seraphim [82]

Answer:

F₄ = 29.819 N

Explanation:

Given

F₁ = (- 25*Cos 50° i + 25*Sin 50° j + 0 k) N

F₂ = (12*Cos 50° i + 12*Sin 50° j + 0 k) N

F₃ = (0 i + 0 j + 4 k) N

Then we have

F₁ + F₂ + F₃ + F₄ = 0

⇒   F₄ = - (F₁ + F₂ + F₃)

⇒   F₄ = - ((- 25*Cos 50° i + 25*Sin 50° j) N + (12*Cos 50° i + 12*Sin 50° j) N + (4 k) N) = (13*Cos 50° i - 37*Sin 50° j - 4 k) N

The magnitude of the force will be

F₄ = √((13*Cos 50°)² + (- 37*Sin 50°)² + (- 4)²) N = 29.819 N

6 0
3 years ago
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