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2 years ago

How do you enter meters per second in mastering physics?

1 answer:

7 0

Type one space between the number and the units, and then enter the units in the correct format. For example, a velocity would be entered as "1.2 m/s", not "1.2m/s".

<h3>What is velocity?</h3>

The pace at which an object's direction changes when it is traveling is referred to as its velocity and is measured by a particular unit of time and from a particular point of view. Velocity is a fundamental concept in kinematics, the branch of classical mechanics that deals with the motion of bodies.

<h3>What is an example of velocity?</h3>

Velocity can be defined as the rate at which something moves in a specific direction. as the speed of a car driving north on a highway or the pace at which a rocket takes off.

To learn more about velocity visit:

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You might be interested in

Displacement vectors of 4 km north, 2km south, 5km north , 5km south combine a total displacement of A 2km south B16 km south C1

Ivahew [28]

The answer your looking for is C

4 0

3 years ago

There are four charges, each with a magnitude of 1.96 µC. Two are positive and two are negative. The charges are fixed to the co

mart [117]

Answer:

Magnitude of the resultant force (Fn₁) on q₁

Fn₁ = 0.142N (directed toward the center of the square)

Explanation:

Theory of electrical forces

Because the particle q₁ is close to three other electrically charged particles, it will experience three electrical forces and the solution of the problem is of a vector nature.

Graphic attached

The directions of the individual forces exerted by q₂, q₃ and q₄ on q₁ are shown in the attached figure.

The force (F₁₄) of q₄ on q₁ is repulsive because the charges have equal signs and the forces (F₁₂) and (F₁₃) of q₂ and q₃ on q₁ are attractive because the charges have opposite signs.

Calculation of the forces exerted on the charge q₁

To calculate the magnitudes of the forces exerted by the charges q₂, q₃, and q₄ on the charge q₁ we apply Coulomb's law:

$F_{12} = \frac{k*q_1*q_2}{r_{12}^2}$ : Magnitude of the electrical force of q₂ over q₁. Equation((1)

$F_{13} = \frac{k*q_1*q_3}{r_{13}^2}$ : Magnitude of the electrical force of q₃ over q₁. Equation (2)

$F_{14} = \frac{k*q_1*q_4}{r_{14}^2}$ : Magnitude of the electrical force of q₄ over q₁. Equation (3)

Equivalences

1µC= 10⁻⁶ C

Known data

q₁=q₄= 1.96 µC = 1.96*10⁻⁶C

q₂=q₃= -1.96 µC = -1.96*10⁻⁶C

r₁₂= r₁₃ = 0.47m: distance between q₁ and q₂ and q₁ and q₄

$r_{14} = \sqrt{0.47^2+ 0.47^2}=0.664m$

k=8.99x10⁹N*m²/C² : Coulomb constant

F₁₂ calculation

We replace data in the equation (1):

$F_{12} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.47^2}$

F₁₂ = 0.156 N Direction of the positive x axis (+x)

F₁₃ calculation

We replace data in the equation (2):

$F_{13} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.47^2}$

F₁₃ = 0.156 N Direction of the negative y axis (-y)

Magnitude of the net electrostatic force between F₁₃ and F₁₂

$F_{n23}= \sqrt{0.156^2+0.156^2} = 0.22N$ (directed toward the center of the square)

F₁₄ calculation

We replace F₁₄ data in the equation (3):

$F_{14} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.664^2}$

F₁₄ = 0.078 N (In the opposite direction to Fn₂₃)

Calculation of the resulting force on q₁: Fn₁

Fn₁ = Fn₂₃ - F₁₄ = 0.22 - 0.078 = 0.142 N

6 0

4 years ago

PLEASE ANSWER NEED HELP!!!!!!!! PLEASE THE CORRECT ANSWER!!!!!!

Lesechka [4]

Answer:

C. unbalanced force

Explanation:

it is called unbalanced force because it has changed direction of the moving object

6 0

3 years ago

Which statements about braking a car are true? A)The greater the kinetic energy of a car, the longer it takes for the car to sto

NikAS [45]

Answer:

option c

Explanation:

Kinetic energy is due to the speed of a body.

$K.E = \frac{1}{2}mv^2$

When speed is doubled, the kinetic energy is quadruple.

From third equation of motion, braking distance is also proportional to square of speed. Thus, when speed is doubled, the braking distance is quadruple.

Thus, option c is correct.

3 0

3 years ago

Read 2 more answers

A car traveling in a straight line has a velocity of 5.0 m/s. After an acceleration of 0.75 m/s/s, the cars velocity is 8.0. In

Bogdan [553]

Vs - velocity on beginning
ve - velocity on ending. You've got:
$v_s = 5 \frac{m}{s} \\ v_e=8 \frac{m}{s} \\ \hbox{Then:} \\ \Delta v=v_e - v_s = 8 \frac{m}{s} - 5\frac{m}{s}=3 \frac{m}{s} \\ a=0,75 \frac{m}{s^2} \\ \hbox{And from formula:} \\ a=\frac{\Delta v}{\Delta t} \qquad \Rightarrow \qquad \Delta t= \frac{\Delta v}{a} \\ \hbox{Substitute:} \\ \Delta t=\frac{3\frac{m}{s}}{0,75 \frac{m}{s^2}}= \frac{3}{\frac{3}{4}} s= 3 \cdot \frac{4}{3} s= 4 s$
So he needed 4 second.

3 0

3 years ago