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4 years ago

What symbol is used when representing momentum in equations?

1 answer:

6 0

Almost in every Physics books, momentum is denoted by P with small arrow(→) pointing towards right, placed on top of P.
I am guessing you asked about linear momentum.

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Initially, a particle is moving at 5.25 m/s at an angle of 35.5Â° above the horizontal. Three seconds later, its velocity is 6.0

ivolga24 [154]

Answer:

a =( -0.32 i ^ - 2,697 j ^) m/s²

Explanation:

This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.

Break down the speeds in two moments

initial

v₀ₓ = v₀ cos θ

v₀ₓ = 5.25 cos 35.5

v₀ₓ = 4.27 m / s

$v_{oy}$ = v₀ sin θ

$v_{oy}$ = 5.25 sin35.5

$v_{oy}$ = 3.05 m / s

Final

vₓ = 6.03 cos (-56.7)

vₓ = 3.31 m / s

$v_{y}$ = v₀ sin θ

$v_{y}$ = 6.03 sin (-56.7)

$v_{y}$ = -5.04 m / s

Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order

a = ( $v_{f}$ - v₀) /t

aₓ = (3.31 -4.27)/3

aₓ = -0.32 m/s²

$a_{y}$ = (-5.04-3.05)/3

$a_{y}$ = -2.697 m/s²

6 0

4 years ago

The sun rose at 6:54 this morning.<br>is this an observation or inference?

jeka57 [31]

Observation: You saw/ observed something and you are stating that.

Inference: You are predicting something to be true. If so and so, then this will occur.

4 0

3 years ago

Read 2 more answers

Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm

8 0

3 years ago

A very long solid insulating cylinder has radius R = 0.1 m and uniform charge density rho0= 10-3 C/m3. Find the electric field a

Galina-37 [17]

Answer:

$E = (0.56 \times 10^8 ) r \ \ N/c$

Explanation:

Given that:

$\rho_o = (10^{-3} ) \ c/m^3$

R = (0.1) m

To find the electric field for r < R by using Gauss Law

${\oint}E^{\to}* da^{\to} = \dfrac{Q_{enclosed}}{\varepsilon_o} --- (1)$

For r < R

$Q_{enclosed}=(\rho) ( \pi r^2 ) l$

$E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}$

$E= \dfrac{\rho ( r)}{2 \varepsilon_o}$

where;

$\varepsilon_o = 8.85 \times 10^{-12}$

$E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}$

$E = (0.56 \times 10^8 ) r \ \ N/c$

4 0

3 years ago

What are some ways to conserve electricity

emmasim [6.3K]

Turn off lights when leaving rooms.

Unplug unused appliances. Even when powered off these appliances use electricity.

Replace regular light bulbs with energy saving bulbs.

7 0

3 years ago

Read 2 more answers