A truck and a car hitting a wall

(PLEASE HELP) Which of the following describes a mixture?

enyata [817]

Answer:

D

Explanation:

3 0

3 years ago

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A +8.75 μC point charge is glued down on a horizontal frictionless table. It is tied to a -6.50 μC point charge by a light, nonc

lisov135 [29]

(a) The tension on the wire when the two charges have opposite signs is 383.5 N.

(b) The tension on the wire if both charges were negative is 3.640.25 N.

The given parameters;

first charge, q₁ = 8.75 μC
second charge, q₂ = -6.5 μC
electric field, E = 1.85 x 10⁸ N/C
distance between the two charges, r = 2.5 cm

(a)

The attractive force between the charges is calculated as follows;

$F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = -819 \ N$

The force on the negative charge due to the electric field is calculated as follows;

$F_2 = Eq_2\\\\F_2 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_2 = 1202.5 \ N$

The tension on the wire is the resultant of the two forces and it is calculated as follows;

$T = F_2 + F_1\\\\T = 1202.5 - 819\\\\T = 383.5 \ N$

(b) when the two charges are negative

The repulsive force between the two charges is calculated as follows;

$F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (-8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = 819 \ N$

The force on the first negative charge due to the electric field is calculated as follows;

$F_2 = Eq_1\\\\F_2 = (1.85 \times 10^8)\times (8.75 \times 10^{-6})\\\\F_2 = 1618.75 \ N$

The force on the second negative charge due to the electric field is calculated as follows;

$F_3 = Eq_2\\\\F_3 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_3 = 1202.5 \ N$

The tension on the wire is the resultant of the three forces and it is calculated as follows;

$T= F_1 + F_2 + F_3\\\\T= 819 + 1618.75 + 1202.5\\\\T = 3,640.25 \ N$

Learn more here:brainly.com/question/19565286

5 0

3 years ago

A car slows down at -5.00 m/s² until it comes to a stop after travelling 15.0 m. What was the initial speed of the car?

lapo4ka [179]

<h2>Answer: 12.24m/s</h2>

According to kinematics this situation is described as a uniformly accelerated rectilinear motion. This means the acceleration while the car is in motion is constant.

Now, among the equations related to this type of motion we have the following that relates the velocity with the acceleration and the distance traveled:

$V_{f}^{2}-V_{o}^{2}=2.a.d$ (1)