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14 days ago

To understand the forces between a bar magnet and 1. a stationary charge, 2. a moving charge, and 3. a ferromagnetic object. A b

ar magnet oriented along the y axis can rotate about an axis parallel to the z axis. Its north pole initially points along j^.

Physics

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An object is released from rest near and above Earth’s surface from a distance of 10m. After applying the appropriate kinematic

serg [3582]

Answer:

v_y = 12.54 m/s

Explanation:

Given values:

- Initial vertical height y_o = 10 m

- Initial velocity v_y,o = 0 m/s

- The object's acceleration in the air = a_y

- The actual time taken to reach the ground t = 3.2 s

Find:

- How to calculate the object's speed when it arrives at the ground?

Solution:

- Apply kinematic equations to find the actual acceleration of the ball when it reaches the ground:

y = y_o + v_y,o*t + 0.5*a_y*t^2

0 = 10 + 0 + 0.5*a_y*(3.2)^2

a_y = - 20 / (3.2)^2 = 1.953125 m/s^2

- Use the total energy conservation principle of the system:

E_p - W_f = E_k

Where, E_p = m*g*y_o

W_f = m*a_y*(y_i - y_f)..... Reflecting air resistance

E_k = 0.5*m*v_y^2

Thus, m*g*y_o - m*a_y*(y_i - y_f) = 0.5*m*v_y^2

g*(10) - (1.953125)*(10) = 0.5*v_y^2

v_y = sqrt(157.1375)

v_y = 12.54 m/s

4 0

3 months ago

A certain carbon monoxide molecule consists of a carbon atom mc = 12 u and an oxygen atom mo = 17 u that are separated by a dist

Maru [3345]

a) x_{cm} = m₂/ (m₁ + m₂) d, b) x_{cm} = 52.97 pm

7 0

1 month ago

The magnitude of the electrical force acting between a +2.4 × 10–8 C charge and a +1.8 × 10–6 C charge that are separated by 0.0

kicyunya [3294]

To solve this problem, Coulomb's law will be applied as follows:
F = k*q1*q2 / r^2 where:
F indicates the force magnitude between the charges
k is a constant = 9.00 * 10^9 N.m^2/C^2
q1 = +2.4 × 10–8 C
q2 = +1.8 × 10–6 C
r represents the distance separating the charges = 0.008 m

By substituting these values, we derive:
F = (9*10^9)(2.4*10^-8)(1.8*10^-6) / (0.008)^2 = 6.075, which rounds to 6.1 Newtons

8 0

2 months ago

Read 2 more answers

In an amusement park rocket ride, cars are suspended from 3.40-m cables attached to rotating arms at a distance of 5.90 m from t

ValentinkaMS [3465]

Answer:

The rotational angular speed is measured at 1.34 rad/s.

Explanation:

Considering the following parameters,

Length = 3.40 m

Distance = 5.90 m

Angle = 45.0°

We are tasked with finding the angular speed of rotation

Using the balance equation

Horizontal component

$T\cos\theta=mg$

$T=\dfrac{mg}{\cos\theta}$

Vertical component

$T\sin\theta=m\omega^2 r$

Substituting the tension value

$mg\tan\theta=m\omega^2(d+L\sin\theta)$

$\omega=\sqrt{\dfrac{g\tan\theta}{(d+L\sin\theta)}}$

Substituting the value into the equation

$\omega=\sqrt{\dfrac{9.8\tan45.0}{5.90+3.40\sin45.0}}$

$\omega=1.34\ rad/s$

Thus, the angular speed of rotation computes to 1.34 rad/s.

7 0

3 months ago

A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w

Keith_Richards [3271]

25.82 m/s Explanation: Given: Force applied by the baseball player; F = 100 N Distance the ball travels; d = 0.5 m Mass of the ball; m = 0.15 kg To find the velocity at which the ball is released, we will equate the work done with the kinetic energy involved. It's important to recognize that work done reflects the energy the baseball player has used. Thus, the relationship can be represented as follows: F × d = ½mv² 100 × 0.5 = ½ × 0.15 × v² Solving gives: v² = (2 × 100 × 0.5) / 0.15 v² = 666.67 v = √666.67 v = 25.82 m/s.

4 0

2 months ago