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777dan777 [17]
3 years ago
12

A particle with charge 8 µC is located on the x-axis at the point −10 cm , and a second particle with charge 3 µC is placed on t

he x-axis at −8 cm . −10−8 −6 −4 −2 2 4 6 8 10 8 µ C 3 µ C − 3 µ C x → (cm) What is the magnitude of the total electrostatic force on a third particle with charge −3 µC placed on the x-axis at 2 cm
Physics
1 answer:
bixtya [17]3 years ago
6 0

Answer:Force on -7 uC charge due to charge placed at x = - 10cm

now we will have

towards left

similarly force due to -5 uC charge placed at x = 6 cm

now we will have

towards left

Now net force on 7 uC charge is given as

towards left

Explanation:

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A 0.335 kg mass is attached to a spring and executes simple harmonic motion with a period of 0.38 s. The total energy of the sys
kramer

Answer:

k=91.54 \frac{N}{m}

Explanation:

The angular speed is defined as the angle traveled in one revolution over the time taken to travel it, that is, the period. Therefore, it is given by:

\omega=\frac{2\pi}{T}\\\omega=\frac{2\pi}{0.38s}\\\omega=16.53\frac{rad}{s}

The angular frequency of the simple harmonic motion of the mass-spring system is defined as follows:

\omega=\sqrt{\frac{k}{m}}

Here, k is the spring's constant and m is the mass of the body attached to the spring. Solving for k:

\omega^2=\frac{k}{m}\\k=m\omega^2\\k=0.335kg(16.53\frac{rad}{s})^2\\k=91.54 \frac{N}{m}

6 0
3 years ago
In the figures, the masses are hung from an elevator ceiling. Assume the velocity of the elevator is constant. Find the tensions
Keith_Richards [23]

The elevator may be moving, but if it is moving at a constant velocity, then the observer viewing the mass-rope system is in an inertial reference frame (non-accelerating) and Newton's laws of motion will apply in this reference frame.

A) Choose the point where the ropes intersect (the black dot above m₁) and set up equations of static equilibrium where the forces are acting on that point:

We'll assume that, because rope 3 is oriented vertically, T₃ also acts vertically.

Sum up the vertical components of the forces acting on the point. We will assign upward acting components as positive and downward acting components as negative.

∑Fy = 0

Eq 1: T₁sin(θ₁) + T₂sin(θ₂) - T₃ = 0

Sum up the horizontal components of the forces acting on the point. We will assign rightward acting components as positive and leftward acting components as negative.

∑Fx = 0

Eq 2: T₂cos(θ₂) - T₁cos(θ₁) = 0

T₃ is caused by the force of gravity acting on m₁ which is very easy to calculate:

T₃ = m₁g

m₁ = 3.00kg

g is the acceleration due to earth's gravity, 9.81m/s²

T₃ = 3.00×9.81

T₃ = 29.4N

Plug in known values into Eq. 1 and Eq. 2:

Eq. 1: T₁sin(38.0) + T₂sin(52.0) - 29.4 = 0

Eq. 2: T₂cos(52.0) - T₁cos(38.0) = 0

We can solve for T₁ and T₂ by use of substitution. First let us rearrange and simplify Eq. 2 like so:

T₂cos(52.0) = T₁cos(38.0)

T₂ = T₁cos(38.0)/cos(52.0)

T₂ = 1.28T₁

Now that we have T₂ isolated, we can substitute T₂ in Eq. 1 with 1.28T₁:

T₁sin(38.0) + 1.28T₁sin(52.0) - 29.4 = 0

Rearrange and simplify, and solve for T₁:

T₁(sin(38.0) + 1.28sin(52.0)) = 29.4

1.62T₁ = 29.4

T₁ = 18.1N

Recall from our previous work:

T₂ = 1.28T₁

Plug in T₁ = 18.1N and solve for T₂:

T₂ = 1.28×18.1

T₂ = 23.2N

B) We'll assume that, because rope 2 is horizontally oriented, T₂ also acts horizontally.

Again, choose the point where the ropes intersect and write equations of static equilibrium involving the forces acting at that point:

Sum up the vertical components of the forces

∑Fy = 0

Eq. 3: T₁sin(θ₃) - T₃ = 0

Sum up the horizontal components of the forces

∑Fx = 0

Eq. 4: T₂ - T₁cos(θ₃) = 0

Right away we can solve for T₃, which is the force of gravity acting on m₂:

T₃ = m₂g, m₂ = 6.00kg, g = 9.81m/s²

T₃ = 6.00×9.81

T₃ = 58.9N

Plug in known values into Eq. 3:

T₁sin(61.0) - 58.9 = 0

We can solve for T₁ now that is is the only unknown value in this equation

0.875T₁ = 58.9

T₁ = 67.3N

Plug in known values into Eq. 4:

T₂ - 67.3cos(61.0) = 0

We can solve for T₂ now that it is the only unknown value in this equation

T₂ = 67.3cos(61.0)

T₂ = 32.6N

6 0
3 years ago
A 1.8-m uniform rod is being used to balance two buckets of paint. One pucket is full, but the other is mostly empty, and has on
Lisa [10]

Answer: 0.36 m

Explanation:

By definition, the x coordinate of the center of mass of the system obeys the following equation:

Xm = (m1x1 + m2x2 + …….+ mnxn) / m1+m2 +……+ mn

Neglecting the mass of the rod, and choosing our origin to be coincident with the location of the full bucket, we can write the following expression for the X coordinate of the center of mass (Assuming that both masses are aligned over the x-axis, so y-coordinates are zero):

Xm = 0.25 mb . 1.8 m / (1+0.25) mb

Simplifying mb, we get:

Xm= 0.36 m (to the right of the full bucket).

6 0
3 years ago
A body of rotational inertia 1.0 kg m2 is acted upon by a torque of 2.0 Nm. The angular acceleration of the body will be
lubasha [3.4K]

Answer:

The angular acceleration is 2 rad/s².

Explanation:

Given that,

Rotational inertia = 1.0 kg m²

Torque = 2.0 Nm

We need to calculate the angular acceleration

Using formula of torque

\tau=I\alpha

\alpha=\dfrac{\tau}{I}

Where, I = moment of inertia

\tau = torque

Put the value into the formula

\alpha=\dfrac{2.0}{1.0}

\alpha=2\ rad/s^2

Hence, The angular acceleration is 2 rad/s².

5 0
3 years ago
When a liquid is cooled, the kinetic energy of the particles . The force of attraction between the particles , the space between
NARA [144]

Answer:

The kinetic energy is less.

The force of attraction increase

The space between the particles decrease.

Explanation:

Temperature is a measure of the kinetic energy of the particles of a substance. When you increase its temperature, also its kinetic energy increases, and vice-versa. That's why te kinetic energy is less.

In solids, the intermolecular forces are stronger than in liquids. The constituent particles are more packed also. That is the reason why the force of attraction increase and the space between the particles decrease.

4 0
3 years ago
Read 2 more answers
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