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goldenfox
3 months ago
11

A dipole of moment 0.5 e·nm is placed in a uniform electric field with a magnitude of 8 times 104 N/C. What is the magnitude of

the torque on the dipole when (a) the dipole is parallel to the electric field,
Physics
1 answer:
Ostrovityanka [3.2K]3 months ago
8 0

Answer:

The resultant torque is zero.

Explanation:

Assuming the dipole consists of two equal but opposite charges e, it can be represented by a rod with one end featuring a charge e and the other end with -e. Since the dipole is aligned with the electric field, both charges experience forces that are parallel to this electric field. Consequently, there are no force components that act perpendicular to the rod, which is necessary for torque to occur.

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A 4.5-m-long wooden board with a 24-kg mass is supported in two places. One support is directly under the center of the board, a
Ostrovityanka [3204]

All the weight of the wooden board rests solely on the support situated at the center of the rod, while the other support positioned at one end experiences no reaction force, resulting in a 0 reaction force.

Thus, the reaction force at the central support corresponds to the weight of the board, whereas the end support has 0 reaction force.

8 0
2 months ago
Read 2 more answers
A large box of mass m sits on a horizontal floor. You attach a lightweight rope to this box, hold the rope at an angle θ above t
inna [3103]

Answer:

The answer to the specified question will be "\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}".

Explanation:

Referring to the question,

\sum F_{x}

⇒  TCos \theta-F_{s}=0

⇒  T_{m}Cos \theta =F_{s}...(equation 1)

\sum F_{y}

⇒  TSin \theta+F_{N}=m_{g}

⇒  M_{g}-TSin \theta=F_{N}...(equation 2)

Now,

From equation 1 and equation 2, we conclude

⇒  T_{m} Cos \theta = \mu_{s}F_{N}

By substituting the value of F_{N}, we derive

⇒  T_{m} Cos\theta = \mu_{s}(M_{g}-T_{m}Sin \theta)

⇒  \mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}

4 0
2 months ago
The flight of a kicked football follows the quadratic function f(x)=−0.02x2+2.2x+2, where f(x) is the vertical distance in feet
Keith_Richards [3271]
The ball covers a horizontal distance of 0.902 meters. The trajectory of a kicked football adheres to a quadratic equation expressed as: f(x), where f(x) indicates the vertical distance in feet, and x signifies how far the ball travels horizontally. To compute the distance the ball will advance before striking the ground, we set the condition f(x) = 0. Upon solving this quadratic equation, we find that the horizontal distance traveled by the ball is: x = -0.902 meters, leading us to conclude that it travels 0.902 meters across the field.
7 0
2 months ago
What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t
Yuliya22 [3333]

Details that are not provided: the problem figure is included.

We can address the exercise by applying Poiseuille's law. This law indicates that for a fluid flowing in a laminar manner within a confined pipe,

\Delta P = \frac{8 \mu L Q}{\pi r^4}

where:

\Delta P represents the pressure difference across the two ends

\mu denotes the viscosity of the fluid

L signifies the length of the pipe

Q=Av indicates the volumetric flow rate, where A=\pi r^2 is the cross-sectional area of the tube and v refers to the fluid's velocity

r stands for the pipe's radius.

This law can be utilized for the needle, allowing us to compute the pressure difference between point P and the needle's end. In this scenario, we have:

\mu=0.001 Pa/s is the dynamic viscosity of water at 20^{\circ}

L=4.0 cm=0.04 m

Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s

and r=1 mm=0.001 m

Substituting these values into the formula yields:

\Delta P = 3200 Pa

This pressure difference specifies the value between point P and the needle's termination. As the end of the needle is under atmospheric pressure, the gauge pressure at point P, relative to atmospheric pressure, is exactly 3200 Pa.

8 0
3 months ago
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