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Studentka2010 [4]

3 years ago

7

Trong thí nghiệm về giao thoa sóng trên mặt nước gồm hai nguồn kết hợp S1S2 cách nhau 15 cm với dao động với tần số 30Hz. Tốc độ

truyền vòng 48 cm/s.
a. Tìm khoảng cách giữa hai cực đại gần nhau nhất.
b Tìm số điểm dao động với biên độ cực tiểu giữa S1S2.
c. Trong vùng giao thoa điểm M trên mái nhất cách S1S2 lần lược các khoảng MS1 =1,5 cm và MS2=7,9 cm. M là cực đại hay cực tiểu và là đường thứ bao nhiêu so với đường trung trực của S1S2.

1 answer:

mojhsa [17]3 years ago

6 0

Answer:

this is a difficult question but I will try to answer it answer for this is 3220 a + b b u s y d l new

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A soccer player kicks a soccer ball with a force of 1.8 N. If the mass of the ball is .43 kg. How fast will the ball accelerate?

Vesnalui [34]

Answer:

The acceleration of the ball is 4.18 [m/s^2]

Explanation:

By Newton's second law we can find the acceleration of the ball

$F = m*a\\where:\\F = force applied [N] or [kg*m/s^2]\\m = mass of the ball [kg]\\a = acceleration [m/s^s]$

Now we have:

$a = F/m\\a = \frac{1.8 [kg*m/s^s]}{0.43[kg]} \\a = 4.18 [kg]$

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3 years ago

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Which statement is true of earths poles?

djverab [1.8K]

Answer choice d is correct

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3 years ago

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A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the o

lys-0071 [83]

Answer:

The work done by the force is 820.745 joules.

Explanation:

Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

$K_{1} + W_{F} = K_{2}$

Where:

$W_{F}$ - Work done by the external force, measured in joules.

$K_{1}$ , $K_{2}$ - Translational potential energy, measured in joules.

The work done by the external force is now cleared within:

$W_{F} = K_{2} - K_{1}$

After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:

$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$

Where:

$m$ - Mass of the object, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the object, measured in meters per second.

Now, each speed is the magnitude of respective velocity vector:

Initial velocity

$v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}$

$v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}$

$v_{1} \approx 25.060\,\frac{m}{s}$

Final velocity

$v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}$

$v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}$

$v_{2} \approx 33.121\,\frac{m}{s}$

Finally, if $m = 3.5\,kg$ , $v_{1} \approx 25.060\,\frac{m}{s}$ and $v_{2} \approx 33.121\,\frac{m}{s}$ , then the work done by the force is:

$W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right]$

$W_{F} = 820.745\,J$

The work done by the force is 820.745 joules.

6 0

3 years ago

How is heat transfer involved in the movement of tectonic plates?

Simora [160]

Mantle convection

Explanation:

The driver of moving plates is the mantle convection.

Convection is a heat transfer process that involves the movement of fluids from one place to another.

How does convection occur in the mantle?

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Over riding the asthenosphere is the brittle lithosphere made up of the upper mantle and the entirety of the crust.
The asthenosphere is at a higher temperature compared to the lithosphere on top.
In order to achieve stability and thermal equilibrium, the asthenosphere rises to the surface whereas the dense and cold lithosphere sinks down.
This way a convective cell is set up.
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Through this the plates moves.

Learn more:

Mantle brainly.com/question/9582362

#learnwithBrainly

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A striped billiard ball moves toward the right with speed 3 m/s. A solid billiard ball with the same mass moves toward the left

Helen [10]

Answer:

Final speed of striped ball is 3 m/s in left direction .

Explanation:

Given :

Two billiard ball with the same mass moves toward the left at the same speed 3 m/s .

Let , us assume right hand side direction to be positive and left hand side direction to be negative .

Also , let speed of ball after collision is (striped ball ) u and (solid ball) v .

It is also given that the collision is elastic .

Therefore , kinetic energy is conserved .

$\dfrac{m(3)^2}{2}+\dfrac{m(3)^2}{2}=\dfrac{mu^2}{2}+\dfrac{mv^2}{2}\\\\u^2+v^2=18$ ...... ( 1 )

Also , by conserving linear momentum .

We get :

$3m-3m=mu+mv\\u=-v$ ...... ( 2 )

Putting value of u from equation 2 to equation 1 .

We get :

$2v^2=18\\v=3\ m/s$

And , u = -3 m/s .

Therefore , final speed of striped ball is 3 m/s in left direction .

Hence , this is the required solution .

4 0

3 years ago