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2 years ago

13

Which is an example of a wave?

2 answers:

larisa [96]2 years ago

7 0

A Is the correct answer

Travka [436]2 years ago

4 0

Answer:

The answer is A

Explanation:

Someone shouting to you from across a room

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the modern periodic law, according to Moseley, classifies elements based on arranging elements in ___ order

Amanda [17]

increasing atomic number

4 0

4 years ago

Which factors control the cooling time of magma within the crust?

Oxana [17]

Answer:

The depth of the intrusion,

the shape and size of a magma body

Explanation:

The depth of the intrusion, the shape and size of a magma body: the greater the surface area for a given volume of intrusion, the faster it cools, and the presence of circulating ground water: water passing through magma absorbs and carries away heat

8 0

3 years ago

Read 2 more answers

A bullet moving at a speed of 152 m/s passes through a plank of wood at 128m/s. Another bullet moving at 97m/s passes through th

Ivahew [28]

Answer:

The speed of the plank is 81.68 m/s

Explanation:

Given that,

Speed of bullet = 152 m/s

Speed of wood = 128 m/s

Speed of another bullet = 97 m/s

We need to calculate the speed of plank

Using conservation of momentum

$m_{1}u_{1}=(m_{1}+m_{2})v$

Where,

u = initial velocity

v = final velocity

$152m_{1}=(m_{1}+m_{2})128$ ....(I)

$m_{1}97=(m_{1}+m_{2})v$ ....(II)

From equation(I) and equation(II)

$\dfrac{152}{97}=\dfrac{128}{v}$

$= \dfrac{128\times97}{152}$

$v=81.68\ m/s$

Hence, The speed of the plank is 81.68 m/s

4 0

4 years ago

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0

3 years ago

Will mark brainliest

sweet-ann [11.9K]

Answer: Wouldn't it just be her blocks all walked in an hour added together?

Explanation: 5+2+3+2=12 so 12 blocks an hour?

3 0

3 years ago