How are transverse and longitudinal waves the same?

What is the smallest particle that can completely represent water?

julsineya [31]

An atom would be your answer, so B!

4 0

3 years ago

The experimenter fom the video rotates on his stool, this time holding his empty hands in his lap. You stand on a desk above him

gtnhenbr [62]

Answer:

Explanation:

The experimenter is rotating on his stool with angular velocity ω ( suppose )

His moment of inertia is I say

We are applying no torque from outside . therefore , the angular momentum will remain the same

Thus angular momentum L = I ω = constant

Thus we can say I₁ ω₁ = I₂ω₂ = constant

here I₁ is the initial moment of inertia and ω₁ is the initial angular velocity

Similarly I₂ is the final moment of inertia and ω₂ is the final angular velocity

When a been bag is dropped on his lap , his moment of inertia increases due to increase in mass

In the above equation, when moment of inertia increases , the angular velocity decreases . So its motion of rotation will decrease .

7 0

3 years ago

Two blocks are sliding along a frictionless track. Block A (mass 4.03 kg) is moving to the right at 3.00 m/s. Block B (mass 4.84

julsineya [31]

Answer:

The total momentum of the system before the collision is 5.334 kg-m/s towards left.

Explanation:

Given that,

Mass of the block A, $m_A=4.03\ kg$

Speed of block A, $v_A=3\ m/s$

Mass of the block B, $m_B=4.8\ kg$

Mass of block B, $u_B=-3.6\ m/s$

Let p is the total momentum of the system before the collision. It is given by :

$p=m_Av_A+m_Bv_B\\\\p=4.03\times 3+4.84\times (-3.6)\\\\p=-5.334\ kg-m/s$

So, the total momentum of the system before the collision is 5.334 kg-m/s towards left. Hence, this is the required solution.

6 0

3 years ago

A ball rolls onto the path of your car as you drive down a quiet neighborhood street. To avoid hitting the child that runs to re

Ilya [14]

Answer:

a) F = -1035.385 N

b) Backwards

c) s = 15.60 m

Explanation:

Given information

$u$ = Initial Speed of Car = 15.0 m/s

$v$ = Final Speed of Car = 9.00 m/s

$t_{b}$ = Breaking Time = 1.30 s

$m$ = Mass of Car = 1040 kg

Part (a)

To find the force exerted on the car we use the following formula

$F = ma$

Where

$F$ = Force = unknown

$m$ = Mass of Car = 1040 kg

$a$ = Acceleration of Car / Deceleration of Car = unknown

To find the force (F) we need to first find the deceleration rate (a)

To find the deceleration rate we use the following formula

$a = \frac{v - u}{t}$

Inputting the given values

$a = \frac{15 - 9}{1.30} \\ a = -4.615$

To find the force

$F = ma \\ F = (1040)(-4.615) \\ F = (1040)(-4.615) \\ F = -1035.385 N$

Part (b)

Since the value of F is negative this means the the force was opposite the direction of motion, hence the force was backwards.

Part (c)

To find the total distance the car moved while braking we use the following formula

$v^2 = u^2 + 2as$

Where

$s$ = distance traveled

Inputting the values given

$(9)^2 = (15)^2 + 2(-4.615)s$

$s = 15.60 m$

6 0

3 years ago

What type of energy conversion occurs at the moment fireworks explode?

Paladinen [302]

chemical energy converts into light, heat and sound energy

6 0

4 years ago

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