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zysi [14]
3 years ago
9

I NEED HELP PLEASE, THANKS! :)

Physics
1 answer:
Rainbow [258]3 years ago
4 0

Explanation:

Usually when we think of waves, we think of transverse waves.  These are waves where points move up and down perpendicular to the motion of the wave.  Examples include water waves, whipping a rope, or even doing the "wave" in a crowd.  You can think of these as "two dimensional" waves.

Longitudinal waves are waves where points move left or right, parallel to the motion of the wave.  In other words, there is compression and expansion of the medium.  Examples include sound waves, or pulses in a slinky.

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Acceleration increases

Explanation:

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he will use kinetic energy to break through the door

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because kinetic energy is movement and force

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3 years ago
Nathan is walking to the store and sees a snake slithering across the sidewalk. He jumps over it with an initial vertical veloci
Travka [436]

Answer:

0.62\:\mathrm{s}

Explanation:

Since the universal SI unit for velocity is meters/second, let's convert ft/s to m/s:

10\:\mathrm{ft/s}=3.048\:\mathrm{m/s}

We can use the following kinematics equation to solve this question:

v_f=v_i+at

What we know:

  • The initial velocity, v_i, is 3.048\:\mathrm{m/s}
  • (physics concept) The final velocity must be equal in magnitude but opposite in direction to the initial velocity (v_f=-3.048\:\mathrm{m/s})
  • Acceleration, a, is acceleration due to gravity at about 9.8\:\mathrm{m/s}

Solving for t:

-3.048=3.048+(-9.8t),\\-6.096=-9.8t,\\t=\frac{-6.096}{-9.8}\approx \boxed{0.62\:\mathrm{s}}

8 0
2 years ago
Two capacitors with capacitances of 3.25 and 5.00 μF, respectively, are connected in parallel. The system is connected to a 55-V
Nataliya [291]

The charge accumulated in 3.25 μF capacitor is 178.75 μC.

Answer:

Explanation:

In parallel connection, the voltage drop across any passive devices like capacitor or resistor will be constant. So the current flow will be varying in case of parallel connections of capacitors or resistors.

As the capacitance is the measure of amount of charge or current generated for a given amount of voltage, it is directly proportional to the charge or current and inversely proportional to voltage.

C = Q/V

Here the charge accumulated in a capacitor of capacitance 3.25 microfarad need to be determined which is in parallel connection with another capacitor. So the voltage through both the capacitor will be equal to the voltage of the battery which is stated as 55 V.

3.25×10^{-6} = Q/55

Q = 3.25 * 55 = 178.75 μC

So the charge accumulated in 3.25 μF capacitor is 178.75 μC.

7 0
3 years ago
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