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3 years ago

In the simulation above, as the projectile travels downward, how does the vertical velocity change?

Physics

2 answers:

NARA [144]3 years ago

7 0

B) is the correct answer

e-lub [12.9K]3 years ago

7 0

As the projectile travels downward, the vertical velocity increases.

Answer: Option D

Explanation:

If the object was thrown up, it will tend to reach the ground to the gravitational pull. So, exhibiting a projectile motion with an initial velocity, the object travels some distance in air with constant horizontal velocity and varying vertical velocity.

When the object is moving downward, the gravitational force will be more leading to increase in the acceleration and vertical velocity of the object. Thus, the vertical velocity will increase due to the increase in gravitational force acting on the projectile.

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3 years ago

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6) Set the battery to a value between 0.0 V and 1.5 V. Now drag the voltage meter toward the capacitor and move the red and blac

KIM [24]

Answer:

A) 1.5 v

B) Top plate is at higher voltage than the bottom plate

Explanation:

Battery value set between 0.0 V and 1.5 V

a) The potential difference between the plates

Δ V = V1( potential at top plate) - V2( potential at lower plate )

potential at top plate = 1.5 V

potential at lower plate = 0.0 V

hence potential difference = 1.5 V

b ) The top plate is always connected to the positive terminal of the DC source ( which is at a higher potential )while the bottom plate is connected to the negative terminal of the DC source ( which is at a lower potential )

hence the Top plate is at higher voltage than the bottom plate

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3 years ago

All Physicsts over here plz help in these questions!!!!!!!!!

mrs_skeptik [129]

Hello!

First one we can use that PE=mgh so we have

4.37*10^5J/(9.12*10^3kg*9.80m/s^2)= 4.89m

Second one we can use Newton’s Second Law
F=ma and in this case F=mg so we have

g= 3.28*10^-2N/6*10^-3kg = 5.47m/s^2

Hope this helps. Any questions please ask. Thank you.

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3 years ago

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Say I have a series circuit with 20v and four 65 ohm resistors, what is the current in each resistor?

Komok [63]

Data:

$E = 20 V$
$R_{1} = 65\Omega$
$R_{2} = 65\Omega$
$R_{3} = 65\Omega$
$R_{4} = 65\Omega$

Now that we have all the values we need properly identified, simply calculate the equivalent total resistance of the circuit and the intensity of the total electric current using the Ohm's Law:

 $R_{T} = R_{1} + R_{2} + R_{3} + R_{4}$
$R_{T} = 65 + 65 + 65+ 65$
$R_{T} = 260\Omega$

Like this:

$I_{T} = \frac{E}{ R_{T} }$

$I_{T} = \frac{20}{ 260 }$
$I_{T} = 0,076923076...$

$\boxed{\boxed{I_{T} \approx 0,07A}}$
Answer:
The intensity of the total electric current
 $\boxed{\boxed{I_{T} \approx 0,07A}}$

P.S:. Since the association is in series, the current of 0.07A is the same for all resistors.

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