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3 years ago

1. Velocity and speed are both calculated by using what formula?

Physics

1 answer:

ValentinkaMS [17]3 years ago

5 0

Explanation:

Velocity (v) is a vector quantity that measures displacement (or change in position, Δs) over the change in time (Δt), represented by the equation v = Δs/Δt. Speed (or rate, r) is a scalar quantity that measures the distance traveled (d) over the change in time (Δt), represented by the equation r = d/Δt.

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To practice Problem-Solving Strategy 21.1 Coulomb's Law. Three charged particles are placed at each of three corners of an equil

Salsk061 [2.6K]

Answer:

The force felt by charge 3 is F=(-5.6*10⁻⁶,3.36⁻⁵)N

Explanation:

As the superposition principle applies to static charges, we can find the net electric force as the sum of the two forces felt by q3.

Looking at the drawing and knowing that they form an equilateral triangle of lenght 4 we can conclude that each internal angle is 60°.

So, the positions in our coordinate system are:

$r_1=(0,0)\\r_2=(4\ cos(60\°),4\ sin(60\°))\\r_3=(4,0)\\$

Now using Coulomb's force:

$F_{ij}=\frac{-kq_iq_j}{d^2}(\vv{r}_j-\vv{r}_i)$

Where d=4, q1 = -7.8*10⁻⁹C, q2 = -15.6 *10⁻⁹C, q3 = 8.0 *10⁻⁹C, k=8.98*10⁹, e0=8.8*10¹⁰:

Replacing we get 2 equations:

$F_{13}=\frac{-kq_1q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(-0.04\ cos(60\°),-0.04\ sin(60\°))\\\\F_{23}=\frac{-kq_2q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(0.04-0.04\ cos(60\°),-0.04\ sin(60\°))\\$

To work with the sam

F=∑F_i=3.5*10⁻⁴(0.023,0.032)+7*10⁻⁴(-0.016,0.032)=

=((3.5*10⁻⁴-7*10⁻⁴)*0.016,(3.5*10⁻⁴+7*10⁻⁴)*0.032)=

F=(-5.6*10⁻⁶,3.36⁻⁵)N

7 0

3 years ago

A child has an ear canal that is 1.3 cm long. At what sound frequencies in the audible range will the child have increased heari

ElenaW [278]

Answer:

The answer to this is

6600 Hz to 19,800 Hz

Explanation:

The shape of the human ear is analogous to a closed ended pipe hence

we have λ = 4L or wavelength = 4 * length of the child ear

The frequency c/λ where c = speed of sound = 343 m/s

hence the child's audible range is multiples of 343/(4*0.13) =6600Hz

or 13200 Hz or 19,800 Hz

The generally quoted range of human hearing is 20 Hz to 20 kHz

7 0

3 years ago

A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min, clockwise (when viewed from above). When swit

Ratling [72]

Answer:

The magnitude of angular acceleration is $232.38\ rad/s^2$ .

Explanation:

Given that,

Initial angular velocity, $\omega_i=3500\ rev/min=366.5\ rad/s$

When it switched off, it comes o rest, $\omega_f=0$

Number of revolution, $\theta=46=289.02\ rad$

We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{-(366.51)^2}{2\times 289.02}\\\\\alpha =-232.38\ rad/s^2$

So, the magnitude of angular acceleration is $232.38\ rad/s^2$ . Hence, this is the required solution.

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3 years ago

Humberto builds two circuits using identical components.Circuit 1: A series circuit with three lightbulbs Circuit 2: A parallel

PolarNik [594]

<span>Humberto builds two circuits using identical components,
and then adds components to each circuit.

Circuit 1:
A series circuit with three lightbulbs.
Then add three more lightbulbs in series.

Circuit 2:
A parallel circuit with three lightbulbs
Then add two more lightbulbs on new branches
in parallel with each original bulb.

After adding the new lightbulbs in Circuit 1:
-- the voltage across each of the original bulbs is less,
-- the current through the whole series circuit is less,
-- the original three bulbs shine dimmer than before, and
-- the total power delivered from the battery is less.
-- The battery lasts longer.

After adding the new lightbulbs in Circuit 2:
</span>-- the voltage across each of the original bulbs is doesn't change,
-- the current through each original bulb doesn't change,
-- the original three bulbs shine just as bright as before,
-- the total currrent drawn by the circuit, and the total current
delivered by the battery, increases, and
-- the total power delivered from the battery increases.
-- The battery runs down sooner.

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3 years ago

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What happens to the pwc when the throttle is released?

Sergio [31]

PWC operators must keep in mind that a jet drive needs moving water through the drive nozzle for maneuverability. In other term, you must have power applied in order to uphold steering control. If the engine shuts off during operation or If you release the throttle to idle, you will lose all steering control. In either situation, the PWC will remain in the direction it was headed before the throttle was released or the engine was shut off. Action of the steering control will have no outcome. If you are approaching a dock, shore, or other vessel at a speed bigger than you can control and you release the throttle to idle or shut off the engine, you won’t have maneuvering capability and the PWC will stay its forward movement.

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4 years ago