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k0ka [10]
3 years ago
6

Determine the answer to the equation 30 km/h × 17 h =

Physics
1 answer:
Jobisdone [24]3 years ago
5 0
30 km/h * 17 h =  30*17  km/h  *h
                         =    510 km
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How to find instantaneous velocity physics?
evablogger [386]
You should have the velocity as a function of time either given explicitly or implicitly (a graph)

v = ds/dt  (differentiating the position vector)

integrating the acceleration.

you can use impulse or work and energy principle and also newton law of motion to find acceleration then velocity


NOT SURE IF THAT WHAT YOU WANT.
 
8 0
3 years ago
A capacitor is connected to an ac generator that has a frequency of 2.3 kHz and produces a rms voltage of 1.5 V. The rms current
miv72 [106K]

Answer:

Explanation:

Given that,

AC frequency of 2.3KHz

f=2.3×10³Hz

Vrms produce is

Vrms=1.5V

Current rms

Irms= 31mA

The capacitor is reconnected to a generator of frequency

f=4.8KHz =4800Hz

The current rms becomes

Irms= 85mA

Vrms=?

Solution

First genrator

The capacitive reactance is given as

Xc=Vrms/Irms

Xc=1.5/31×10^-3

Xc=48.39 ohms

Now, to know the capacitance of the capacitor

Xc=1/2πfC

Then,

C=1/2πfXc

So,

C=1/2×π×2300×48.39

C=1.43×10^-6C

C=1.43μF

Note: the capacitance of the capacitor did not change,

Now for generator two.

The reactance are given as

Xc=1/2πfC

Xc=1/2×π×4800×1.43×10^-6

Xc=23.19ohms

Then,

Vrms2=Irms2 ×Xc

Vrms2=85×10^-3×23.19ohms

Vrms2=1.97V

Vrms2=1.97Volts

7 0
3 years ago
Sound waves are also called compression waves. This means that as the wave travels through air, the ____ in creases and decrease
Alja [10]
The speed creases and decreases
5 0
3 years ago
A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li
Kitty [74]

Answer:

E_{net} = 6.44 \times 10^5 N/C

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

E = \frac{2k \lambda}{r}

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}

now plug in all data

\lambda_1 = 4.68 \muC/m

\lambda_2 = 2.48 \mu C/m

r = 0.200 m

now we have

E_{net} = \frac{2k}{r}(4.68 + 2.48)

E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})

E_{net} = 6.44 \times 10^5 N/C

8 0
3 years ago
I need help ASAP! It's urgent ​
djverab [1.8K]

Answer:

hope this helps you

3 0
3 years ago
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