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3 years ago

How much Tim in minutes will it take a car driving at 90km/hr to travel 27 kilometers

Physics

1 answer:

astraxan [27]3 years ago

6 0

Answer:

18min

Explanation:

v=d/t

t=d/v= 27/90 =0.3hrs =18min

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A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap

viktelen [127]

Answer:

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

$V_r + V_L + V_c = V_{net}$

now we will have

$iR + L\frac{di}{dt} + \frac{q}{C} = 12 V$

now we have

$1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12$

So we will have

$q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}$

at t = 0 we have

q = 0

$0 = 3\times 10^{-6} + c_1 + c_2$

also we know that

at t = 0 i = 0

$0 = -4000 c_1 - 1000c_2$

$c_2 = -4c_1$

$c_1 = 1 \times 10^{-6}$

$c_2 = -4 \times 10^{-6}$

so we have

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

5 0

3 years ago

You need to measure the lengthof you driveway.which measuring tool would produce accurate results that would expect to be the sa

Aleonysh [2.5K]

C a meter stick with only centimeters

D a ruler with millimeters and centimeters

D would be to the nearest half milimeter. Take some time to measure with a 2 inch ruler. Would you really need to know the length to half a mil ?

3 0

3 years ago

Use the general formulas for gravitational force and centripetal force to derive the relationship between speed (v) and orbital

german

Solution :

We know that :

Formula for Gravitational force is given by :

$F_g=\frac{Gmn}{r^2}$

where, G is the gravitational constant

M is the mass of the bigger body

m is the mass of the smaller body

r is the distance between the two bodies.

And the formula for the centripetal force is given by :

$F_c=\frac{mv^2}{r}$

where, m is the mass of the rotating body

v is the velocity

r is the radius of rotation of the body.

We know that mathematically, the gravitational force is equal to the centripetal force of the body.

Therefore,

$F_g=F_c$

$\frac{GMm}{r^2}=\frac{mv^2}{r}$

$\sqrt{\frac{GM}{r}}=v$

Hence derived.

5 0

3 years ago

URGENT PLEASE HELP!!!!

AlexFokin [52]

It just completely stops kinetic energy is movement so technically it would stop.

6 0

3 years ago

Someone please help? :)

Gnoma [55]

Answer:

(A) $F_N - mg\cos\theta$

Explanation:

The net force perpendicular to the surface of the incline is the sum of the gravity force component, which is mgcos(theta), and the reactionary normal force caused by the surface of the incline. The sum is F_N - mgcos(theta) and is usually 0 which is why the object is not moving perpendicularly to the surface of the incline.

8 0

4 years ago