Calculate the acceleration of the car which moves 36m/s in 8s

Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50

defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

8 0

3 years ago

John’s mass is 95.6 kg, and Barbara’s is 55.3 kg. He is standing on the x axis at xJ = +10.9 m, while she is standing on the x a

Anna11 [10]

Answer:Shifted towards Left by distance of 2.243 m

Explanation:

Given

Mass of john $m_1=95.6 kg$

Mass of barbara $m_2=55.3 kg$

John is standing at $x=10.9 m$

Barbara is standing at $x=2.50 m$

$x_{com}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

$x_{com}=\frac{95.6\times 10.9+55.3\times 2.5}{95.6+55.3}$

$x_{com}=\frac{1180.29}{150.9}$

$x_{com}=7.821 m$

Now if they change their Position then

$x'_{com}=\frac{95.6\times 2.5+55.3\times 10.9}{95.6+55.3}$

$x'_{com}=\frac{841.77}{150.9}$

$x'_{com}=5.578$

Thus we can see that center of mass shifted towards left by a distance of $2.243 m$ because heavier is shifted towards left

8 0

3 years ago

Starting from the front door of your ranch house, you walk 50.0 m due east to your windmill, and then you turn around and slowly

Hunter-Best [27]

Answer:

Average velocity

$v=\frac{d}{t}\\ v=\frac{10m}{70s}\\v=.1428 \frac{m}{s}$

Average speed,

$S=\frac{D}{t}\\ S=\frac{90}{70}\\ S=1.29\frac{m}{s}$

Explanation:

(a)Average velocity

We have to find the average velocity. We know that velocity is defined as the rate of change of displacement with respect to time.

To find the average velocity we have to find the total displacement.

since displacement along east direction is 50m

and displacement along west=40m

so total displacement,

$d=50m-40m\\d=10m$

total time,

$t=28 s+42 s\\t=70 s$

therefore, average velocity

$v=\frac{d}{t}\\ v=\frac{10m}{70s}\\v=.1428 \frac{m}{s}$

(b)Average Speed:

Average speed is defined as the ratio of total distance to the total time

it means

Average speed= total distance/total time

here total distance,

$D= 50m+40m\\D=90m$

and total time,

$t= 28s+40s\\t=70s$

therefore,

Average speed,

$S=\frac{D}{t}\\ S=\frac{90}{70}\\ S=1.29\frac{m}{s}$

8 0

3 years ago

A sea breeze occurs because warm air rises on land and moves toward the ocean to cool. The cool air then moves from the ocean to

Norma-Jean [14]

Im going to go with a. True

7 0

2 years ago

Read 2 more answers

A bullet fired horizontally hits the ground in 0.5 sec. If it had been fired with a much higher speed in the same direction, and

stira [4]

Answer:

3. 0.5 sec.

Explanation:

A bullet fired horizontally follows a projectile motion, which consists of two independent motions:

- A horizontal motion with constant speed

- A vertical motion with constant acceleration, g = 9.8 m/s^2, towards the ground

The time taken for the bullet to reach the ground can be calculated just by considering the vertical motion:

$y(t) = h + v_{0y} t - \frac{1}{2}gt^2$

where y is the vertical position at time t, h is the initial height, and $v_{0y}$ is the initial vertical velocity of the bullet.

Since the bullet is fired horizontally, $v_{0y}=0$ . So the equation becomes

$y(t) = h - \frac{1}{2}gt^2$

And the time that the bullet takes to reach the ground can be found by requiring y=0 and solving for t:

$t=\sqrt{\frac{2h}{g}}$

As we can see, in this equation there is no dependance on the initial speed of the bullet: therefore, if the bullet is fired still horizontally but with a different speed, it will still take the same time (0.5 s) to reach the ground.

4 0

3 years ago