What type of circuit is illustrated? PLEASE HELPPPP

An action that has the ability to change an object’s state of motion is

Aleksandr-060686 [28]

Answer:

An unbalanced force

Explanation:

7 0

2 years ago

In a 35 mm single lens reflex camera (SLR) the distance from the lens to the film is varied in order to focus on objects at vary

Evgesh-ka [11]

Answer:

range of movement is 1.49 mm

Explanation:

given data

focal length = 45 mm

distance = 1.4 m

distance from the lens = 35 mm

distance from infinity down = 1.4 m =

to find out

range of movement

solution

we will apply here lens equation that is

1/f = 1/p + 1/q

here f = 45 and p = infinity to 1400 mm

we find here image distance that is q

1/45 = 0 + 1/q ......1

q = 45 mm

and

1/45 = 1/1400 + 1/q ......2

q = 46.49

so range of movement

that is 46.49 - 45

range of movement is 1.49 mm

8 0

3 years ago

A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$

Therefore

$V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437$

Charge, Q is given by CV hence for the first capacitor charge will be $Q_1=C_1V$

Here, $Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C$

8 0

3 years ago

An object accelerates from rest to 93 m/s over a distance of 49 m. What acceleration did it experience?

Maksim231197 [3]

Answer:

$88.3ms {}^{ - 2}$

or 88.3m/s^2

Explanation:

Using suvat where we list everything that we are given

s=49m

u=0m/s

v=93m/s

a=?

t=we are not given this value, so we don't use

using a formula that doesn't involve time:

$v {}^{2} = u {}^{2} + 2as$

rearranging to find acceleration by subtracting u^2 on both sides

$v {}^{2} - u {}^{2} = 2as$

then dividing 2s on both sides

$\frac{v {}^{2} - u {}^{2} }{2s} = a$

$\frac{(93) {}^{2} - (0) {}^{2} }{2 \times(49) } = 88.3ms {}^{ - 2}$

so the acceleration is 88.3ms^-2 (1dp)

3 0

3 years ago

Estimate how long the sun would last if it were merely a huge fire that was releasing chemical energy. Assume that the sun begin

stepan [7]

Answer:

≅ 17000 years or 1.7 x 10⁴ years

Explanation:

time= total energy/power

= (10⁸J/kg)(2x10³⁰ kg) / 3.8 x 10²⁶ J/s

= 526,315,789,473 s

= 16689 years

≅ 17000 years or 1.7 x 10⁴ years

7 0

3 years ago