Whitt is considered a prokaryotic cell

Could someone please tell me what I'm doing wrong?

7nadin3 [17]

Answer:

M1 = 16.9 mA

M2 = 0 A

Explanation:

As the ratio of the two sets of series resistors is almost exactly identical, there is no voltage difference across M2 to cause current flow

269/(269 + 439) = 0.3799...

500/(500 + 815) = 0.38022

M2 = 0

M1 sees only the current flowing through the far left resistors in series

A = V/R = 12/(269 + 439) = 0.016949... ≈ 16.9 mA

7 0

3 years ago

The drawing shows three identical springs hanging from the ceiling. Nothing is attached to the first spring, whereas a 4.5 n blo

Naya [18.7K]

<Continuation of the question>

(a) the spring constant (in N/m) and

(b) the weight of the block hanging from the third spring.

the distance for the first spring is 20cm, the second 35cm, the third 50cm.

Answer:

a) To find the spring constant, we'll use the formula

F=kx, if we make k the subject we'll get

k=F/x, where F=4.5N, x = 35cm - 20cm = 15cm = 0.15m

k=F/x = 4.5N/0.15m = 30N/m is the Answer

b) to find the weight of block hanging on third spring

we use the formula F=kx

where k = 30N/m, x=50cm-20cm=30cm=0.30m

F=kx = (30N/m)*(0.30m) = 9N is the Answer

4 0

3 years ago

A man cleaning his apartment pushes a vacuum cleaner with a force of magnitude 84.5 N. The force makes an angle of 33.9 ◦ with t

Sav [38]

Answer:

183.75641 Joules

Explanation:

F = Force of the vacuum cleaner = 84.5 N

s = Displacement of the vacuum cleaner = 2.62 m

$\theta$ = Angle the force makes with the horizontal = 33.9°

Work done is given by

$W=F\times scos\theta\\\Rightarrow W=84.5\times 2.62\times cos33.9\\\Rightarrow W=183.75641\ J$

The work done by the force of the vacuum cleaner is 183.75641 Joules

6 0

3 years ago

In a car lift used in a service station, compressed air exerts a force on a small piston of circular cross section having a radi

Vitek1552 [10]

Answer:

(a) 1,569.63 N

(b) 195,933.99 Pa

(c) As pressure and volume are equal for each piston, workdone must also be equal

(d) 1,647.47 kg

Explanation:

Let the cross-sectional area (CSA) of small piston = A₁

Let the cross-sectional area (CSA) of the bigger piston = A₂

Let the Force applied at the smaller piston = F₁

Let the Force applied at the bigger piston = F₂

The principle of hydraulic lift assumes the that the fluid is in-compressible, resulting to a constant pressure system.

F₁/A₁ =F₂/A₂----------------------------------------------------------- (1)

(a) F₁= F₂ xA₁ /A₂

F₂ = 13,300 N

A₁ = π r₁²

=π x (0.0505)²

= 0.008011 m²

A₂= π r₂²

=π x (0.147)²

= 0.06788 m²

Substituting into (1)

F₁ = 13,300 x 0.008011/0.06788

= 1,569.6272

≈ 1,569.63 N

(b) Air pressure = Force/Area

= F₁/A₁

= 1,569.6272/ 0.008011

= 195,933.99 Pa

(c) The pressure is constant for both pistons according to Pascal Law.

Workdone = force x distance----------------------------------------- (2)

force = pressure × area

distance = volume/area from

Substituting into (2)

Workdone = pressure × volume.

As pressure and volume are equal for each piston, work must also be equal

(d) F₂ = F₁ x A₂/ A₁---------------------------------------------------- (3)

A₁ = π r₁²

=π x (0.079)²

= 0.01960 m²

A₂= π r₂²

=π x (0.353)²

= 0.3914 m²

Substituting into (2)

F₂= 825 x 0.3914/ 0.01960

= 16,474.7448

≈ 16,474.74 N

= 1,647.47 kg

3 0

3 years ago

A piece of taffy slams into and sticks to another identical piece of taffy that is at rest. The momentum of the two pieces stuck

-Dominant- [34]

Answer:

C. 50%

Explanation:

Lets consider that the mass of taffy is m and its initial velocity is u and final velocity is v. Momentum is conserved so we can write it as,

$mu = 2mv\\v = \frac{u}{2} \\$

The initial kinetic energy = $\frac{mu^{2} }{2}$

As the Kinetic energy partly converted into heat.

The final kinetic energy = $\frac{m(\frac{u}{2})^{2} }{2}$ = $\frac{mu^{2} }{4}$

Change in the kinetic energy = $\frac{mu^{2} }{2} - \frac{mu^{2} }{4} = \frac{mu^{2} }{4}$

now we can determine the fraction of kinetic energy that has turned into heat

= $\frac{Heat energy}{initial K.E}$ = $\frac{\frac{mu^{2} }{4} }{\frac{mu^{2} }{2} } = 1/2$

= 50%

3 0

4 years ago