What is one defining feature of a prokaryotic cell

The energy delivered to the resistive coil is dissipated as heat at a rate equal to the power input of the circuit. However, not

Nuetrik [128]

Answer:

P(bat) = V²r/(R+r)²

Explanation:

Let the resistance of the coil be R

Internal resistance of the battery be r

Emf of the battery = V

Power dissipated in the internal resistance of the battery is normally given as P = I²r

where I is the current flowing in the circuit.

From Ohm's law,

V = I R(eq)

R(eq) = (R + r)

I = V/(R+r)

P = I²r

P = [V/(R+r)]²r

P = V²r/(R+r)²

Hope this Helps!!!

6 0

3 years ago

In preparation for an electrophoresis procedure, enzymes are added to DNA in order to

lapo4ka [179]

It is 2.) Cut the DNA into fragments

3 0

3 years ago

If displacement covered by a particle is zero then distance cover by it<br>

sleet_krkn [62]

Answer:

When displacement is zero, the particle may be at rest, therefore, distance travelled = 0.

Again, when displacement is zero, the final position matches with the initial position after some time, but the distance travelled will not be zero.

7 0

3 years ago

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

4 years ago

These nerve cells are needed in muscle and tendons to detect degree of stress and stretching as a sensory activity. A. Ganglia.

SOVA2 [1]

Answer:

D.Proprioceptors

Explanation:

"Proprioceptors is Sensory receptors found in muscle an tendons that detect their degree of stretch"

4 0

3 years ago