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vichka [17]
3 years ago
13

Help me do this question ​

Physics
1 answer:
Zepler [3.9K]3 years ago
3 0

Answer:

c20800

Explanation:

go to bear khana bro your book looks cheap go and study in durbar kanda school and take somee cash and do ash ah boy

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The difference between generalized and specialized transduction is:
nikitadnepr [17]

When foreign DNA is introduced in a cell, that is called transduction.

<span>Specialized transduction is when restricted set of genes speciifc,DNA fragments are introduced. On the other hand in generalized or random transduction transfer the genes is accidently.</span>



8 0
4 years ago
A cart is moving at 55 m/s at an angle of 25° to the ground. Determine the horizontal component.
Elodia [21]

Answer:

<em>The horizontal component of the velocity is 49.85 m/s.</em>

Explanation:

<u>Rectangular Components of a Vector</u>

A 2D vector can be expressed in several forms. The rectangular form gives its two components, one for each axis (x,y). The polar form gives the components as the pair (r,θ) being r the magnitude and θ the angle.

When the magnitude and angle of the vector are given, the rectangular components are calculated as follows:

v_x=v\cos\theta

v_y=v\sin\theta

Where v is the magnitude of the vector and θ is the angle with respect to the x positive direction.

The cart is moving at v=55 m/s at θ=25°, thus:

v_x=55\cos 25^\circ

v_x=49.85\ m/s

The horizontal component of the velocity is 49.85 m/s.

7 0
3 years ago
Earth has seasons because _____.
sladkih [1.3K]

Answer:

c, its axis is tilted

maybe

As it works its way around the sun, its tilted axis exposes different parts of earth.

C would be it because the roation of Earth on its axis doesn't have anything to do with the exposer of the revolution on its axis

7 0
3 years ago
Read 2 more answers
A 1 pF capacitor is connected in parallel with a 2 pF capacitor, the parallel combination then being connected in series with a
asambeis [7]

Hi there!

We can approach this problem in many ways, but to show you how I arrive to the final conclusion, I will begin by solving the circuit with an assigned value for the power source.

Let's use a power source value of 6V (produces nice numbers).

Recall the following rules.

Capacitors in series:

  • Voltage ADDS up.
  • Charge is EQUAL across each.
  • Total capacitance uses the reciprocal rule.

Capacitors in parallel:

  • Voltage is EQUAL across each.
  • Charge ADDS up.
  • Total capacitance is simply the sum.

Solving for total capacitance:
C_P = 1 + 2 = 3 pF\\\\C_T = \frac{1}{\frac{1}{3} + \frac{1}{3}} = 1.5 pF

Using rules for capacitors in series and parallel, the total capacitance of the circuit is 1.5 pF.

Thus, the total charge is:
Q = C_TV\\Q = 1.5 pF * 6 V = 9 pC

9 pC will go through the parallel combination and the individual capacitor in series with the combination.

Since the voltage adds up, we can find the amount of voltage across the 3pF capacitor with the remaining going through the branches of the parallel combination.

V = \frac{Q}{C}\\\\V = \frac{9pC}{3pF} = 3V

Therefore, 3V goes through each branch since 6V - 3V = 3V.

Solving for the charge for each capacitor:
Q  = CV\\Q = (1 pF)(3V) = 3pC\\\\Q = (2pF)(3V) = 6pC\\\\Q = (3pF)(3V) = 9pC

<u>Thus, the capacitor with the greatest charge is the 3 pF capacitor. </u>

To explain without all of the work above, the equivalent capacitance of the parallel combination (1 pF + 2pF = 3pF) is equivalent to the capacitance of the capacitor in series (3pF). Thus, the voltage across the parallel capacitors (since voltage is the same across branches in parallel) and the series capacitor is equal. However, since charge SUMS UP for capacitors in parallel, they would have less charge than the single capacitor in series.

5 0
2 years ago
Two large, flat metal plates are separated by a distance that is very small compared to their height and width. The conductors a
mote1985 [20]

Answer:

E=0

Explanation:

Electric field due to each thin sheet of charge=\sigma/2\varepsilon

let us say the right plate has positive charge density \varepsilonand left sheet has a negative charge density -\varepsilon .

In the region between the plates,the electric field due to each plate is in same direction,

E=\sigma/2\varepsilon-(-\sigma/2\varepsilon)

E=\sigma/\varepsilon

in the region outside the plates, the field due to the plates is in opposite directions

E=-\sigma/2\varepsilon-(-\sigma/2\varepsilon)

E=-\sigma/2\varepsilon+\sigma/2\varepsilon

E=0

4 0
3 years ago
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