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3 years ago

What is one defining feature of a prokaryotic cell

Physics

1 answer:

trapecia [35]3 years ago

5 0

A. nucleus
hope this helps if so please mark me brianliest

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The electric field between the plates of a parallel-plate capacitor is horizontal, uniform, and has a magnitude E. Asmall object

kirza4 [7]

Answer:

$m = 0.041 kg$

Explanation:

As we know that the small particle is in equilibrium at an angle of 16 degree with the vertical

so here we can use force balance in vertical and horizontal direction

$T cos\theta = mg$

$T sin\theta = qE$

now from above equation we have

$T = \sqrt{(mg)^2 + (qE)^2}$

also by division of above two equations we have

$\frac{qE}{mg} = tan\theta$

$qE = mg tan\theta$

now from above equation again

$T = \sqrt{(mg)^2 + (mg)^2 tan^2\theta}$

$T = mg sec\theta$

$0.420 = m(9.81) sec 16$

$m = 0.041 kg$

3 0

3 years ago

Acceleration directed toward the center of a circular path that produces uniform circular motion is

tester [92]

<span>Th answer is Centripetal</span>

4 0

4 years ago

Read 2 more answers

The normal boiling point of water is 100.0 °c and its molar enthalpy of vaporization is 40.67 kj/mol. what is the change in entr

SashulF [63]

ΔS= nΔHvap/T,

Where, ΔS = Change in entropy, n = moles of water = 39.3/18 = 2.188 moles, ΔHvap = 40.67kJ/mol = 40670 J/mol, T = Temperature (K) = 100+272.15 = 373.15 K

Therefore,
ΔS = (2.188*40670)/373.15 = 237.96 J/K

6 0

3 years ago

How can we get a clue as to the temperature of a star

Juli2301 [7.4K]

The colour of the star and the brightness.

8 0

3 years ago

a length of wire is cut into five equal pieces. the five pieces are then connected in parallel, with the resulting resistance be

Natali [406]

The equivalent resistance of n resistors connected in parallel is given by
$\frac{1}{R_{eq}} = \frac{1}{R_1}+ \frac{1}{R_2}+...+ \frac{1}{R_n}$ (1)

In our problem, the resulting resistance of the 5 pieces connected in parallel is $R_{Eq}=2.00 \Omega$ , and since the 5 pieces are identical, their resistance R is identical, so we can rewrite (1) as
$\frac{1}{R_{Eq} }= \frac{1}{2 \Omega}= \frac{1}{R}+ \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{5}{R}$
From which we find $R= 5 \cdot 2 \Omega = 10 \Omega$ .

So, each piece of wire has a resistance of $10 \Omega$ . Before the wire was cut, the five pieces were connected as they were in series. The equivalent resistance of a series of n resistors is given by
$R_{Eq}=R_1 + R_2 + ...+R_n$
So if we apply it at our case, we have
$R_{eq}=R+R+R+R+R=5 R= 5\cdot 10 \Omega= 50 \Omega$

therefore, the resistance of the original wire was $50 \Omega$ .

5 0

3 years ago