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Molodets [167]
3 years ago
11

What is the answer to this question?

Physics
1 answer:
sesenic [268]3 years ago
8 0

Answer:                Unknown

Explanation:         ???

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The electric potential at the surface of a charged conductor _______.
frez [133]

Answer:

The electric potential at the surface of a charged conductor<u> is always such that the potential is zero at all points inside the conductor.</u>

Explanation:

Each point on the surface of a balanced charged conductor has the same electrical potential.

The surface on any charged conductor in electrostatic equilibrium is an equipotential surface. Since the electric field is equal to zero inside the conductor, the electric potential at any point inside and on the surface is equivalent to its value.

6 0
3 years ago
A basketball player jumps straight up for a ball. To do this, he lowers his body 0.310 m and then accelerates through this dista
Nastasia [14]

Answer:A)u =4.295m/s  , B)a = 29.746m/s²   C) F=3,153N

Explanation:

Using the kinematic expression  

v² = u² - 2as

where

u = initial velocity

v = final velocity

s = distance

g = acceleration due to gravity .

Given that he reaches a height of 0.940 m above the floor,

the final velocity  = 0

Here, acceleration due to gravity is acting in  opposite the initial direction of motion. So, a=-9.81 m/s.

v² = u² + 2as

0² - u² = 2 (- 9.81) × 0.940

- u² = 2 × - 9.81 × 0.920

- u² = -18.4428

cancelling the minus in both sides , we have that  

u² = 18.4428

u = √18.4428

u =4.295m/s

(b) His acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.310 m. m/s2

Using v² = u² + 2as

where u = initial speed of basketball player before lengthening = 0 m/s,

v = final speed of basketball player after lengthening =  4.295m/s,

a = acceleration while  straightening his legs

s = distance moved during lengthening = 0.310m

v² = u² + 2as  

 a = (v² - u²)/2s

a = (4.29m/s)² - (0 m/s)²)/(2 × 0.310m)

a = (18.4428 m²/s² - 0 m²/s²)/(0.62 m)

a = (18.4428 m²/s²/(0.62 m)

a = 29.746m/s²

c) The force (in N) he exerts on the floor to do this, given that his mass is 106 kg. N

Force= mass x acceleration.

F = 106 kg X 29.746m/s²

 F = 3,153.076 rounded to  3,153N

8 0
3 years ago
Physics approach to study macromoelcues at nanoscales <br> in detail plx
erastova [34]

Answer:

Abstracto

Los ácidos nucleicos y las proteínas comprenden una red de biomacromoléculas que almacenan y transmiten información que sustenta la vida de la célula. El estudio de estos mecanismos es un campo llamado biología molecular. El desarrollo de esta ciencia siempre ha ido acompañado de avances técnicos que permiten romper barreras metodológicas para probar hipótesis novedosas. Entre los métodos disponibles para los biólogos moleculares, destacan cinco: electroforesis, secuenciación, clonación, transferencia y reacción en cadena de la polimerasa. Su impacto llega a la genética, la medicina y la biotecnología. Aquí, se revisan la relevancia histórica, los fundamentos técnicos y las tendencias actuales de estos cinco métodos esenciales. La revisión pretende ser útil tanto para estudiantes como para científicos profesionales que buscan adquirir conocimientos avanzados sobre el valor de estos métodos para investigar los mecanismos moleculares que sostienen la vida.

8 0
2 years ago
Prove the correctness of this equation s=vt + 1/2at ​
Kaylis [27]

Answer:

The formula is dimensionally correct.

Explanation:

Given

s = ut + \frac{1}{2}at^2

Required

Prove its correctness

Write out the dimension of each:

s = M^0LT^0 --- displacement

ut = M^0LT^{-1} *  T --- velocity * time

\frac{1}{2}at^2 = M^0LT^{-2} * T^2 --- acceleration * square of time

The expression becomes:

s = ut + \frac{1}{2}at^2

M^0LT^0 = M^0LT^{-1} *  T + M^0LT^{-2} * T^2

Apply law of indices

M^0LT^0 = M^0LT^{-1+1} + M^0LT^{-2+2}

M^0LT^0 = M^0LT^{0} + M^0LT^{0}

M^0LT^0 = M^0LT^{0}

Both sides of the equation are equal

5 0
2 years ago
A wheel with a tire mounted on it rotates at the constant rate of 2.73 revolutions per second. Find the radial acceleration of a
Lostsunrise [7]

Answer:

110.9 m/s²

Explanation:

Given:

Distance of the tack from the rotational axis (r) = 37.7 cm

Constant rate of rotation (N) = 2.73 revolutions per second

Now, we know that,

1 revolution = 2\pi radians

So, 2.73 revolutions = 2.73\times 2\pi=17.153\ radians

Therefore, the angular velocity of the tack is, \omega=17.153\ rad/s

Now, radial acceleration of the tack is given as:

a_r=\omega^2 r

Plug in the given values and solve for a_r. This gives,

a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]

Therefore, the radial acceleration of the tack is 110.9 m/s².

4 0
3 years ago
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