An object weighs 32 newtons. What is its mass if a gravitometer indicates that g = 8.25 m/s?

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

Two equal point charges QQQ are separated by a distance ddd. One of the charges is released and moves away from the other due on

lys-0071 [83]

Answer:

The kinetic energy K of the moving charge is K = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

Explanation:

The potential energy due to two charges q₁ and q₂ at a distance d from each other is given by U = kq₁q₂/r.

Now, for the two charges q₁ = q₂ = Q separated by a distance d, the initial potential energy is U₁ = kQ²/d. The initial kinetic energy of the system K₁ = 0 since there is no motion of the charges initially. When the moving charge is at a distance of r = 3d, the potential energy of the system is U₂ = kQ²/3d and the kinetic energy is K₂.

From the law of conservation of energy, U₁ + K₁ = U₂ + K₂

So, kQ²/d + 0 = kQ²/3d + K

K₂ = kQ²/d - kQ²/3d = 2kQ²/3d

So, the kinetic energy K₂ of the moving charge is K₂ = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

4 0

3 years ago

A _____ provides rapid calculus removal by converting very-high-frequency sound waves into mechanical energy and reduces operato

sveticcg [70]

Answer:

The Ultrasonic Scaler

Explanation:

8 0

2 years ago

Find the work w1 done on the block by the force of magnitude f1 = 60.0 n as the block moves from xi = -3.00 cm to xf = 1.00 cm

Norma-Jean [14]

By definition, the work done by a force is given by:
$W1 = F1 * d
$
Where,
F: magnitude of force
d: distance traveled.
Substituting values we have:
$W1 = F1 * (xf - xi)

W1 = 60 * (0.01 - (-0.03))
$
$W1 = 60 * (0.01 + 0.03)

W1 = 60 * (0.04)

W1 = 2.40 J
$
Answer:
the work w1 done on the block by the force of magnitude f1 = 60.0 n is:
W1 = 2.40 J

6 0

3 years ago

The spring of modulus k = 200 n /m is compressed a distance of 300 mm and suddenly released with the system at rest. determine t

DiKsa [7]

I attached the missing picture.
Let's analyze the situation as spring goes from stretched to unstretched state.
When you stretch the string you have to use force against ( you are doing work) this energy is then stored in the spring in the form of potential energy. When we release the spring the energy is being used to push the two carts. When the spring reaches its unstretched length its whole initial potential energy has been used on the carts, and this is the moment when two carts have maximum velocity.
The potential energy of compressed ( stretched) spring is:
$E_p=\frac{1}{2}kx^2$
The kinetic energy of two carts is:
$E_{k1}+E_{k2}=m_1\frac{v_1^2}{2}+m_2\frac{v_2^2}{2}$
So we have:
$E_p=E_{k1}+E_{k2}\\ \frac{1}{2}kx^2=m_1\frac{v_1^2}{2}+m_2\frac{v_2^2}{2}$
Momentum also has to be conserved:
$m_1v_1-m_2v_2=0\\ m_1v_1=m_2v_2\\ v_1=\frac{m_2}{m_1}v_2$
Momentum before the release of the spring is zero so it has to stay zero. We plug this back into the expresion we got from law of conservation of energy and we get:
$v_2^2=\frac{m_1^2}{m_2^2-m_1^2}kx^2=4.05\\
v_2=\sqrt{4.05}=2.012\frac{m}{s}$
Now we go back to the momentum equation:
$v_1=\frac{m_2}{m_1}v_2\\
v_1=4.69\frac{m}{s}$

8 0

3 years ago