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3 years ago

Buoyancy increases with the increase in the density of a) Submerged body b) Fluid

Physics

1 answer:

AlekseyPX3 years ago

4 0

Answer:

Submerged body

Explanation:

If buoyancy is greater than weight then object will float.
If buoyancy is less then weight then object will sink.
If buoyancy=weight then objects remains stable

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An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

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3 years ago

A fighter jet accelerates from 10 m/s to 75 m/s in 7.0 s. The acceleration of the jet is

liraira [26]

Acceleration = final velocity - inital / time
a = 75-10 / 7
a = 65 / 7
a = 9.29 m/s^2

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3 years ago

A 170 kg astronaut (including space suit) acquires a speed of 2.25 m/s by pushing off with his legs from a 2600 kg space capsule

saw5 [17]

Explanation:

Mass of the astronaut, m₁ = 170 kg

Speed of astronaut, v₁ = 2.25 m/s

mass of space capsule, m₂ = 2600 kg

Let v₂ is the speed of the space capsule. It can be calculated using the conservation of momentum as :

initial momentum = final momentum

Since, initial momentum is zero. So,

$m_1v_1+m_2v_2=0$

$170\ kg\times 2.25\ m/s+2600\ kg\times v_2=0$

$v_2=-0.17\ m/s$

So, the change in speed of the space capsule is 0.17 m/s. Hence, this is the required solution.

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3 years ago

En un rio una Onda viaja con una velocidad de propagación de 50 m/s con una longitud de Onda de 40 metros. Hallar la frecuencia

sattari [20]

Answer:

Frequencia = 1.25 Hz

Explanation:

<u>Dados los siguientes datos;</u>

Velocidad = 50 m/s
Longitud de onda = 40 metros

Para encontrar la frecuencia de la onda;

Matemáticamente, la velocidad de una onda viene dada por la fórmula;

$Velocidad = Longitud \; de \; onda * Frequencia$

Haciendo de la frecuencia el tema de la fórmula, tenemos;

$Frequencia = \frac {Velocidad}{Longitud \; de \; onda}$

Sustituyendo en la fórmula, tenemos;

$Frequencia = \frac {50}{40}$

<em>Frequencia = 1.25 Hz</em>

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3 years ago

In the context of energy transfers with hot and cold reservoirs, the sign convention is that _______________.

Likurg_2 [28]

Answer:

B. QC > 0; QH < 0

Explanation:

Given that there are two reservoir of energy.

Sign convention for heat and work :

1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

From hot reservoir heat is going out that is why it is taken as negative

$Q_H$

From cold reservoir heat is coming inside the reservoir that is why it is taken as positive

$Q_C>0$

That is why the answer will be

$Q_H$ , $Q_C>0$

8 0

3 years ago

Buoyancy increases with the increase in the density of a) Submerged body b) Fluid​

Buoyancy increases with the increase in the density of a) Submerged body b) Fluid