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3 years ago

Please any one help me

Physics

1 answer:

Hatshy [7]3 years ago

4 0

Answer:

Which one are we supposed to do

Explanation:

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A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/s. The wheel can be considered

mart [117]

Answer:

$\omega_f=12.2954 \,rad.s^{-1}$ i.e.

$Frequency=1.9569\,rev.s^{-1}$

Explanation:

Given:

angular speed, $\omega_i =2\times 2\pi \,rad.s^{-1}$

mass of the disk, $M=4.9\,kg$

radius of the disk, $R=0.2\,m$

mass of the chunk, $m=2.7\,kg$

radius of the chunk, $r=0.04\,m$

We know that the angular momentum is given by:

$L=I.\omega$

and moment of inertia for a disc:

$I=\frac{1}{2} m.r^2$

According to the conservation of angular momentum, the final angular momentum is equal to the initial angular momentum.

$\frac{1}{2} \times M.R^2\times \omega= \frac{1}{2} \times (M.R^2+m.r^2) \omega_f$

$4.9\times 0.2^2\times (2\times 2\pi)=(4.9\times 0.2^2+2.7\times 0.04^2 )\tiems \omega_f$

$\omega_f=12.2954 \,rad.s^{-1}$

$Frequency=1.9569\,rev.s^{-1}$

4 0

4 years ago

Two balls are kicked with the same initial speeds. Ball A was kicked at the angle 20° above horizontal and ball B was kicked at

inysia [295]

Answer:Impossible to answer without knowing their actual initial speeds

Explanation:

Given

Ball A launched at angle of $20^{\circ}$

Ball B is launched at angle of $75^{\circ}$

At the highest point Ball vertical velocity will be and there will be only horizontal velocity i.e. $v_acos20$ and $v_bcos75$

where $v_a and v_b$ is the velocity of A & B

thus to find which one is greater we need to know their initial velocity.

It is impossible to know which one have higher velocity at highest point.

7 0

3 years ago

A substance is boiled repeatedly and stirred, but the solute never mixes with the solvent. Which best describes why

maks197457 [2]

Either Solute or solvent may have hydrogen bond in it but another one has opposite. So, in situation of un-like characteristics, they can't dissolve by any means!

Hope this helps!

3 0

3 years ago

A solar sail allows a spacecraft to use radiation pressure for propulsion, similar to the way wind propels a sailboat. The sails

Neporo4naja [7]

Answer:

F = 2 I A / c

Explanation:

The radiation pressure on a reflective surface is

P = 2 S / c

Where S is the Poynting Vector and c the speed of light

Furthermore pressure is defined as the ratio of force to area

P = F / A

Let's replace

F / A = 2 S / c

F = 2 S A / c

The poynting vector is the power per unit area that is equal to the intensity

S = I

F = 2 I A / c

7 0

3 years ago

Three metal fishing weights, each with a mass of 1.00x102 g and at a temperature of 100.0°C, are placed in 1.00x102 g of water a

worty [1.4K]

Answer:

Approximately $0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}$ assuming no heat exchange between the mixture and the surroundings.

Explanation:

Consider an object of specific heat capacity $c$ and mass $m$ . Increasing the temperature of this object by $\Delta T$ would require $Q = c\, m \, \Delta T$ .

Look up the specific heat of water: $c(\text{water}) = 4.182\; {\rm J \cdot g^{-1} \cdot K^{-1}}$ .

It is given that the mass of the water in this mixture is $m(\text{water}) = 1.00 \times 10^{2}\; {\rm g}$ .

Temperature change of the water: $\Delta T(\text{water}) = (45 - 35)\; {\rm K} = 10\; {\rm K}$ .

Thus, the water in this mixture would have absorbed :

$\begin{aligned}Q &= c\, m\, \Delta T \\ &= 4.182\; {\rm J \cdot g^{-1}\cdot K^{-1}} \\ &\quad \times 1.00 \times 10^{2}\; {\rm g} \times 10\; {\rm K} \\ &= 4.182 \times 10^{3}\; {\rm J}\end{aligned}$ .

Thus, the energy that water absorbed was: $Q(\text{water}) = 4.182 \times 10^{3}\; {\rm J}$ .

Assuming that there was no heat exchange between the mixture and its surroundings. The energy that the water in this mixture absorbed, $Q(\text{water})$ , would be the opposite of the energy that the metal in this mixture released.

Thus: $Q(\text{metal}) = -Q(\text{water}) = -4.182 \times 10^{3}\; {\rm J}$ (negative because the metal in this mixture released energy rather than absorbing energy.)

Mass of the metal in this mixture: $m(\text{metal}) = 3 \times 1.00 \times 10^{2}\; {\rm g} = 3.00 \times 10^{2}\; {\rm g}$ .

Temperature change of the metal in this mixture: $\Delta T(\text{metal}) = (100 - 45)\; {\rm K} = 55\; {\rm K}$ .

Rearrange the equation $Q = c\, m \, \Delta T$ to obtain an expression for the specific heat capacity: $c = Q / (m\, \Delta T)$ . The (average) specific heat capacity of the metal pieces in this mixture would be:

$\begin{aligned}c &= \frac{Q}{m\, \Delta T} \\ &= \frac{-4.182 \times 10^{3}\; {\rm J}}{3.00 \times 10^{2}\; {\rm g} \times (-55\; {\rm K})} \\ &\approx 0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}\end{aligned}$ .

6 0

3 years ago