A substance has a mass of 2795 g and a volume of 312 cm

Ryan swings a pail of water in a vertical circle 1.0 m in radius at a constant speed. If the water is NOT to spill on him:

nexus9112 [7]

Part 1

If water does not spill at the top point of the circular motion then for the minimum speed condition we can say normal force will be zero at the top position

$F_g = ma$

$mg = m\frac{v^2}{R}$

$g = \frac{v^2}{R}$

$v = \sqrt{Rg}$

given that

R = 1 m

g = 9.8 m/s^2

now from above equation we have

$v = \sqrt{1(9.8)} = 3.13 m/s$

Part b)

for minimum value of angular speed we will have

$\omega = \frac{v}{R}$

$\omega = \frac{3.13}{1}$

$\omega = 3.13 rad/s$

3 0

3 years ago

A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f

VashaNatasha [74]

Answer:

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 = X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + $(\frac{L-y}{2})$ ) g sinФ - F2 x ( $\frac{L}{2}$ ) x gsinФ = 0

plug in the equations of F1 and F2 into the above equation and after simplification, we get:

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . 5/9 x w

Hence,

$(L^{2} - y^{2} )$ = $\frac{5L^{2} }{9}$

$\frac{L^{2} - y^{2} }{L^{2} }$ = $\frac{5}{9}$

Taking $L^{2}$ common and solving for $\frac{y}{L}$ , we will get

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

8 0

3 years ago

Explain why dogs pant during hot summer days using the evaporation concept?

damaskus [11]

All, or almost all, warm-blooded creatures get rid of excess heat by evaporating moisture from their bodies. It's a great system, because evaporation takes a lot of heat. That's the reason people perspire when we're active and build up a lot of heat inside. The evaporation of sweat from our skin carries away heat with it. Dogs do not sweat on their skin. The only place they can evaporate moisture is through their mouth. Panting speeds up the evaporation by blowing air across the moisture.

5 0

3 years ago

In aircraft design, the pressure coefficient Cp is usually measured during wind tunnel testing of an aircraft component to predi

Elza [17]

Answer:

Check the explanation

Explanation:

From given data, it can be noted that 95% of given confidently data, means 5% of data is uncertain. According to the question, we have to calculate uncertainty in Cp .

Kindly check the attached image below for the step by step explanation to the question above.