Examples of increase in pressure due to increase in applied force<br>

Ahat [919]

Hi,Find answers from Task 5

1.(X+4)+(X)+(X+4)+(X)=50cm

4x+8=50cm

4x=42

X=10.5cm

Length=10.5+4=14.5cm

Width=10.5cm

Area= length × width=(10.5/100) × (14.5/100) =0.0152m2

2. Volume of a sphere= 4/3 ×π×r³

4/3 ×π×r³=3.2×10^-6 m³

r³=3.2×10^-6 m³/1.33×π

r³=7.64134761e-7

r=0.00914m

Surface area of the blood drop= 4πr²

=4×3.142×0.00914×0.00914=0.00105m²

3.

Equation of an ideal gas = PV =n RT

Equation for pressure, = P= n RT/V

Equation for the volume of an ideal gas= V= n RT/P

If the volume of gas doubles ,V(new)= 2n RT/P

Equation for temperature of an ideal gas, T = PV/n R

If temperature of gas triples, T (new)= 3PV/n R

New Equation for Pressure, = n× R× (3PV/n R)/(2n RT/P)

Pressure factor increase= P(new)/P(old) ={ n× R× (3PV/n R)/(2n RT/P)}/{ n RT/V}

=3PV²/2n RT

5 0

2 years ago

Energy can be stored in a magnetic field. True or false?

Rzqust [24]

Your answer is here!

Yes It is true

8 0

3 years ago

Read 2 more answers

You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). Your mass is 50 kg. Your friend

Bogdan [553]

Answer:

a. $\tau=200J$ b. $\alpha=0.44 \frac{rad}{s^2}$ c. $\alpha=0.33\frac{rad}{s^2}$ d. The angular acceleration when sitting in the middle is larger.

Explanation:

a. The magnitude of the torque is given by $\tau=rF\sin \theta$ , being r the radius, F the force aplied and $\theta$ the angle between the vector force and the vector radius. Since $\theta=90^{\circ}, \, \sin\theta=1$ and so $\tau=rF=2m100N=200Nm=200J$ .

b. Since the relation $\tau=I\alpha$ hols, being I the moment of inertia, the angular acceleration can be calculated by $\alpha=\frac{\tau}{I}$ . Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is $I=\frac{1}{2}Mr^2$ , being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by $I_p=m_pr_p^{2}$ , being $m_p$ the mass of the person and $r_p^{2}$ the distance from the person to the center. Given all of this, we have

$\alpha=\frac{\tau}{I}=\frac{\tau}{I_{disk}+I_{person}}=\frac{Fr}{\frac{1}{2}Mr^2+m_pr_p^{2}}=\frac{200Nm}{\frac{1}{2}200kg*4m^2+50kg*1m^2}=\frac{200\frac{kgm^2}{s^2}}{450Nm^2}\approx 0.44\frac{rad}{s^2}$ .

c. Similar equation to b, but changing $r_p=2m$ , so

$alpha=\dfrac{200\frac{kgm^2}{s^2}}{\frac{1}{2}200*4kg\,m^2+50*4 kg\,m^2}=\dfrac{200}{600}\dfrac{1}{s^2}\approx 0.33 \frac{rad}{s^2}$ .

d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.

5 0

3 years ago

A ball rolls for 8 seconds and travels 24 meters. How fast was it traveling?

belka [17]

Answer:

The speed of the ball was, v = 3 m/s

Explanation:

Given data,

The time period of the ball, t = 8 s

The distance the ball rolled, d = 24 m

The velocity of an object is defined as the object's displacement to the time taken. The formula for the velocity is,

v = d / t m/s

Substituting the given values in the above equation,

v = 24 / 8

= 3 m/s

Hence, the speed of the ball was, v = 3 m/s

8 0

3 years ago

A swimmer is capable of swimming 0.42 m/s in still water. part a if she aims her body directly across a 66-m-wide river whose cu

satela [25.4K]

<span>If the swimmer is swimming perpendicular to the current, it will take her 66m / 0.42 m/s = 157.14 seconds to cross the river. At the same time, the current will be taking her downstream at a rate of 0.32 m/s. So, when she reaches the opposite bank, her total downstream distance traveled will have been 0.32*157.14 = 50.28 meters.</span>

3 0

3 years ago

Examples of increase in pressure due to increase in applied force​

Examples of increase in pressure due to increase in applied force