All categories

3 years ago

How does area affect the pressure?

1 answer:

5 0

The smaller the area the greater the pressure, while the bigger the smaller the pressure. So they are inversely proportional

You might be interested in

Breanna is standing beside a merry-go-round pushing 19° from the tangential direction and is able to accelerate the ride and her

leva [86]

To solve the problem it is necessary to apply the concepts given in the kinematic equations of angular motion that include force, acceleration and work.

Torque in a body is defined as,

$\tau_l = F*d$

And in angular movement like

$\tau_a = I*\alpha$

Where,

F= Force

d= Distance

I = Inertia

$\alpha =$ Acceleration Angular

PART A) For the given case we have the torque we have it in component mode, so the component in the X axis is the net for the calculation.

$\tau= F*cos(19)*d$

On the other hand we have the speed data expressed in RPM, as well

$\omega_f = 10rpm = 10\frac{1rev}{1min}(\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 1.0471rad/s$

Acceleration can be calculated by

$\alpha = \frac{\omega_f}{t}$

$\alpha = \frac{1.0471}{9}$

$\alpha = 0.11rad/s^2$

In the case of Inertia we know that it is equivalent to

$I = \frac{1}{2}mr^2 = \frac{1}{2}(750)*(2.3)^2$

$I = 1983.75kg.m^2$

Matching the two types of torque we have to,

$\tau_l=\tau_a$

$Fd=I\alpha$

$Fcos(19)*2.3=1983.75(0.11)$

$F=100.34N$

PART B) The work performed would be calculated from the relationship between angular velocity and moment of inertia, that is,

$W = \frac{1}{2}I\omega_f^2$

$W= \frac{1}{2}(1983.75)(1.0471)^2$

$W=1087.51J$

7 0

2 years ago

How many protons, neutrons, and electrons does a neutral atom of this element have? (round atomic mass to nearest whole number)

adelina 88 [10]

Answer:

The element "AI" has:

Protons: 13 Neutrons: 14 Electrons: 13

Have a great day.

8 0

3 years ago

Read 2 more answers

If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55.5 m/s in the opposite directi

Lelu [443]

Answer:

ΔP = 14.5 Ns

I = 14.5 Ns

ΔF = 5.8 x 10³ N = 5.8 KN

Explanation:

The mass of the ball is given as 0.145 kg in the complete question. So, the change in momentum will be:

ΔP = mv₂ - mv₁

ΔP = m(v₂ - v₁)

where,

ΔP = Change in Momentum = ?

m = mass of ball = 0.145 kg

v₂ = velocity of batted ball = 55.5 m/s

v₁ = velocity of pitched ball = - 44.5 m/s (due to opposite direction)

Therefore,

ΔP = (0.145 kg)(55.5 m/s + 44.5 m/s)

ΔP = 14.5 Ns

The impulse applied to a body is equal to the change in its momentum. Therefore,

Impulse = I = ΔP

I = 14.5 Ns

the average force can be found as:

I = ΔF*t

ΔF = I/t

where,

ΔF = Average Force = ?

t = time of contact = 2.5 ms = 2.5 x 10⁻³ s

Therefore,

ΔF = 14.5 N.s/(2.5 x 10⁻³ s)

ΔF = 5.8 x 10³ N = 5.8 KN

4 0

3 years ago

A car is traveling in a race. The car went from the initial velocity of 35 m/s to the final velocity of 65 m/s in 5 seconds

melisa1 [442]

Answer:

Explanation:

65-35 = 30

30 divided by 5 = 6

The answer will be 6 m/s

6 0

2 years ago

A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

$B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T$

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0

3 years ago

How does area affect the pressure?​

How does area affect the pressure?