Ask question

Ask question

All categories

3 years ago

6

What is the magnitude of the total acceleration of point A after 2 seconds? The bar starts from rest and has a constant angular

acceleration of 1.0 rad/s2 clockwise.

1 answer:

monitta3 years ago

7 0

Answer:

a_total = 2 √ (α² + w⁴) , a_total = 2,236 m

Explanation:

The total acceleration of a body, if we use the Pythagorean theorem is

a_total² = a_T²2 + $a_{c}$ ²

where

the centripetal acceleration is

a_{c} = v² / r = w r²

tangential acceleration

a_T = dv / dt

angular and linear acceleration are related

a_T = α r

we substitute in the first equation

a_total = √ [(α r)² + (w r² )²]

a_total = 2 √ (α² + w⁴)

Let's find the angular velocity for t = 2 s if we start from rest wo = 0

w = w₀ + α t

w = 0 + 1.0 2

w = 2.0rad / s

we substitute

a_total = r √(1² + 2²) = r √5

a_total = r 2,236

In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m

a_total = 2,236 m

You might be interested in

A cart is driven by a large propeller or fan, which can accelerate or decelerate the cart. The cart starts out at the position x

mash [69]

Answer:

The acceleration of the cart is 1.0 m\s^2 in the negative direction.

Explanation:

Using the equation of motion:

Vf^2 = Vi^2 + 2*a*x

2*a*x = Vf^2 - Vi^2

a = (Vf^2 - Vi^2)/ 2*x

Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.

Let x = Xf -Xi

Where Xf is the final position of the cart and Xi the initial position of the cart.

x = 12.5 - 0

x = 12.5

The cart comes to a stop before changing direction

Vf = 0 m/s

a = (0^2 - 5^2)/ 2*12.5

a = - 1 m/s^2

The cart is decelerating

Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.

5 0

3 years ago

What is the net work done<br> on the object over the distance shown?

enyata [817]

Answer:

0

Explanation:

the positive cancels out the negative area

3 0

2 years ago

What is the proper hand position when driving straight ahead

vekshin1

NHTSA (National Highway Traffic Safety Administration) now recommends the technique known as 9 and 3. Place your left hand on the left portion of the steering wheel in a location approximate to where the nine would be if the wheel was a clock. Your right hand should be placed on the right portion of the wheel where the three would be located.

7 0

3 years ago

On a hot summer day, the density of air at atmospheric pressure at 31.5°C is 1.2074 kg/m3.

mario62 [17]

To solve this problem, it is necessary to apply the ideal Gas equations, as well as the calculation equations of the weight difference, which under the comparison of two values.

By definition we know that the ideal gas equation is given by the equation,

PV = nRT

Where,

P = Pressure

V = Volume

R = Gas ideal constant

T = Temperature

n = number of moles

Our values are given by

$P = 1.1013*10^5Pa$

$T = 31.5\°C = 304.5K$

$\rho = 1.2074kg/m^3$

PART A) Using this previous equation we can find the number of moles per Volume, that is

$PV = nRT$

$\frac{n}{V} = \frac{P}{RT}$

Replacing with our values

$\frac{n}{V} = \frac{1.013*10^5}{(8.314)(304.5)}$

$\frac{n}{V} = 40.014mol/m^3$

PART B ) We can calculate the number of moles of 1m^3 through Avogadro number, then

$M = nA$

$M = 40.014*(2.88*10^{-2})$

$M = 1.1524Kg$

Therefore in $1m^3$ there are 1.1524Kg of Gas.

PART C ) Density can be defined as the proportion of mass in a specific quantity of Volume, then

$\rho_2 = \frac{M}{V}$

$\rho_2 = \frac{1.1524}{1}$

$\rho_2 = 1.1524kg/m^3$

The difference of percentage then is

$\Delta \rho = \frac{\rho_1-\rho_2}{\rho_2}*100\%$

$\Delta \rho = \frac{1.2074-1.1524}{1.1524}*100\%$

$\Delta \rho = 0.0477*100\%$

$\Delta \rho = 4.77\%$

YES, because as the percentage is less than 10%, the calculated value agrees with the stated value.

5 0

3 years ago

Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( 3.0×106V/m ) if

xxMikexx [17]

(a) The electric field strength between two parallel conducting plates does not exceed the breakdown strength for air ( $3 \times 10^{6} V / m$ )

(b) The plates can be close together to 1.7 mm with this applied voltage

<u>Explanation:</u>

Given data:

Dielectric strength of air = $3 \times 10^{6} V / m$

Distance between the plates = 2.00 mm = $2.00 \times 10^{-3} \mathrm{m}$

Potential difference, V = $5.0 \times 10^{3} V$

We need to find

a) whether the electric field strength between two parallel conducting plates exceed the breakdown strength for air or not

b) the minimum distance at which the plates can be close together with this applied voltage.

The voltage difference (V) between two points would be equal to the product of electric field (E) and distance separation (d). The equation form is and apply all given value,

$E=\frac{V}{d}=\frac{5.0 \times 10^{3}}{2.00 \times 10^{-3}}=2.5 \times 10^{6} \mathrm{V} / \mathrm{m}$

From the above, concluding that The electric field strength between two parallel conducting plates ( $2.5 \times 10^{6} \mathrm{V} / \mathrm{m}$ ) does not exceed the breakdown strength for air ( $3 \times 10^{6} V / m$ )

b) To find how close together can the plates be with this applied voltage:

The formula would be,

$d_{\min }=\frac{V}{E_{\max }}$

Apply all known values, we get

$d_{\min }=\frac{5.0 \times 10^{3}}{3 \times 10^{6}}=1.7 \times 10^{-3} \mathrm{m}=1.7 \mathrm{mm}$

3 0

3 years ago