This is what causes acceleration when two forces are acting opposite from each other

A disk of radius 10 cm speeds up from rest. it turns 60 radians reaching an angular velocity of 15 rad/s. what was the angular a

stepan [7]

Answer:

a) α = 1.875 $\frac{rad}{s^{2} }$

b) t = 8 s

Explanation:

Given:

ω1 = 0 $\frac{rad}{s}$

ω2 = 15 $\frac{rad}{s}$

theta (angular displacement) = 60 rad

*side note: you can replace regular, linear variables in kinematic equations with angular variables (must entirely replace equations with angular variables)*

a) α = ?

(ω2)^2 = (ω1)^2 + 2α(theta)

$15^{2}$ = $0^{2}$ + 2(α)(60)

225 = 120α

α = 1.875 $\frac{rad}{s^{2} }$

b)

α = (ω2-ω1)/t

t = (ω2-ω1)/α = (15-0)/1.875 = 8

t = 8 s

4 0

2 years ago

The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve

Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

$\varepsilon _t=\dfrac{D-d}{d}$

$\varepsilon _t=\dfrac{32-30}{30}$

$\varepsilon _t=0.0667$

We know that

$n=-\dfrac{\epsilon _t}{\varepsilon _{long}}$

$\varepsilon _{long}=-\dfrac{\epsilon _t}{n}$

$\varepsilon _{long}=-\dfrac{0.0667}{0.45}$

$\varepsilon _{long}=-0.148$

So the axial pressure

$P=E\times \varepsilon _{long}$

$P=5\times 0.148$

P=740 KPa

The movement in the sleeve

$\Delta =\varepsilon _{long}\times L$

$\Delta =0.148\times 50$

Δ=7.4 mm

6 0

3 years ago

During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshi

vladimir2022 [97]

Answer:

At a deceleration of 60g, or 60 times the acceleration due to gravity a person will travel a distance of 0.38 m before coing to a complete stop

Explanation:

The maximum acceleration of the airbag = 60 g, and the duration of the acceleration = 36 ms or 36/1000 s or 0.036 s

To find out how far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60g

we write out the equation of motion thus.

S = ut + 0.5at²

wgere

S = distance to come to complete stop

u = final velocoty = 0 m/s

a = acceleration = 60g = 60 × 9.81

t = time = 36 ms

as can be seen, the above equation calls up the given variable as a function of the required variable thus

S = 0×0.036 + 0.5×60×9.81×0.036² = 0.38 m

At 60g, a person will travel a distance of 0.38 m before coing to a complete stop

7 0

3 years ago

To test the hypothesis that the population mean mu=3. 6, a sample size n=14 yields a sample mean 4. 007 and sample standard devi

PolarNik [594]

The P value for the given data set is 25127. For finding P value, we have to must find the Z value.

<h3>How to get the z scores?</h3>

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

The Z value is calculated as;

$Z = \dfrac{X - \mu}{\sigma})$

Z = (X - μ) / σ

Z = (4.007 - 3.6) / 0.607

Z = 0.67051

The P value for the given data set is 25127.

Learn more about z-score here:

brainly.com/question/21262765

#SPJ1

5 0

2 years ago

Write a numerical expression for the emissive intensity (in W/m^2.sr) coming out of a tiny hole in an enclosure of surface tempe

stiks02 [169]

Answer:

6.0 × $10^{11}$ W/ $m^{2}$

Explanation:

From Wien's displacement formula;

Q = e A $T^{4}$

Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.

The emissive intensity = $\frac{Q}{A}$ = e $T^{4}$

Given from the question that: e = 0.6 and T = 1000K, thus;

emissive intensity = 0.6 × $(1000)^{4}$

= 0.6 × 1.0 × $10^{12}$

= 6.0 × $10^{11}$ $\frac{W}{m^{2} }$

Therefore, the emissive intensity coming out of the surface is 6.0 × $10^{11}$ W/ $m^{2}$ .

3 0

2 years ago