Pyramid math (algebra 1)

An electric motor rotating a workshop grinding wheel at a rate of 151 rev/min is switched off. Assume constant angular decelerat

tamaranim1 [39]

Answer:

Explanation:

Initial angular velocity ω₀ = 151 x 2π / 60

= 15.8 rad /s

final velocity = 0

Angular deceleration α = 2.23 rad / s

ω² = ω₀² - 2 α θ

0 = 15.8² - 2 x 2.23 θ

= 55.99 rad

one revolution = 2π radian

55.99 radian = 55.99 / 2 π no of terns

= 9 approx .

8 0

2 years ago

My Notes Ask Your Teacher 6. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclin

Tresset [83]

Answer:

Kf > Ka = Kb > Kc > Kd > Ke

Explanation:

We can apply

E₀ = E₁

where

E₀: Mechanical energy at the beginning of the motion (top of the incline)

E₁: Mechanical energy at the end (bottom of the incline)

then

K₀ + U₀ = K₁ + U₁

If v₀ = 0 ⇒ K₀

and h₁ = 0 ⇒ U₁ = 0

we get

U₀ = K₁

U₀ = m*g*h₀ = K₁

we apply the same equation in each case

a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J

d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J

f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J

finally, we can say that

Kf > Ka = Kb > Kc > Kd > Ke

8 0

2 years ago

A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p

777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

$v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s$

Now, we can calculate the angular acceleration (w0=0rad/s)

$\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}$

$\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}$

with this value we can compute the angular velocity

$\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}$

and the tangential velocity of point B, and then the acceleration of point B:

$v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}$

hope this helps!!

6 0

2 years ago

Read 2 more answers

Which region of the early universe was most likely to become a galaxy?

kramer

Answer:

This is likely possible for a region whose matter density is higher than the normal average.

Explanation:

A galaxy is a collection of lumps in space which are clumped together and interact with each other. There are a lot of speculations on how galaxies were birthed. some believe its formed by a collection of massive gas, dust which eventually collapsed under their own gravitational pull. others says its formed by the combination of large lumps of matter which accumulated forming thee galaxies. The possibility of a galaxy forming is dependent on how massive the matter in the region of the universe is.

3 0

2 years ago

An irrigation canal has a rectangular cross section. At one point whare the canal is 16.0 m wide, and the water is 3.8 m deep, t

Irina-Kira [14]

Answer:

The depth of the water at this point is 0.938 m.

Explanation:

Given that,

At one point

Wide= 16.0 m

Deep = 3.8 m

Water flow = 2.8 cm/s

At a second point downstream

Width of canal = 16.5 m

Water flow = 11.0 cm/s

We need to calculate the depth

Using Bernoulli theorem

$A_{1}V_{1}=A_{2}V_{2}$

Put the value into the formula

$16.0\times3.8\times2.8=16.5\times x\times 11.0$

$x=\dfrac{16.0\times3.8\times2.8}{16.5\times11.0}$

$x=0.938\ m$

Hence, The depth of the water at this point is 0.938 m.

7 0

2 years ago