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riadik2000 [5.3K]
3 years ago
14

Zoning laws establish _______.

Physics
1 answer:
Aleonysh [2.5K]3 years ago
5 0

Answer:

Zoning ordinances detail whether specific geographic zones are acceptable for residential or commercial purposes. Zoning ordinances may also regulate: - size

- placement

- density

- height of structures

HOPE IT HELPS :)

PLEASE MARK IT THE BRAINLIEST!

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It is the normal line
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An X-Ray machine delivers a radiation dose of 5mRem/hr. at 3ft from the machine. How far will the X-Ray technician have to move
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4.7ft

Explanation:

Pls see attached file

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3 years ago
A 1.5 kg spherical ball is has a radius of 50 cm is rotating with angular velocity of 12 revolutions per minute. Determine the r
kykrilka [37]

Answer:

K.E = 0.0075 J

Explanation:

Given data:

Mass of the ball = 1.5 kg

radius, r = 50 cm = 0.5 m

Angular speed, ω = 12 rev/min = (12/60) rev/sec = 0.2 rev/sec

Now,

the kinetic energy is given as:

K.E = K.E=\frac{1}{2}I\omega^2

where,

I is the moment of inertia = mr²

on substituting the values, we get

K.E=\frac{1}{2}\times1.5\times0.5^2\times0.2^2

or

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3 0
3 years ago
A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M a
aleksklad [387]

Answer:

a)  y = 0.98 t², t=1s y= 0.98 m,  

b) he two blocks must move the same distance

c) v = 1.96 m / s,  d)  a = -1.96 m / s², e)  x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

             N-W = 0

             N = M g

X axis

             T- fr = Ma

the friction force has the expression

             fr = μ N

             fr = μ Mg

small block

             w- T = m a

             

we write the system of equations

             T - fr = M a

             mg - T = m a

we add and resolved

             mg-  μ Mg = (M + m) a

             a = g \ \frac{m - \mu M}{m+M}

             a = 9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}

             a = 9.8 (6/30)

             a = 1.96 m / s²

a) now we can use the kinematic relations

             y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

             y = ½ a t²

             y = ½ 1.96 t²

             y = 0.98 t²

for t = 1s y = 0.98 m

       t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

           v = vo + a t

           v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

       t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

           -fr = M a

          - μ M g = M a

          a = - μ g

          a = - 0.2 9.8

          a = -1.96 m / s²

e) the distance to stop the block is

         v² = vo² - 2 a x

        0 = vo² - 2a x

        x = vo² / 2a

        x = 1.96² / 2 1.96

        x = 0.98 m

the time to travel this distance is

        v = vo - a t

        t = vo / a

        t = 1.96 /1.96

        t = 1 s

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What distance will a car cover while uniformly accelerating from 12m/s to 26m/in 14 seconds​
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