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myrzilka [38]
3 years ago
14

Which of the following is not a possible environmental consequence of urban sprawl?

Physics
1 answer:
vodomira [7]3 years ago
5 0
E none of the above because they’re all effects of urban sprawl
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In this unit, the amount of solute is measured in?
Hoochie [10]
Try liters if you haven’t done it yet. I’m so sorry if i’m incorrect.
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Riders on a ferris wheel move in a circle with a speed of 4.0 m/s. As they go around, they
Vanyuwa [196]
Answer: D = 16m

Explanation: given values: a = 2 m/s2, v = 4 m/s
In this case we have to determine the diameter of the Ferris wheel.
Diameter of circle is given as: D = 2.r.
First we have to find radius of wheel. The best way to find that is using the centripetal acceleration equation: a = v2/r
Plug in values in above equation to find radius: 2 m/s2 = (4 m/s)2/r 2 m/s2 = (16 m2/s2)/r r = (16 m2/s2)/2 m/s2
r = 8.0m
Diameter of Ferris wheel is:
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D = 2.8m
D = 16m
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2 years ago
What are the starting materials for nuclear fission
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3 years ago
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A yoyo with a mass of m = 150 g is released from rest as shown in the figure.
avanturin [10]

(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The  weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

<h3> Linear acceleration of the yoyo</h3>

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

  • I is moment of inertia
  • α is angular acceleration
  • T is tension in the rope
  • r is inner radius
  • R is outer radius
  • f is frictional force

rT - Rf = Iα  ----- (1)

T - f = Ma  -------- (2)

a = Rα

where;

  • a is the linear acceleration of the yoyo

Torque equation for frictional force;

f = (\frac{r}{R} T) - (\frac{I}{R^2} )a

solve (1) and (2)

a = \frac{TR(R - r)}{I + MR^2}

since the yoyo is pulled in vertical direction, T = mg a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2

<h3>Angular acceleration of the yoyo</h3>

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

<h3>Weight of the yoyo</h3>

W = mg

W = 0.15 x 9.8 = 1.47 N

<h3>Tension in the rope </h3>

T = mg = 1.47 N

<h3>Angular speed of the yoyo </h3>

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

Learn more about angular speed here: brainly.com/question/6860269

#SPJ1

3 0
1 year ago
A man pushes his lawnmower with a velocity of +0.75 m/s relative to the ground. A girl rides by on her bike with a velocity of +
SashulF [63]

Answer:

B. +5.75 m/s

Explanation:

When there are two bodies, a and b, whose velocities measured by a third observer (in this case, the ground) are V_a and V_b respectively, the relative velocity of B with respect to A is given by:

V_{ba}=V_b-V_a

Thus, the velocity of the girl relative to the lawnmower is:

V_{ba}=6.5\frac{m}{s}-0.75\frac{m}{s}\\V_{ba}=5.75\frac{m}{s}

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