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2 years ago

A clock has radius of 0.5m. The outermost point on its minute hand travels along the edge. What is its tangential speed?

1 answer:

6 0

The angular speed of the minute hand is the ratio between the angle covered and the time taken. The angle covered by the minute hand is $2 \pi$ , while the time needed is $t=60 min= 3600 s$ , so the angular speed is
$\omega = \frac{2 \pi}{3600 s}=0.0017 rad/s$

And the tangential speed is given by
$v=\omega r=(0.0017 rad/s)(0.5 m)=8.72 \cdot 10^{-4} m/s$

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Water runs through a plumbing with a flow of 0.750m3/s and arrives to every exit of a fountain. At what speed will the water com

Lubov Fominskaja [6]

Divide the flow rate (0.750 m³/s) by the cross-sectional area of each pipe:

diameter = 40 mm ==> area = <em>π</em> (0.04 m)² ≈ 0.00503 m²

diameter = 120 mm ==> area = <em>π</em> (0.12 m)² ≈ 0.0452 m²

Then the speed at the end of the 40 mm pipe is

(0.750 m³/s) / (0.00503 m²) ≈ 149.208 m/s ≈ 149 m/s

(0.750 m³/s) / (0.0452 m²) ≈ 16.579 m/s ≈ 16.6 m/s

7 0

1 year ago

A Turtle and a Snail are 360 meters apart, and they start to move towards each other at 3 p.m. If the Turtle is 11 times as fast

Neko [114]

Answer:

Snail's speed = $\frac{30m}{2400s}$ = 0.0125m/s

Turtle's speed = $\frac{330m}{2400s}$ = 0.1375m/s

Explanation:

Let the snail's speed be x m/s

The turtle's speed then is 11x m/s

Speed = Distance ÷ Time

Since speed and distance are directly proportional;

The ratio of the distances snail and turtle cover before they meet is x:11x respectively.

Simplified, the ratio of snail distance : turtle distance = 1:11

So snail covers a distance of $\frac{1}{12}$ × 360 = 30m

And turtle covers a distance of $\frac{11}{12}$ × 360 = 330m

The time each took before they met is 40 × 60 = 2400 seconds

Snail's speed = $\frac{30m}{2400s}$ = 0.0125m/s

Turtle's speed = $\frac{330m}{2400s}$ = 0.1375m/s

8 0

2 years ago

Read 2 more answers

A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f

konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

$A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m$

$E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s$

8 0

2 years ago

A star orbiting a black hole in a clockwise direction at a radial distance of 1.0 × 106 km is acted upon by a counterclockwise f

snow_tiger [21]

Answer:

$2.5 \times 10^{11} N-m$ upwards

Explanation:

Torque is the vector cross product of the force and radial distance.

$\tau = rF sin \theta$

$\tau = (1.0\times 10^6 \times 10^3) m \times 250 N\times sin 90^o \\\Rightarrow \tau= 2.5 \times 10^{11} N-m$

The direction of the torque would be perpendicular to the direction of the force and radial distance. The direction of the force is counter-clockwise. The direction of the torque would be upwards.

4 0

2 years ago

The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determi

ankoles [38]

Answer:

F = - 59.375 N

Explanation:

GIVEN DATA:

Initial velocity = 11 m/s

final velocity = 1.5 m/s

let force be F

work done = mass* F = 4*F

we know that

Change in kinetic energy = work done

kinetic energy = $= \frac{1}{2}*m*(v_{2}^{2}-v_{1}^{2})$

kinetic energy = $= \frac{1}{2}*4*(1.5^{2}-11^{2})$ = -237.5 kg m/s2

-237.5 = 4*F

F = - 59.375 N

7 0

2 years ago