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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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The number of significant figures on the measurement 0.050010 kg id

Alex17521 [72]

Its been some time so i might be wrong but i think the answer is 3 either or 2

6 0

2 years ago

At time t, gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With rig

Elena-2011 [213]

Complete Question

The complete Question is shown on the first uploaded image

Answer:

The torque acting on the particle is $\tau = 48t \r k$

The magnitude of the angular momentum increases relative to the origin

Explanation:

From the equation we are told that

The position of the particle is $\= r = 4.0 t^2 \r i - (2.0 t - 6.0 t^2 ) \r j$

The mass of the particle is $m = 3.0 kg$

The time is t

The torque acting on the particle is mathematically represented as

$\tau = \frac{ d \r l }{dt}$

where $\r l$ is change in angular momentum which is mathematically represented as

$\r l = m (\r r \ \ X \ \ \r v)$

Where X mean cross- product

$\r v$ is the velocity which is mathematically represented as

$\r v = \frac{d \r r }{dt}$

Substituting for $\r r$

$\r v = \frac{d }{dt} [ 4 t^2 \r i - (2t + 6t^2 ) \r j]$

$\r v = 8t \r i - (2 + 12 t) \r j$

Now the cross product of $\r r \ and \ \r v$ is mathematically evaluated as

$\r r \ \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right]$

$= 0 \r i + 0 \r j + (- 8t^2 -48t^3 + 16t^2 + 48t^3 ) \r k$

$\r r \ \ X \ \ \r v = 8t^2 \r k$

So the angular momentum becomes

$\r l = m (8t^2 \r k)$

Substituting for m

$\r l = 3 * (8t^2 \r k)$

$\r l =24t^2 \r k$

Substituting into equation for torque

$\tau = \frac{d}{dt} [24t^2 \r k]$

$\tau = 48t \r k$

The magnitude of the angular momentum can be evaluated mathematically as

$|\r l| = \sqrt{(24 t^2) ^2}$

$|\r l| = 24 t^2$

From the is equation we see that the magnitude of the angular momentum is varies directly with square of the time so it would relative to the origin because at the origin t= 0s and we move out from origin t increases hence angular momentum increases also

4 0

2 years ago

Some gliders are launched from the ground by means of a winch, which rapidly reels in a towing cable attached to the glider. Wha

MArishka [77]

Answer:

$P = 1.64 \times 10^4 Watt$

Explanation:

Here we know that the glider is accelerated uniformly from rest to final speed of 25.7 m/s in total distance of d = 46.9 m

so we will have

$v_f = 25.7 m/s$

$v_i = 0$

d = 46.9

so for uniformly accelerated motion we have

$d = \frac{v_f + v_i}{2} t$

$46.9 = \frac{25.7 + 0}{2}t$

$t = 3.65 s$

now we will find the total work done given as change in kinetic energy

$W = \frac{1}{2}mv^2$

$W = \frac{1}{2}(181)(25.7^2)$

$W = 5.97 \times 10^4 J$

now power is given as

$P = \frac{W}{t}$

$P = \frac{5.97 \times 10^4}{3.65}$

$P = 1.64 \times 10^4 Watt$

6 0

2 years ago

A pole-vaulter is nearly motionless as he clears the bar, set 4.2 m above the ground. he then falls onto a thick pad. the top of

scZoUnD [109]

Refer to the diagram shown below.

Neglect wind resistance, and use g = 9.8 m/s².

The pole vaulter falls with an initial vertical velocity of u = 0.
If the velocity upon hitting the pad is v, then
v² = 2*(9.8 m/s²)*(4.2 m) = 82.32 (m/s)²
v = 9.037 m/s

The pole vaulter comes to res after the pad compresses by 50 cm (or 0.5 m).
If the average acceleration (actually deceleration) is (a m/s²), then
0 = (9.037 m/s)² + 2*(a m/s²)*(0.5 m)
a = - 82.32/(2*0.5) = - 82 m/s²

Answer: - 82 m/s² (or a deceleration of 82 m/s²)

8 0

2 years ago

Read 2 more answers

If the cold temperature reservoir of a Carnot engine is held at a constant 306 K, what temperature should the hot reservoir be k

Paraphin [41]

Efficiency η of a Carnot engine is defined to be:
η = 1 - Tc / Th = (Th - Tc) / Th 
where 
Tc is the absolute temperature of the cold reservoir, and 
Th is the absolute temperature of the hot reservoir. 

In this case, given is η=22% and Th - Tc = 75K 
Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). 

Th = (Th - Tc) / η 
Th = 75 / 0.22 = 341 K (rounded to closest number) 
Tc = Th - 75 = 266 K 

Lower temperature is Tc = 266 K 
Higher temperature is Th = 341 K

6 0

2 years ago