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3 years ago

5

A bicycle has wheels of 0.70 m diameter. The bicyclist accelerates from rest with constant acceleration to 22 km/h in 10.8 s. Wh

at is the angular acceleration of the wheels?

1 answer:

Anna11 [10]3 years ago

3 0

Answer:

$\alpha=0.16\frac{rad}{s^2}$

Explanation:

Angular acceleration is defined as the variation of angular speed with respect to time:

$\alpha=\frac{\omega}{t}$

The relation between the angular speed and the linear speed is given by:

$\omega=\frac{v}{r}$

Replacing (2) in (1):

$\alpha=\frac{v}{rt}$

We need to convert $\frac{km}{h}$ to $\frac{m}{s}$ :

$22\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=6.11\frac{m}{s}$

Recall that:

$r=\frac{d}{2}=\frac{0.7m}{2}=3.5m$

Replacing:

$\alpha=\frac{6.11\frac{m}{s}}{(3,5m)(10.8s)}\\\alpha=0.16\frac{rad}{s^2}$

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A Man Moved first a Distance of 1000 m in 25 second and 2.5 km in 50 second along a in straight line?

11Alexandr11 [23.1K]

Answer:

Average speed = 46.67 m/s

Explanation:

Given that the time taken in covering first 1000 m = 25 seconds.

The time taken in covering next 2.5 km = 50 seconds.

Total distance covered = 1000 m + 2500 m = 3500 m

Total time taken = 25+50=75 seconds

Average speed = Total distance covered / total time taken

= 3500/75 = 46.67 m/s

3 0

2 years ago

Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person's ankle. The cord is

Damm [24]

Complete question is;

Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person's ankle. The cord is 40 feet long, but can stretch up to 120 feet. They both start from the same location. Allyson moves 10 ft/sec and Adrian moves 9 ft/sec in the directions indicated. Adrian stops moving at time t = 5.5 sec, but Allyson keeps on moving 10 ft/sec in the indicated direction. (If a coordinate system is used, assume that the girls' starting position is located at

(x, y) = (0, 0) and that Allyson and Adrian move in the positive y and negative x directions, respectively. Let one unit equal one foot.)

Compute the length of the bungee cord at t = 7 seconds. (Round your answer to three decimal places.)

Answer:

Length of bungee cord = 85.734 ft

Explanation:

We are told that Adrian moves 9ft/sec. Thus, at 5.5 seconds, distance he moved is; 9 ft/sec × 5.5sec = 49.5 ft in the negative x (-x) direction. Therefore, the coordinate is (-49.5, 0).

Now, Allyson has moved 10ft/sec. Thus, at 7 seconds, distance he moved would be; 10 ft/sec x 7sec = 70 feet in the positive (+y) direction. Therefore, the coordinate is (0, 70).

Now, since they started from the origin, it means (0, 0) is a coordinate. Thus, we now have 3 coordinates which are; (0, 0), (0, 70) & (-49.5,0). These 3 coordinates would therefore combine to form a right triangle.

The hypotenuse is the distance between Allyson and Adrian.

Thus, from pythagoras theorem, we can find the distance between them which is same as the length of the cord.

Thus;

(-49.5)² + 70² = D².

D² = 2450.25 + 4900

D = √7350.25

D = 85.734 ft

4 0

2 years ago

You and a partner sit on the floor and stretch out a coiled spring to a length of 7.2 meters. You shake the coil so you

vekshin1

Answer:

Approximately $5.9\; {\rm m\cdot s^{-1}}$ (assuming that the partner is holding the other end of the coil stationary.)

Explanation:

In a standing wave, an antinode is a point that moves with maximal amplitude, while a node is a point that does not move at all. There is an antinode between every two adjacent nodes. Likewise, there is a node between every two adjacent antinodes.

The side of the spring that is being shaken moving with maximal amplitude. Hence, that point on this spring would also be an antinode. In contrast, the side of the spring that is held still (does not move at all) would be a node.

There would be a node between:

the antinode at the end of the spring that is being shaken, and
the antinode between the two ends of this spring.

Overall, the nodes and antinodes on this spring would be:

node at the end that is being held still,
antinode (as mentioned in the question),
node (inferred, not mentioned in the question), and
antinode at the end that is being shaken.

The distance between two adjacent nodes is equal to one-half (that is, $(1/2)$ ) the wavelength of the wave. The distance between a node and an adjacent antinode is one-quarter (that is, $(1/4)$ ) of the wavelength of the wave.

Thus, if the wavelength of the wave in this question is $\lambda$ , the length of this spring would be:

$\displaystyle \frac{1}{2}\, \lambda + \frac{1}{4}\, \lambda = \frac{3}{4}\, \lambda$ .

The question states that the length of this coiled spring is $7.2\; {\rm m}$ . In other words, $(3/4) \, \lambda = 7.2\; {\rm m}$ . The wavelength of this wave would be $(7.2\; {\rm m}) / (3/4) = 9.6\; {\rm m}$ .

The frequency $f$ of this wave is the number of cycles in unit time:

$\begin{aligned} f &= \frac{10}{16.3\; {\rm s}} \approx 0.613\; {\rm s^{-1}}\end{aligned}$ .

Hence, the speed $v$ of this wave would be:

$\begin{aligned} v &= \lambda\, f \\ &=9.6\; {\rm m} \times 0.613\; {\rm s^{-1}} \\ &\approx 5.9\; {\rm m \cdot s^{-1}}\end{aligned}$ .

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2 years ago

A tennis ball is dropped from 1.13 m above the

Alex73 [517]

Answer:

-4.71 m/s

Explanation:

Given:

y₀ = 1.13 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.13 m)

v = -4.71 m/s

7 0

2 years ago

It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (For example, w

Charra [1.4K]

Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

q > m

As we know the formula:

F = Eq

And we also know that :

F = Bqv

F = $\frac{mv^{2} }{r}$

Bqv = $\frac{mv^{2} }{r}$

or Eq = $\frac{mv^{2} }{r}$

Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

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3 years ago