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Alina [70]
4 years ago
14

When you ride a bicycle, in what direction is the angular velocity of the wheels? When you ride a bicycle, in what direction is

the angular velocity of the wheels? to your right forwards up to your left backwards g
Physics
1 answer:
Hitman42 [59]4 years ago
3 0

Complete question is;

When you ride a bicycle, in what direction is the angular velocity of the wheels? A) to your left B) to your right C) forwards D) backwards

Answer:

Option A - to your left

Explanation:

While an object rotates, each particle will have a different velocity:

the 'Speed' component will vary with radius while the 'Direction' component will vary with angle.

All of the velocity vectors are aligned in the same plane.

We can be solve this by choosing a single vector normal to ALL of the possible velocity vectors of the rotating object in that plane.

This convention used is known as "Right-hand rule". The angular velocity vector points along the wheel's axle. For instance, if you Imagine wrapping your right hand around the axle so that your fingers point in the direction of rotation, with your thumb sticking out. You will notice that your thumb points to the left.

Thus;

By right-hand rule, a wheel rotating on a forward - moving bicycle has an angular velocity vector pointing to the rider's left.

So, option A is the correct answer

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A ball rolls downhill with a constant acceleration of 4m/s squared. If it started from rest,it’s velocity at the end of 3 meters
vladimir2022 [97]

Answer:

4.9 m/s

Explanation:

Since the motion of the ball is a uniformly accelerated motion (constant acceleration), we can solve the problem by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the ball in this problem,

u = 0 (it starts from rest)

a=4 m/s^2 is the acceleration

s = 3 m is the distance covered

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(4)(3)}=4.9 m/s

3 0
4 years ago
How to get stud multipliers in lego star wars the skywalker saga?.
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7 0
3 years ago
A ball of mass M is suspended by a thin string (of negligible mass) from the ceiling of an elevator.uploaded image
lilavasa [31]

Answer:

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor.  T > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor. T > mg

(c) The elevator is traveling downward and its downward velocity is decreasing as it nears a stop at a lower floor. T < mg

(d) The elevator is traveling downward at a constant velocity. T = mg

(e) The elevator is traveling downward and its downward velocity is increasing. T < mg

(f) The elevator is stationary and remains at rest. T = mg

Explanation:

To answer this question, consider all the forces acting on the elevator.

The mass of the ball acting downwards due to gravity = mg

The tension on the string depends on upward or downwards force on the ball. T = m(a+g)

where a is acceleration and increase in velocity causes increase in acceleration, and vice versa. (a = v/t)

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor.

If the upward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a+g) > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor.

If the upward velocity is increasing, its acceleration is also increasing.

Then, T = m(a+g) > mg

(c) The elevator is traveling downward and its downward velocity is decreasing as it nears a stop at a lower floor.

If the downward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a-g) < mg

(d) The elevator is traveling downward at a constant velocity

At constant velocity, acceleration is zero, because acceleration is the rate of change of velocity.

T = m(0+g) = mg

(e) The elevator is traveling downward and its downward velocity is increasing

If the downward velocity is increasing, its acceleration is also increasing

T = m(a-g) < mg

(f) The elevator is stationary and remains at rest.

if the elevator is at rest, its acceleration is zero

T = m(0+g) = mg

6 0
3 years ago
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
3 years ago
Explain what happens to the pitch of a cell phone ringing when the amplitude of a sound wave increases.
rewona [7]
As the amplitude of a sound wave increases the pitch of the ringing would be much higher (like if you were to inhale helium.. just with a phone)
3 0
3 years ago
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