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3 years ago

The radii of the sprocket assemblies and the wheel of the bicycle in the figure are:

1 answer:

7 0

To solve this task we have to make a proportion, but firstly we have to set up all the main points : so, the distance is s=r(B), that has its r=radius,B=angle in rad velocity v=ds/dt= w(r)
Do not forget about w = angular speed in rad/s and $w1 = 1 revolution/sec = 2Pi (rad/s)$
Now we can go to proportion
$v1=v2$
$w1*r1 = w2r2$ $w2 = w1 * r1/r2 = 2w1 = 4Pi (rad/s)$
$w2 = w3 (which is the angular velocity of the rear wheel) 
$
SOLVING FOR A : $v3 = w3 * r3 = 4pi * 14 (inch/s) = 14.66 ft/sec$
$v3 = 14.66 ft/sec(1 mile/5280 ft)( 3600 sec/h)$ $= 9.99$ or something about 10 mph --- SOLVING FOR B.
I'm sure it helps!

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What is the value of acceleration due to gravity, g, on Earth?

qwelly [4]

Explanation:

It varies with altitude, but at sea level, it's 9.8 m/s².

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4 years ago

At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There

katen-ka-za [31]

a) Constant

b) Constant

Explanation:

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

$\tau_{net} = I \alpha$ (1)

where

$\tau_{net}$ is the net torque acting on the object in rotation

I is the moment of inertia of the object

$\alpha$ is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

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

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time interval

So we can rewrite eq.(1) as

$\tau_{net}=I\frac{\Delta \omega}{\Delta t}$

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

$\tau_{net}=0$

From the previous equation, this implies that

$\Delta \omega =0$

Which means that the angular velocity at that instant does not change anymore.

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

$\tau_{net}=0$

Therefore, this means that

$\Delta \omega =0$

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

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3 years ago

A table-tennis ball is thrown at a stationary bowling ball. The table-tennis ball makes a one-dimensional elastic collision and

Alecsey [184]

Option b. A smaller magnitude of momentum and more kinetic energy.

<h3>What is a momentum?</h3>

In Newtonian physics, an object's linear momentum, translational momentum, or simply momentum is defined as the product of its mass and velocity.
It has both a magnitude and a direction, making it a vector quantity. The object's momentum, p, is defined as: p=mv if m is the object's mass and v is its velocity (also a vector quantity).
The kilogram metre per second (kg m/s), or newton-second in the International System of Units (SI), is the unit used to measure momentum.
The rate of change of a body's momentum is equal to the net force exerted on it, according to Newton's second law of motion.

To know more about momentum, refer:

brainly.com/question/1042017

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2 years ago

An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 43.8 N,

maw [93]

A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant k is

43.8 N = k (0.155 m) ==> k = (43.8 N) / (0.155 m) ≈ 283 N/m

The total work done on the spring to stretch it to 15.5 cm from equilibrium is

1/2 (283 N/m) (0.155 m)² ≈ 3.39 J

The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be

1/2 (283 N/m) (0.259 m)² ≈ 9.48 J

Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.

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3 years ago

Which levels of government establish and implement educational requirements for minors

romanna [79]

Answer:

Federal Role in Education is primarily a State and local responsibility in the United States. It is States and communities, as well as public and private organizations of all kinds, that establish schools and colleges, develop curricula, and determine requirements for enrollment and graduation.

Explanation:

The Role of Public Education in Supporting American Democracy. Since the founding of public education in the United States, public schools have been charged not only with giving future workers skills for the private marketplace, but also with preparing students to be citizens in a democracy.

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3 years ago