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kkurt [141]
4 years ago
13

Use the terms "force", "weight", "mass", and "inertia" to explain why it is easier to tackle a 220 lb football player than a 288

lb football player.
Physics
2 answers:
Tomtit [17]4 years ago
7 0
<span><u>Answer </u>
The mass of 220 lb football has less than 288 lb football. So, it will be easier to move it since it will require less force. The heavy football will have a bigger momentum. Since 288 lb has more weight than 220 lb, it will have bigger inertia making it difficult for the players to stop it.
This makes it easier to tackle 220 lb football than 288 lb football. 
</span>
Flura [38]4 years ago
6 0

There are several scientific aspects that require to be comprehended beforehand;

Force = mass * acceleration

Weight = mass * gravity (10m/s)

Momentum = mass * velocity

Inertia is a property of mass which is the tendency of an object to remain at constant velocity unless acted upon by an outside force.

Based on these formulae, it is evident that if the 220lb player and 288lb player are moving at the same velocity, the 288lb player has more inertia and momentum than the 22olb. Therefore, it would require more force (hence energy) to change the direction or the speed (in tackling) of the player with 288lb compared with what would be required on the 220lb player.

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A bicycle wheel is rotating at 49 rpm when the cyclist begins to pedal harder, giving the wheel a constant angular acceleration
raketka [301]
In 9 sec, it increases the angular velocity by (0.45 x 9) rad/s which will give us 4.05 rad/s 
Now get the angular velocity and divide it 2pi = 4.05 by 2(pi) to give how many revolutions 4.05 rad is equivalent to = 0.6446 rps 
Them, multiply this by 60 to get it from rps to rpm increase (0.6446 x 60)=38.676 rpm 
Add this and the starting revolution frequency of 49 rpm to give: (49 rpm + 38.676 rpm) = 87.6760 rpm
6 0
4 years ago
Read 2 more answers
The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ
LenaWriter [7]

The orbital period of the satellite[T] is given as 2*10^{6} S.

The radius of the satellite is given [R] 2.5*10^{12} m.

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way  the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V]   written as -

                                V=\frac{2\pi R}{T}

where T is the time period of the satellite and R is the radius of the satellite.

                        V=\frac{2*3.14*10^{12} }{2*10^{6} }

                               = 7.85*10^{6} m/s

There is also another way through which we can get  the solution as explained below-

We know that the tangential speed of a satellite V=\sqrt{\frac{GM}{R^{2} } }

where G is the gravitational constant and M is the mas of central object.

But we know that g=\frac{GM}{R^{2} }

                               ⇒GM=gR^{2}  where g is the acceleration due to gravity of that central object.


Hence    V=\sqrt{\frac{gR^{2} }{R} }

               ⇒   V=\sqrt{gR}

By knowing the value of g due to that central object we can also calculate its tangential speed.

                           

 




7 0
3 years ago
Read 2 more answers
An optical disk drive in a computer can spin a disk at up to 10,000 rpm. If a particular disk is spun at 7570 rpm while it is be
anastassius [24]

Answer:

The magnitude of the average angular acceleration is calculated as 1822.36\ rad/s^{2}

Explanation:

Maximum speed that can be attained by the disk, N_{m} = 10,000 rpm

Speed of spinning of the disk, N = 7570 rpm

Time taken to come to rest, t = 0.435 s

Now,

The initial angular velocity is given by:

\omega = \frac{2\pi N}{60} = 792.73\ rads^{-1}

Final angular velocity, \omega' = 0\ rads^{- 1}

The average angular acceleration of the disk can be computed by using the kinematic eqn:

\omega' = \omega + \alpha t

0 = 792.73 + 0.435\alpha

\alpha = - 1822.36\ rads^{- 2}

5 0
4 years ago
Read 2 more answers
How can u use physics in the real world
Blizzard [7]
Physics can be used in a variety of ways, a specific example would be rollercoaster. Physics can be used to determine the speed at which a rollercoaster can travel without going off the rails or when it needs to begin to slow down to make a sharp turn.
3 0
4 years ago
Two guitars are playing together producing a beat frequency. The first guitar is playing at a frequency of 867 Hz and the beat i
IgorLugansk [536]

Answer:

892 Hz or 842 Hz

Explanation:

Given

The frequency of first guitar is 867 Hz

Beat is produced for every 0·04 s

∴ Beat has a time period of 0·04 s

<h3>Beat frequency is defined as the inverse of the time period of the beat </h3>

Beat frequency = 1 ÷ (Time period) = 1 ÷ 0·04 = 25 Hz

Let the frequency of the second guitar be f Hz

<h3>Beat frequency is defined as the absolute difference between the two frequencies</h3>

∴ Beat frequency = (867 - f) or (f - 867)

First case

25 = 867 - f ⇒ f = 842 Hz

Second case

25 = f - 867 ⇒ f = 892 Hz

∴ Possible frequencies of second guitar could be 892 Hz or 842 Hz

4 0
3 years ago
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