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3 years ago

Discuss how the motion of the stunt cyclist relates to the forces acting on the stunt cyclist during the collision

Physics

1 answer:

ivolga24 [154]3 years ago

6 0

Answer:

hello your question is incomplete hence I will give you a general answer on how a stunt cyclist performs his stunts successfully

answer ;

The stunt cyclist trying to perform some sort of stunt with the bicycle will be faced with some forces like the static fiction between the tires of the bicycle and wall and also a centripetal force.

In order to overcome the centripetal force the minimal grip of the bicycle must be equal to the weight and the speed at which the cyclist must go can be represented as :

$V_{min} = \sqrt{\frac{r.g}{u} }$

Explanation:

The stunt cyclist trying to perform some sort of stunt with the bicycle will be faced with some forces like the static fiction between the tires of the bicycle and wall and also a centripetal force.

In order to overcome the centripetal force the minimal grip of the bicycle must be equal to the weight and the speed at which the cyclist must go can be represented as :

$V_{min} = \sqrt{\frac{r.g}{u} }$

where ; r = radius of the tunnel where the stunt is to be performed

g = gravitational speed

u = coefficient of static friction

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Divide that answer by the accepted value.

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3 years ago

Why do scientist study fossils?

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3 years ago

Two 1.0 cm * 2.0 cm rectangular electrodes are 1.0 mm apart. What charge must be placed on each electrode to create a uniform el

kvv77 [185]

Answer:

The number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.

Explanation:

<u>Step 1:</u> calculate the charge on each electrode

Given;

Electric field strength = 2.0 X 10⁶ N/C

The distance between the electrode = 1mm = 1 X 10⁻³ m

Electric field strength (E) = Force (F)/Charge (q)

$E =\frac{Kq}{r^2}$

where;

E is the electric field strength = 2.0 X 10⁶ N/C

K is coulomb's constant = 8.99 X 10⁹ Nm²/C²

r is the distance between the electrodes = 1 X 10⁻³ m

q is the charge in each electrode = ?

$q = \frac{Er^2}{K} = \frac{(2X10^6)(1X10^{-3})^2}{8.99 X10 ^9}$ = 0.2225 X 10⁻⁹ C

The charge on each electrode is 0.2225 X 10⁻⁹ C

<u>Step 2:</u> calculate the number of electrons to be moved from one electrode to the other.

1 electron contains 1.602 X 10⁻¹⁹ C

So, 0.2225 X 10⁻⁹ C will contain how many electrons ?

= (0.2225 X 10⁻⁹)/(1.602 X 10⁻¹⁹)

= 1.4 X 10⁹ electrons

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3 years ago

The transfer of heat is through direct contact of particles is called

mixas84 [53]

The answer would be Conduction!

Hope this helps!!

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3 years ago

Hi.

docker41 [41]

<h3>Answer :</h3>

Let the final temperature be "T".

For the piece of copper :

mass, $\sf{m_c=40\ g.}$

specific heat capacity, $\sf{c_c=0.4\ J\,g^{-1}\,K^{-1}.}$

initial temperature, $\sf{T_c=200^{\circ}C.}$

Then the heat of copper :

$\sf{\dashrightarrow Q_c=m_cc_c\,\Delta\!T_c}$

$\sf{\dashrightarrow Q_c =16(T-200)\ J}$

For copper calorimeter :

mass, $\sf{m_{cc} =60\ g.}$

specific heat capacity, $\sf{c_{cc} =0.4\ J\,g^{-1}\,K^{-1}.}$

initial temperature, $\sf{T_{cc} =25^{\circ}C.}$

Then the heat of copper calorimeter :

$\sf{\dashrightarrow Q_{cc} =m_{cc}c_{cc}\,\Delta\!T_{cc}}$

$\sf{\dashrightarrow Q_{cc} =24(T-25)\ J}$

For water :

mass, $\sf{m_w=50\ g. }$

specific heat capacity, $\sf{c_w= 4.2\ J\,g^{-1}\,K^{-1}.}$

initial temperature, $\sf{T_w=25^{\circ}C.}$

Then heat of water :

$\sf{\dashrightarrow Q_w=m_wc_w\,\Delta\!T_w}$

$\sf{\dashrightarrow Q_w=210(T-25)\ J}$

By energy conservation, the sum of all these energies should be zero as there were no heat energy change before the process, i.e.,

$\sf{\dashrightarrow Q_c+Q_{cc}+Q_w=0}$

$\sf{\dashrightarrow16(T-200)+24(T-25)+210(T-25)=0}$

$\sf{\dashrightarrow 250T- 9050=0}$

$\sf{\dashrightarrow T=36.2^{\circ}C}$

$\large \underline{\underline{\boxed{\sf T=36.2^{\circ}C}}}$

<u>____________________________</u>

[Note: in case of considering temperature difference it's not required to convert the temperatures from $\sf{^{\circ}C}$ to K or K to $\sf{^{\circ}C}$ .]

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3 years ago