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sattari [20]
2 years ago
12

Which of the following is a purpose of rules in sports and games? (5 points)

Physics
1 answer:
ValentinkaMS [17]2 years ago
4 0

Answer:

To keep everyone safe

Explanation:

Without rules people could get hurt

You might be interested in
Explain the difference between speed and velocity and indicate if these are scalar or vector quantities
Natali5045456 [20]
Speed can never be negative because it does not depend in which direction the car moves whereas, velocity will change if a car turns from due North to East.
Quantities which can be described only by their magnitudes are called scalars and those which are described by both, magnitude and direction are vectors
3 0
3 years ago
Usain Bolt's world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bo
ZanzabumX [31]

a) 893 N

b) 8.5 m/s

c) 3816 W

d) 69780 J

e) 8030 W

Explanation:

a)

The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:

F_{net}=ma

where

m is Bolt's mass

a is the acceleration

In the first 0.890 s of motion, we have

m = 94.0 kg (Bolt's mass)

a=9.50 m/s^2 (acceleration)

So, the net force is

F_{net}=(94.0)(9.50)=893 N

And according to Newton's third law of motion, this force is equivalent to the force exerted by Bolt on the ground (because they form an action-reaction pair).

b)

Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:

v=u+at

where

v is the  final speed

u is the initial speed

a is the acceleration

t is the time

In the first phase of Bolt's race we have:

u = 0 m/s (he starts from rest)

a=9.50 m/s^2 (acceleration)

t = 0.890 s (duration of the first phase)

Solving for v,

v=0+(9.50)(0.890)=8.5 m/s

c)

First of all, we can calculate the work done by Bolt to accelerate to a speed of

v = 8.5 m/s

According to the work-energy theorem, the work done is equal to the change in kinetic energy, so

W=K_f - K_i = \frac{1}{2}mv^2-0

where

m = 94.0 kg is Bolt's mass

v = 8.5 m/s is Bolt's final speed after the first phase

K_i = 0 J is the initial kinetic energy

So the work done is

W=\frac{1}{2}(94.0)(8.5)^2=3396 J

The power expended is given by

P=\frac{W}{t}

where

t = 0.890 s is the time elapsed

Substituting,

P=\frac{3396}{0.890}=3816 W

d)

First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.

In the first 0.890 s, the force exerted was

F_1=893 N

We know that the average force for the whole race is

F_{avg}=820 N

Which can be rewritten as

F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}

And solving for F_2, we find the average force exerted by Bolt on the ground during the second phase:

F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}\\F_2=\frac{(0.890+8.69)F_{avg}-0.890F_1}{8.69}=812.5 N

The net force exerted by Bolt during the second phase can be written as

F_{net}=F_2-D (1)

where D is the air drag.

The net force can also be rewritten as

F_{net}=ma

where

a=\frac{v-u}{t} is the acceleration in the second phase, with

u = 8.5 m/s is the initial speed

v = 12.4 m/s is the final speed

t = 8.69 t is the time elapsed

Substituting,

a=\frac{12.4-8.5}{8.69}=0.45 m/s^2

So we can now find the average drag force from (1):

D=F_2-F_{net}=F_2-ma=812.5 - (94.0)(0.45)=770.2 N

So the increase in Bolt's internal energy is just equal to the work done by the drag force, so:

\Delta E=W=Ds

where

d is Bolt's displacement in the second part, which can be found by using suvat equation:

s=\frac{v^2-u^2}{2a}=\frac{12.4^2-8.5^2}{2(0.45)}=90.6 m

And so,

\Delta E=Ds=(770.2)(90.6)=69780 J

e)

The power that Bolt must expend just to voercome the drag force is given by

P=\frac{\Delta E}{t}

where

\Delta E is the increase in internal energy due to the air drag

t is the time elapsed

Here we have:

\Delta E=69780 J

t = 8.69 s is the time elapsed

Substituting,

P=\frac{69780}{8.69}=8030 W

And we see that it is about twice larger than the power calculated in part c.

3 0
3 years ago
A block of weight w sits on a frictionless inclined plane, which makes an angle theta with respect to the horizontal, as shown.
zavuch27 [327]

Answer:

a.

Explanation:

6 0
3 years ago
The sound source of a ship’s sonar system operates at a frequency of 22.0 kHz . The speed of sound in water (assumed to be at a
Degger [83]

Answer:

147.456077993 Hz

Explanation:

f_0 = Frequency of the sonar = 22 kHz

v_w = Velocity of the whale = 4.95 m/s

v = Speed of sound in water = 1482 m/s

The difference in frequency is given by

\Delta f=f_0\times\dfrac{2v_{w}}{v-v_w}\\\Rightarrow \Delta f=22000\times\dfrac{2\times 4.95}{1482-4.95}\\\Rightarrow \Delta f=147.456077993\ Hz

The difference in frequency is 147.456077993 Hz

6 0
3 years ago
A certain copper wire has a resistance of 13.0 Ω . At some point along its length the wire was cut so that the resistance of one
alekssr [168]

Answer with Explanation:

Let r be the resistance of short piece of copper wire.

Resistance of copper wire=R=13\Omega

Resistance is directly proportional to length.

If a wire has greater resistance then,the wire will be greater in length.

Therefore,resistance of long piece of wire=7r

Total resistance of copper  wire=Sum of resistance of two piece of wires

r+7r=13

8r=13

r=\frac{13}{8}ohm

Resistance of long piece of wire=7\times\frac{13}{8}=\frac{91}{8}\Omega

Resistance of short piece of wire =\frac{13}{8}\Omega

Resistivity of wire and cross section area of wire remains same .

Let L be the total  length  of wire and L' be the length of short  piece of wire.

We know that

R=\frac{\rho L}{A}=\frac{\rho}{A}L=KL

\frac{R}{L}=K

Where K=\frac{\rho}{A}=Constant

Using the formula

\frac{13}{L}=\frac{\frac{13}{8}}{L'}

\frac{L'}{L}=\frac{13}{8}\times \frac{1}{13}=\frac{1}{8}

L'=\frac{L}{8}

Length of short piece of wire=L'=\frac{L}{8}

Length of long piece of  wire=L-L'=L-\frac{L}{8}=\frac{8L-L}{8}=\frac{7}{8}L

% of length of short piece of   wire=\frac{\frac{L}{8}}{L}\times 100=12.5%%

The resistance of the short piece=\frac{13}{8}\Omega

The resistance of the long piece=\frac{91}{8}\Omega

8 0
3 years ago
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