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ElenaW [278]
3 years ago
14

Describe how personal fitness contributes to physical,mental/emotional,social health

Physics
1 answer:
Semenov [28]3 years ago
3 0
When you work out it helps clear your thoughts and think of nothing. It makes you feel good.
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A 10 kg ball has a momentum of 40 kg.m/s.what is the ball's speed?
iren [92.7K]
M = 10kg
P = 40 kg.m/s
V= 4m/s

Formula v = p/m

V= p/m
=4okg.m/s / 10kg
=4m/s
7 0
3 years ago
A sculpture is suspended in equilibrium by two cables, one from a wall and the other
aleksandrvk [35]

Answer:

T_1=6655.295917 \approx 6655.3N

Explanation:

From the question we are told that

Angle of cable 2 \theta=37.0\textdegree

Weight of sculpture W=5000 N

Generally the Tension from cable 2 T_2 is mathematically given by

   T_2sin37\textdegree=5000N

   T_2=5000N/sin37\textdegree

   T_2=8308.2N

Generally the Tension from Cable 1 T_1 is mathematically given by

   T_1=T_2 cos37\textdegree

   T_1=8308.2* cos 37\textdegree

   T_1=6655.295917 \approx 6655.3N

7 0
3 years ago
Most scientists believe the Big Bang Theory explains which of the following questions?
kari74 [83]
A how our planets and moons formed
5 0
3 years ago
Read 2 more answers
Bats use a process called echolocation to find their food. This involves giving out sound waves that hit possible prey or food.
cupoosta [38]
The type of waves used by bats are sound waves. Most of the species use their larynx to produce ultrasound waves in the frequency range of 20 to 200 kilohertz.
These sound waves are echoed, reflected, by surroundings, in this case food or prey. These reflections are received by the specialized receptor cells in the ears of bats. The reflections are analyzed by the brain to make an image.
Fun fact: The brain cells of bats are also specialized to better analyze the frequency of ultrasound used by the bat.
8 0
3 years ago
Read 2 more answers
A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim
Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

F_D = -\frac{1}{2}\rho V^2 C_d A

Where

\rho is the density of the flow

V = Velocity

C_d= Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

F=ma=m\frac{dV}{dt}

Equating both equations we have:

m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A

m(dV)=-\frac{1}{2}\rho C_d A (dt)

\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)

Integrating

\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)

-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t

Here,

V_f = 60mph = 26.82m/s

V_i = 120.7m/s

m= 1600lbf = 725.747Kg

\rho = 1.21 kg/m^3

C_d = 0.3

\Delta t=7s

Replacing:

\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)

-0.029 = -5.4997r^2

r = 2.2963m

d= r*2 = 4.592m \approx 15.065ft

4 0
3 years ago
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