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miss Akunina [59]
3 years ago
15

A dog has a speed of 7 m/s and a mass of 45 kg. What is the dog's kinetic energy?

Physics
2 answers:
Sonbull [250]3 years ago
5 0
K.E = 1/2 m*v^2
K.E = 1/2 * 45 * (7)^2
K.w = 1102.5 J
Elan Coil [88]3 years ago
3 0
The kinetic energy should be 1,102.5 joules.
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A car moving at a speed of 36 km/h reaches the foot of a smooth
boyakko [2]

Answer:

d = 10.2 m

Explanation:

When the car travels up the inclined plane, its kinetic energy will be used to do the work in climbing up. So according to the law of conservation of energy, we can write that:

Kinetic\ Energy\ of\ the \ Car = Work\ Done\ while\ moving\ up\ the\ plane\\\frac{1}{2}mv^{2} = Fd

where,

m = mass of car

v = speed of car at the start of plane = (36 km/h)(1000 m/1 km)(1 h/3600 s)

v = 10 m/s

F = force on the car in direction of inclination = W Sin θ

W = weight of car = mg

θ = Angle of inclinition = 30°

d = distance covered up the ramp = ?

Therefore,

\frac{1}{2}mv^{2} = mgdSin\theta\\\frac{1}{2}v^{2} = gdSin\theta\\\frac{1}{2}(10\ m/s)^{2} = d(9.81\ m/s^{2}) Sin\ 30^{0}

<u>d = 10.2 m</u>

4 0
2 years ago
In a practical machine, the power output is smaller than the power input. Considering the law of energy conservation, what can e
Grace [21]

The law of energy conservation states that energy cannot be created or destroyed, so choices B to D are immediately invalid. Choice A can explain this occurrence: <u>A. Some of the energy is used to combat friction, and thus is transformed from mechanical energy to heat.</u>

5 0
3 years ago
Read 2 more answers
An inelastic collision of two objects is characterized by the following.
serious [3.7K]

Options:

(a) Total kinetic energy of the system remains constant.

(b) Total momentum of the system is conserved.

(c) Both A and B are true.

(d) Neither A nor B are true.

Answer:

(b) Total momentum of the system is conserved.

Explanation:

An inelastic collision is a type of collision in which momentum is conserved and kinetic energy is not conserved. That is, there is loss of kinetic energy.

In an inelastic collision:

Total momentum before collision = Total momentum after collision

An example of inelastic collision is seen in the ballistic pendulum, The ballistic pendulum is a device in which a projectile such as a bullet is fired into a suspended heavy wooden stationary block.

8 0
3 years ago
Before beginning a long trip on a hot day, a driver inflates anautomobile tire to a gauge pressure of 2.70 atm at 300 K. At the
anygoal [31]

Answer:

The value is the temperature of the air inside the tire T_{2} = 340.54 K

% of the original mass of air in the tire should be released 99.706 %

Explanation:

Initial gauge pressure = 2.7 atm

Absolute pressure at inlet P_{1} = 2.7 + 1 = 3.7 atm

Absolute pressure at outlet P_{2} = 3.2 + 1 = 4.2 atm

Temperature at inlet T_{1} = 300 K

(a) Volume of the system is constant so  pressure is directly proportional to the temperature.

\frac{T_{2} }{T_{1} } = \frac{P_{2} }{P_{1} }

\frac{T_{2} }{300}  = \frac{4.2}{3.7}

T_{2} = 340.54 K

This is the value is the temperature of the air inside the tire

(b). Since volume of the tyre is constant & pressure reaches the original value.

From ideal gas equation P V = m R T

Since P , V & R is constant. So

m T = constant

m_{1}  T_{1} =  m_{2}  T_{2}

\frac{m_{2} }{m_{1}  } = \frac{T_{1} }{T_{2} }

\frac{m_{2} }{m_{1}  } = \frac{300}{354.54}

\frac{m_{2} }{m_{1}  } =0.00294

value of  the original mass of air in the tire should be released is  \frac{m_{2} - m_{1}}{m_{1}}

\frac{0.00294-1}{1}

⇒ -0.99706

% of the original mass of air in the tire should be released 99.706 %.

8 0
2 years ago
Water with a volume flow rate of 0.001 m3/s, flows inside a horizontal pipe with diameter of 1.2 m. If the pipe length is 10m an
galben [10]

Answer:

\triangle P=1.95*10^{-4}

Explanation:

Mass m=0.001

Diameter d=1.2m

Length l=10m

Generally the equation for Volume flow rate is mathematically given by

 Q=AV

 V=\frac{Q}{\pi/4D^2}

 V=\frac{0.001}{\pi/4(1.2)^2}

 V=8.84*10^{-4}

Generally the equation for Friction factor is mathematically given by

 F=\frac{64}{Re}

Where Re

Re=Reynolds Number

 Re=\frac{pVD}{\mu}

 Re=\frac{1000*8.84*10^{-4}*1.2}{1.002*10^{-3}}

 Re=1040

Therefore

 F=\frac{64}{Re}

 F=\frac{64}{1040}

 F=0.06

Generally the equation for Friction factor is mathematically given by

 Head loss=\frac{fLv^2}{2dg}

 H=\frac{0.06*10*(8.9*10^-4)^2}{2*1.2*9.81}

 H=19.9*10^{-9}

Where

H=\frac{\triangle P}{\rho g}

\triangle P=\frac{19.9*10^{-9}}{10^3*(9.81)}

\triangle P=H*\rho g

\triangle P=1.95*10^{-4}

 

6 0
3 years ago
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