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3 years ago

A motorist driving at 25 meters/second decelerates to

Physics

2 answers:

prisoha [69]3 years ago

6 0

Answer:

Kinetic energy = (1/2) (mass) (speed)²

Before slowing down, the car's speed is 25 m/s,

and its kinetic energy is ...

(1/2) (1,500 kg) (25 m/s)²

= (1/2) (1,500 kg) (625 m²/s²)

= 468,750 joules .

After slowing down, the car's speed is 15 m/s,

(1/2) (1,500 kg) (15 m/s)²

= (1/2) (1,500 kg) (225 m²/s²)

= 168,750 joules.

The car lost (468,750 - 168,750) = 300,000 joules

and you heard it from the KING

timama [110]3 years ago

5 0

Kinetic energy = (1/2) (mass) (speed)²

Before slowing down, the car's speed is 25 m/s,
and its kinetic energy is ...

   (1/2) (1,500 kg) (25 m/s)²

      = (1/2) (1,500 kg) (625 m²/s²)

      =    468,750 joules .

After slowing down, the car's speed is 15 m/s,
and its kinetic energy is ...

   (1/2) (1,500 kg) (15 m/s)²

      = (1/2) (1,500 kg) (225 m²/s²)

      = 168,750 joules.

The car lost (468,750 - 168,750) = 300,000 joules of K.E.

The law of Conservation of Energy says:

That 300,000 joules had to go somewhere.

If it's a standard, gas-powered car, then the kinetic energy got
put into the brakes. The energy turned into heat, and the heat
was carried off in the air.

If it's a more modern electric or hybrid car, then the kinetic energy
spun the wheel motors, turning them temporarily into electrical
generators. The generators converted the kinetic energy into
electrical energy, which got put back into the car's batteries, and
could be used again. That's why electric cars use less gas.

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A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s

Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8 $^{-3}$

Explanation:

$m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg$

$v_{bullet}=341\frac{m}{s}$

Momentum of the motion the first part of the motion have a momentum that is:

$P_{1}=m_{bullet}*v_{bullet}$

$P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529$

The final momentum is the motion before the action so:

a).

$P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}$

$P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}$

$P_{1}=P_{2}$

$2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}$

b).

kinetic energy

$K=\frac{1}{2}*m*(v)^{2}$

Kinetic energy after

$Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J$

Kinetic energy before

$Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J$

Ratio = $\frac{Ka}{Kb}$

$R=\frac{1.14}{406.4}\\R=2.8x10^{-3}$

3 0

4 years ago

25 Pts. - URGENT

Alex_Xolod [135]

Answer:

I'm pretty sure that it's -225, hope this helps

3 0

3 years ago

7. An engineer is using a wire that has a resistance of 1.5 . This resistance is too high for the application he is designing. T

Veronika [31]

Answer and Explanation:

We know that resistance $R=\frac{\rho l}{A}$ from the given equation of resistance it is clear that resistance depends on resistivity length and area of the material but we can not change the length because it is given that the length must be 2.5 cm long.

So we can do two two things to reduce the resistance

increase the cross sectional area
decrease the resistivity of the material

7 0

3 years ago

A water hose is used to fill a large cylindrical storage tank of

ludmilkaskok [199]

Answer:

maximum possible speed by solving above equation for 7D is

$v_{max} = \sqrt{\frac{49}{5}gD}$

minimum possible value of speed for solving x = 6D is given as

$v_{min} = \sqrt{9gD}$

Explanation:

Let the nozzle of the hose be at the origin. Then the nearest part of the rim of the tank is at (, ) = (6, 2) and the furthest part of the rim is at (, ) = (7, 2).

The trajectory of the water can be found as follows:

$x = (v_o cos45) t$