When an ideal gas does positive work in its surroundings, which of the ga's quantities increases?

On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s

dsp73

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

= 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2

= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

4 0

3 years ago

The spacing between the plates of a 1.0 μF capacitor is 0.050 mm. a) What is the surface area of the plates? b) How much charge

nataly862011 [7]

Answer:

(a) surface area of the plate will be equal to $1.129m^2$

(b) Charge on the capacitor is equal to $1.5\times 10^{-6}C$

Explanation:

We have given spacing between the plates d = 0.05 mm = $5\times 10^{-5}m$

Value of capacitance $C=1\mu F=10^{-6}F$

(A) Capacitance of a parallel plate capacitor is equal to $C=\frac{\epsilon _0A}{d}$

So $10^{-6}=\frac{8.85\times 10^{-12}\times A}{10^{-5}}$

$A=1.129m^2$

So surface area of the plate will be equal to $1.129m^2$

(B) It is given that capacitor is charged by 1.5 volt

So voltage V = 1.5 volt

Charge on the capacitor is equal to $Q=CV$

So $Q=1.5\times 10^{-6}C$

5 0

3 years ago

I WILL GIVE BRAINLIST just please help with these problems 3

statuscvo [17]

Answer:

16. 68.18 Km/h

17. 3 miles.

Explanation:

16. Determination of the speed

Distance travelled = 150 Km

Time = 2.2 hours

Speed =?

Speed is simply defined as the distance travelled with time. Mathematically, it is expressed as:

Speed = Distance / time

With the above formula, we can obtain the speed as follow:

Distance travelled = 150 Km

Time = 2.2 hours

Speed =?

Speed = Distance /time

Speed = 150 / 2.2

Speed = 68.18 Km/h

17. Determination of the distance.

Speed = 3 mph

Time = 1 hour

Distance =?

Speed = Distance /time

3 = distance / 1

Distance = 3 miles

3 0

3 years ago

In an internal combustion engine, heat flow into a gas causes it to ......

ivann1987 [24]

In an internal combustion engine, heat flow into a gas causes it to expand.
The application of direct force to specific parts of the engine will produce expansion of the high-temperature and high-pressure gases. Which will transform the chemical energy from the fuel (such as gasoline or oi) into mechanical energy.

5 0

3 years ago

For the circuit in the previous part, what happens to the maximum current if the frequency is doubled and the inductance is halv

tamaranim1 [39]

Answer:

Following are the responses to these question:

Explanation:

Since the $max^{m}$ is the current of ckt which depend on the reactance which inductor that also enables the ckt and inductor resistance $(X_L)$ for capacities