A train travelling at 18m/s accelerates uniformly at 2m/s² for 6 seconds. Calculate the final speed of the train.

A gas undergoes two processes. In the first, the volume remains constant at 0.200 m3m3 and the pressure increases from 2.50×105

SVETLANKA909090 [29]

Answer:

$W = -4.95 \times 10^4\ J$

Explanation:

given,

In first case Volume remains constant.

Work done in the first case is zero.

In Second case Volume change

V₁ = 0.2 m³

V₂ = 0.11 m³

Pressure, P = 5.5 x 10⁵ Pa

Work done = Pressure x change in volume

W = P ΔV

$W = P(V_2-V_1)$

$W = 5.5\times10^5\times (0.11 - 0.2)$

$W = -4.95 \times 10^4\ J$

Hence, Work done when volume changes is equal to $W = -4.95 \times 10^4\ J$

5 0

2 years ago

If you lose control of your vehicle and collide with a fixed object, such as a tree, at 60 m.p.h., the force of impact is the sa

My name is Ann [436]

You can compare the velocity of the car, 60 mph, with the velocity that a mass would acquire when falls from certain height.

First, convert 60 mph to m/s:

60 miles/h * 1.60 km/mile * 1000 m/km * 1h/3600s = 26.67 m/s

Second, calculate from what height a body in free fall reachs 26.67 m/s velocity when hits the floor.

free fall => Vf^2 = 2g*H => H = Vf^2 / (2g)

H = (26.67m/s)^2 / (2*9.8 m/s) = 36.2 m

If you consider that the height between the floors of a building is approximately 3.6 m, you get 36.2 m / 3.6 m/floor = 10 floors.

Then, you conclude that the force of impact is the same as driving you vehicle off a 10 story building.

7 0

3 years ago

Sonar is a device that uses reflected sound waves to measure underwater depths. If a sonar signal has a frequency of 288 Hz and

mart [117]

Answer:

5.03 m

Explanation:

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency of the wave

For the sonar signal in this problem,

$f=288 Hz$

$v=1.45\cdot 10^3 m/s$

Substituting into the equation, we find the wavelength:

$\lambda=\frac{1.45\cdot 10^3 m/s}{288 Hz}=5.03 m$

3 0

3 years ago

A delivery truck leaves a warehouse and travels 2.60 km north. The truck makes a left turn and travels 1.25 km west before makin

yulyashka [42]

Answer:

4.19 km and 107.35 degrees north of east

Explanation:

So in the end, the truck is (2.6 + 1.4 = 4km) north and 1.25 km west from the warehouse. We can use the Pythagorean formula to calculate the magnitude and direction α of the truck displacement from the warehouse:

$s = \sqrt{s_n^2 + s_w^2} = \sqrt{4^2 + 1.25^2} = \sqrt{16 + 1.5625} = \sqrt{17.5625} = 4.19$ km