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2 years ago

7

Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a

750-ft ramp at a high speed v0 and travels 540 ft in 6 s at constant deceleration before its speed is reduced to v0/2. Assuming the same constant deceleration, determine (a) the additional time required for the truck to stop, (b) the additional distance traveled by the truck.

1 answer:

Furkat [3]2 years ago

6 0

Answer:

bfghhg

Explanation:

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You are testing a new amusement park roller coaster with an empty car with a mass of 130 kg. One part of the track is a vertical

vlada-n [284]

Answer:

Work done by friction along the motion is given as

$W_f = -5857.8 J$

Explanation:

As per work energy theorem we can say

Work done by all forces = change in kinetic energy of the system

so here car is moving from bottom to top

so here the change in kinetic energy is total work done on the car

so here we will have

$W_f + W_g = \frac{1}{2}m(v_f^2 - v_i^2)$

$W_f - mgH = \frac{1}{2}m(v_f^2 - v_i^2)$

now plug in all data in it

$W_f - (130)(9.81)(2\times 12) = \frac{1}{2}(130)(8^2 - 25^2)$

$W_f = 30607.2 - 36465$

$W_f = -5857.8 J$

6 0

1 year ago

Calculate the mass of air in a room of floor dimensions =10M×12M and height 4m(Density of air =1.26kg/m cubic

Alinara [238K]

The volume of the room is the product of its dimensions:

$10\times 12 \times 4 = 480\text{ m}^3$

Now, from the equation

$d=\dfrac{m}{V}$

where d is the density, m is the mass and V is the volume, we deduce

$m=dV$

So, multiply the density and the volume to get the mass of air in the room.

7 0

2 years ago

3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag

aleksandr82 [10.1K]

Answer:

(a) $W=-19.25J$

(b) $W=-52.8J$

Explanation:

Hello.

(a) In this case, since the initial volume is 18.5 dm³ and the final volume is 21 dm³ (18.5 +2.5), we can compute the work at constant pressure as shown below:

$W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J$

Which is negative as it expands against the given pressure.

(b) Moreover, of the process is carried out reversibly, the pressure can change, therefore, we need to compute the work via:

$W=nRTln(\frac{V_1}{V_2} )$

Whereas the moles are computed from the given mass of argon:

$n=6.56g*\frac{1mol}{39.95g}=0.164mol$

Thus, the work is:

$W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J$

Regards.

4 0

1 year ago

A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th

ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

A.

In the first question, we have to caculate the constant of the spring with this equation:

$m*g=k*x$

Getting the k:

$k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]$

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

$x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\$

The energy of a spring:

$E=\frac{1}{2}*k*x^{2}$

For the first case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]$

For the second case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]$

If you take the relation E2/E1 = 4.

B.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

$E=\frac{1}{2}*k*x^{2}$ ⇒ $k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]$

C.

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

$E=W*h=500[N]*10[m]=5000[J]$

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

D.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

7 0

1 year ago

Given an electron beam whose electrons have kinetic energy of 10.0 kev , what is the minimum wavelength λmin of light radiated b

kolbaska11 [484]

To answer the problem we would be using this formula which isE = hc/L where E is the energy, h is Planck's constant, c is the speed of light and L is the wavelength
L = hc/E = 4.136×10−15 eV·s (2.998x10^8 m/s)/10^4 eV

= 1.240x10^-10 m

= 1.240x10^-1 nm

3 0

2 years ago

Read 2 more answers