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statuscvo [17]
3 years ago
13

The engine of a 2000 kg Mercedes going up Pike’s Peak delivers energy to its drive wheel at the rate 100 kW . Neglecting air res

istance, what is the largest speed the car can sustain on the steep Pike’s Peak mountain highway, where the road is 30◦ to the horizontal? The acceleration due to gravity is 10 m/s 2 .
Physics
1 answer:
kykrilka [37]3 years ago
8 0

Answer:

v = 10 m/s

Explanation:

given,

Mass of Mercedes engine = 2000 Kg

Power delivered = 100 kW

angle made with horizontal = 30°

acceleration due to gravity =  10 m/s²

largest speed car can sustain = ?

we know,

Power = Force x velocity

P = F x v

P = mg sinθ x v

P = mg sin 30° x v

P = 0.5 mg  x v

v = \dfrac{100 \times 10^3}{0.5\times 2000 \times 10}

v = 10 m/s

hence, the maximum velocity is equal to v = 10 m/s

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Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. Wh
Alinara [238K]

Answer:

a) k_{avg}=6.22\times 10^{-21}

b) k_{avg}=8.61\times 10^{-21}

c)  k_{mol}=3.74\times 10^{3}J/mol

d)   k_{mol}=5.1\times 10^{3}J/mol

Explanation:

Average translation kinetic energy (k_{avg}) is given as

k_{avg}=\frac{3}{2}\times kT    ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8  

k_{avg}=6.22\times 10^{-21}J

b) at T = 143° C

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416  

k_{avg}=8.61\times 10^{-21}J

c ) The translational kinetic energy per mole of an ideal gas is given as:

       k_{mol}=A_{v}\times k_{avg}

here   A_{v} = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

        k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}

          k_{mol}=3.74\times 10^{3}J/mol

d) now at T = 143° C

        k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}

          k_{mol}=5.1\times 10^{3}J/mol

8 0
3 years ago
If you stretch a rubber hose and pluck it, you can observe a pulse traveling up and down the hose.(i) What happens to the speed
Monica [59]

The thing that happens to the speed of the pulse when you stretch the hose more tightly is that it increases.

<h3>What is wage speed?</h3>

It should be noted that wave speed simply means the distance that a wave travels during a particular time.

It should be noted that higher tension leads to an increase in the speed of the wave.

Therefore, the thing that happens to the speed of the pulse when you stretch the hose more tightly is that it increases.

Learn more about speed on:

brainly.com/question/13943409

#SPJ4

4 0
1 year ago
A ‘thermal tap’ used in a certain apparatus consists of a silica rod which
abruzzese [7]

Correct question is;

A thermal tap used in a certain apparatus consists of a silica rod which fits tightly inside an aluminium tube whose internal diameter is 8mm at 0°C.When the temperature is raised ,the fits is no longer exact. Calculate what change in temperature is necessary to produce a channel whose cross-sectional is equal to that of the tube of 1mm. (linear expansivity of silica = 8 × 10^(-6) /K and linear expansivity of aluminium = 26 × 10^(-6) /K).

Answer:

ΔT = 268.67K

Explanation:

We are given;

d1 = 8mm

d2 = 1mm

At standard temperature and pressure conditions, the temperature is 273K.

Thus; Initial temperature; T1 = 273K,

Using the combined gas law, we have;

P1×V1/T1 = P2×V2/T2

The pressure is constant and so P1 = P2. They will cancel out in the combined gas law to give:

V1/T1 = V2/T2

Now, volume of the tube is given by the formula;V = Area × height = Ah

Thus;

V1 = (πd1²/4)h

V2 = (π(d2)²/4)h

Thus;

(πd1²/4)h/T1 = (π(d2)²/4)h/T2

π, h and 4 will cancel out to give;

d1²/T1 = (d2)²/T2

T2 = ((d2)² × T1)/d1²

T2 = (1² × T1)/8²

T2 = 273/64

T2 = 4.23K

Therefore, Change in temperature is; ΔT = T2 - T1

ΔT = 273 - 4.23

ΔT = 268.67K

Thus, the temperature decreased to 268.67K

6 0
3 years ago
A person can jump 1.5m on the earth. How high could the person jump on a planet having the twice the mass of the earth and twice
MrMuchimi
F=mg=Gm1m2/r^2
g=Gm2/r^2
g=2Gm2/(2r)^2=2Gm2/4r^2=Gm2/2r^2
So since there is half times the gravity on this unknown planet that has twice earth's mass and twice it's radius, then the person can jump twice as high. 1.5*2= 3m high

5 0
4 years ago
Read 2 more answers
A car is going 8 meters per second on an access road into a highway
TiliK225 [7]

Answer:

20.96 m/s^2 (or 21)

Explanation:

Using the formula (final velocity - initial velocity)/time = acceleration, we can plug in values and manipulate the problem to give us the answer.

At first, we know a car is going 8 m/s, that is its initial velocity.

Then, we know the acceleration, which is 1.8 m/s/s

We also know the time, 7.2 second.

Plugging all of these values in shows us that we need to solve for final velocity. We can do so by manipulating the formula.

(final velocity - initial velocity) = time * acceleration

final velocity = time*acceleration + initial velocity

After plugging the found values in, we get 20.96 m/s/s, or 21 m/s

3 0
3 years ago
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