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ICE Princess25 [194]
2 years ago
7

In a previous problem, you calculated how much energy was needed to accelerate a 2,100-kg Tesla Model S from 0 to 100 km/hr, and

found that the value was not dependent on how long it took. However, the power is! Unlike gas engines, electric engines supply equal power regardless of their speed. A Tesla Model S is capable of generating 345 kW of power. Assuming no losses of energy to friction, what is the minimum length of time it takes to accelerate a Tesla Model S from rest to 100 km/hr?
Physics
1 answer:
Naddika [18.5K]2 years ago
5 0

Answer:

t=2.348s

Explanation:

First we need to calculate the Potential Energy of a Tesla to reach the 100km/h

So we know that,

W=KE_f - KE_i = \frac{1}{2}m(V_f^2-V_i^2)=\frac{1}{2}(2100)(27.78^2)W=8.1*10^5J

Here we can to calculate the time through,

t=\frac{W}{P}

t=\frac{8.1*10^5}{345*10^3}

t=2.348s

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A pressure that will support a column of Hg to a height of 256 mm would support a column of water to what height? The density of
Paul [167]

Answer:

<em>The height of water in the column = 348.14 cm</em>

Explanation:

<em>Pressure:</em><em>This is defined as the ratio of the force acting normally ( perpendicular) to the area of surface in contact. The S.I unit of  pressure is N/m²</em>

<em>p = Dgh............... Equation 1</em>

<em>Where p = pressure, D = density, g = acceleration due to gravity, h = height.</em>

<em>From the question, the same pressure will support the column of mercury and water.</em>

<em>p₁ = p₂</em>

<em>Where p₁ = pressure of mercury, p₂ = pressure of water</em>

D₁gh₁ = D₂gh₂.................. Equation 2

making h₂ the subject of equation 2

h₂ = D₁gh/D₂g............... Equation 3

Where D₁ and D₂ = Density of mercury and water respectively, h₁ and h₂ = height of mercury and water respectively

Given: D₁ = 13.6 g/cm³, D₂ = 1.00 g/cm³, h₁ = 256 mm = 25.6 cm.

Constant: g = 9.8 m/s²

Substituting these values into Equation 3,

h₂ = (13.6×9.8×25.6)/1×9.8

<em>h₂ = 348.14 cm</em>

<em>The height of water in the column = 348.14 cm</em>

6 0
2 years ago
The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at
denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m  _____ eqn (1)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s  _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)  

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

<u>K.E of Oxygen = 4000 J</u>

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

<u>v = 2149.24 m/s</u>

<u></u>

8 0
3 years ago
Imagine you have just witnessed a small avalanche on a mountain while skiing, and two slushy snowballs just crashed together in
inn [45]
We have that the momentum p is given by the formula p=mv where m is the mass and v is the velocity. Since for A p=-14kgm/s and m=7, we have that the velocity is -14/7=-2m/s. Hence its speed is 2 m/s.
For b we have that p=15kgm/s and v=3m/s. Because m=p/v, we have m=3kg.
We also have that the momentum is conserved in this system. Hence, the net sum of the momentum of the 2 snowballs equals the momentum of the single giant ball. Hence, p(total)=p(combined)=-14+15=1kgm/s (momentum is a vector; the positive sign means that it tends to the positive direction).

6 0
3 years ago
Read 2 more answers
A truck covers 47.0 m in 8.60 s while smoothly slowing down to final speed of 2.30 m/s. (a) Find its original speed.
Kruka [31]

Explanation:

Given that,

Distance, s = 47 m

Time taken, t = 8.6 s

Final speed of the truck, v = 2.3 m/s

Let u is the initial speed of the truck and a is its acceleration such that :

a=\dfrac{v-u}{t}.............(1)

Now, the second equation of motion is :

s=ut+\dfrac{1}{2}at^2

Put the value of a in above equation as :

s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2

s=\dfrac{t(u+v)}{2}

u=\dfrac{2s}{t}-v

u=\dfrac{2\times 47}{8.6}-2.3

u = 8.63 m/s

So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.

8 0
3 years ago
What type of friction is using chalk in the summer to draw on the ground in Copley square?
Lera25 [3.4K]
The second option rolling friction
7 0
3 years ago
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