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Stels [109]
3 years ago
8

A 1500 kg car is approaching a hill that has a height of 12 m. As the car reaches the bottom of the hill it runs out of gas and

has a constant speed of 10 m/s. Will the car make it to the top of the hill?
Physics
1 answer:
ICE Princess25 [194]3 years ago
3 0

Answer:

No, the car will not make it to the top of the hill.

Explanation:

Let ΔX be how long the slope of the hill is, Δx be how far the car will travel along the slope of the hill, Ф be the angle the slope of the hill makes with the horizontal(bottom of the hill), ki be the kinetic energy of the car at the bottom of the hill and vi be the velocity of the car at the bottom of the hill and kf be the kinetic energy of the car when it stop moving at vf.

Since Ф is the angle between the horizontal and the slope, the relationship between the angle and the slope and the height of the hill is given by

sinФ = 12/ΔX

Which gives you the slope as

ΔX = 12/sinФ

Therefore for the car to reach the top of the hill it will have to travel ΔX.

Ignoring friction the total work done is given by

W = ΔK

W = (kf - ki)

Since the car will come to a stop, kf = 0 J

W = -ki

m×g×sinФ×Δx = 1/2×m×vi^2

(9.8)×sinФ×Δx = 1/2×(10)^2

sinФΔx = 5.1

Δx = 5.1/sinФ

ΔX>>Δx Ф ∈ (0° , 90°)

(Note that the maximum angle Ф is 90° because the slope of a hill can never be greater ≥ 90° because that would then mean the car cannot travel uphill.)

Since the car can never travel the distance of the slope, it can never make it to the top of the hill.

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For your final exam in electronics, you’re asked to build an LC circuit that oscillates at 10 kHz. In addition, the maximum curr
Marina CMI [18]

Answer:

 L= 2 mH

C=1.26\times 10^{-7}\ F            

Explanation:

Given that

Frequency , f= 10 kHz

Maximum current ,I = 0.1 A

Maximum energy stored ,E= 1 x 10⁻⁵ J

The maximum energy stored in the inductor is given as follows

E=\dfrac{1}{2}LI^2

Where ,L= Inductance

I=Current

E=Energy

Now by putting the values in the above equation

10^{-5}=\dfrac{1}{2}\times L\times 0.1^2

L=\dfrac{2\times 10^{-5}}{0.1^2}\ H

L=0.002 H

L= 2 mH

We know that frequency f is given as

2\pi f=\dfrac{1}{\sqrt{LC}}

C=Capacitance  , f=frequency ,L=Inductance

Now by putting the values

2\pi \times 10\times 10^3=\dfrac{1}{\sqrt{0.002\times C} }

62831.85=\dfrac{1}{\sqrt{0.002\times C}}

\sqrt{0.002\times C=\dfrac{1}{62831.85}

0.002\times C=0.0000159^2

C=\dfrac{0.0000159^2}{0.002}\ F

C=1.26\times 10^{-7}\ F

Therefore the inductance and capacitance will be 2 mH and 1.26 x 10⁻⁷ F respectively.

6 0
3 years ago
Two samples of water are mixed together.
horrorfan [7]

Let the cold water go up x degrees.

Let the hot water go down 100 - x degrees.

The formula for heat exchange is m*c*delta t

Givens

Ice

deltat = x

m = 0.50 kg

c = 4.18

Hot water

deltat = 100 - x

m = 1.5 kg

c = 4.18

Formula

The heat up = heat down

0.50 * c * x = 1.5 * c * (100 - x)            Divide both sides by c

Solution

0.50 *x = 1.5*(100 - x)                          Remove the brackets.

0.5x = 150 - 1.5x                                  Add 1.5x to both sides.

0.5x + 1.5x = 150 - 1.5x + 1.5x             Combine like terms  

2x = 150                                               Divide by 2

x = 75

Answer

A

6 0
3 years ago
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