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4 years ago

Im bad at work problems can any one help with this problem ?

1 answer:

4 0

We will put the number of trips in the first column, the miles driven in the second column and gallons of fuel used in the third column.

8 7,680 1,010
7 9,940 1,330
12 14,640 1,790
12 13,920 2,050

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A toy rocket that weights 10 N blasts straight up from ground level with an initial kinetic energy of 40J. at the exact top of i

marishachu [46]

Answer:

14 m

Explanation:

Applying,

P.E = WH............ Equation 1

Where P.E = potential energy at the highest vertical height, W = weight of the toy rocket, H = Vertical height.

Note: At the highest vertical height, the kinetic energy is 0 J

Therefore,

Mechanical energy(M.E) = Potential Energy(P.E)

make H the subject of the equation

H = P.E/W......... Equation 2

From the question,

Given: W = 10 N, P.E = 140 J

Substitute these values in equation 2

H = 140/10

H = 14 m

8 0

3 years ago

Assume you have a rocket in Earth orbit and want to go to Mars. The required change in velocity is ΔV≈9.6km/s . There are two op

Nostrana [21]

Answer: Part 1: Propellant Fraction (MR) = 8.76

Part 2: Propellant Fraction (MR) = 1.63

Explanation: The Ideal Rocket Equation is given by:

Δv = $v_{ex}.ln(\frac{m_{f}}{m_{e}} )$

Where:

$v_{ex}$ is relationship between exhaust velocity and specific impulse

$\frac{m_{f}}{m_{e}}$ is the porpellant fraction, also written as MR.

The relationship $v_{ex}$ is: $v_{ex} = g_{0}.Isp$

To determine the fraction:

Δv = $v_{ex}.ln(\frac{m_{f}}{m_{e}} )$

$ln(MR) = \frac{v}{v_{ex}}$

Knowing that change in velocity is Δv = 9.6km/s and $g_{0}$ = 9.81m/s²

Note: Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.

Part 1: Isp = 450s

$ln(MR) = \frac{v}{v_{ex}}$

ln(MR) = $\frac{9.6.10^{3}}{9.81.450}$

ln (MR) = 2.17

MR = $e^{2.17}$

MR = 8.76

Part 2: Isp = 2000s

$ln(MR) = \frac{v}{v_{ex}}$

ln (MR) = $\frac{9.6.10^{3}}{9.81.2.10^{3}}$

ln (MR) = 0.49

MR = $e^{0.49}$

MR = 1.63

8 0

3 years ago

A 220 g mass is on a frictionless horizontal surface at the end of a spring that has force constant of 7.0

Talja [164]

The concept of conservation of energy and harmonic motion allows to find the result for the power where the kinetic and potential energy are equal is:

x = 0.135 cm

Given parameters

The mass m = 220 g = 0.220 kg

The spring cosntnate3 k = 7.0 N / m

Initial displacement A = 5.2 cm = 5.2 10-2 m

To find

The position where the kinetic and potential energy are equal

A simple harmonic movement is a movement where the restoring force is proportional to the displacement, the result of this movement is described by the expression.

x = A cos wt + fi

w² = $\frac{k}{m}$

Where x is the displacement from the equilibrium position, A the initial amplitude of the system, w the angular velocity t the time, fi a phase constant determined by the initial conditions, k the spring constant and m the mass.

The speed is defined by the variation of the position with respect to time.

v = $\frac{dx}{dt}$

let's evaluate

v = - A w sin (wt + Ф)

Since the body releases for a time t = 0 the velocity is zero, therefore the expression remains.

0 = - A w sin Ф

For the equality to be correct, the sine function must be zero, this implies that the phase constant is zero

x = A cos wt

Let's find the point where the kinetic and potential energy are equal.

K = U

½ m v² = m g x

we substitute

½ A² w² sin² wt = g A cos wt

sin² wt = $\frac{2g}{A}$ cos wt

let's calculate

w = $\sqrt{\frac{7}{0.220} }$

w = 5.64 rad / s

sin² 5.64t = 2 9.8 / 0.052 cos 5.64t

sin² 5.64t = 376.92 cos 5.64 t

1 - cos² 5.64t = 376.92 cos 5.64t

cos² 5.64t -376.92 cos564t -1 = 0

we make the change of variable

x = cos 5.64t

x²- 376.92 x - 1 = 0

x = 0.026

cos 5.64t = 0.026

Let's find the displacement for this time

x = 5.2 10-2 0.026

x = 1.35 10-3 m

In conclusion Using the concepts of conservation of energy and harmonic motion we can find the result for the could where the kientic and potential enegies are equal is:

x = 0.135 cm

Learn more here: brainly.com/question/15707891

8 0

3 years ago

A 50-cm-long spring is suspended from the ceiling. A 330g mass is connected to the end and held at rest with the spring unstretc

Anvisha [2.4K]

Explanation:

Given that,

Length of the spring, l = 50 cm

Mass, m = 330 g = 0.33 kg

(A) The mass is released and falls, stretching the spring by 28 cm before coming to rest at its lowest point. On applying second law of Newton at 14 cm below the lowest point we get :

$kx=mg\\\\k=\dfrac{mg}{x}\\\\k=\dfrac{0.33\times 9.8}{0.14}\\\\k=23.1\ N/m$

(B) The amplitude of the oscillation is half of the total distance covered. So, amplitude is 14 cm.

(C) The frequency of the oscillation is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{23.1}{0.33}} \\\\f=1.33\ Hz$

5 0

4 years ago

A 30 g bullet moving a horizontal velocity of 500 m/s comes to a stop 12 cm within a solid wall. (a) what is the change in the b

Sidana [21]

M = 30 g = 0.03 kg, the mass of the bullet
v = 500 m/s, the velocity of the bullet

By definition, the KE (kinetic energy) of the bullet is
KE = (1/2)*m*v²
= 0.5*(0.03 kg)*(500 m/s)² = 3750 J
Because the bullet comes to rest, the change in mechanical energy is 3750 J.

The work done by the wall to stop the bullet in 12 cm is
W = (1/2)*(F N)*(0.12 m) = 0.06F J

If energy losses in the form of heat or sound waves are ignored, then
W = KE.
That is,
0.06F = 3750
F = 62500 N = 62.5 kN

Answer:
(a) 3750 J
(b) 62.5 kN

7 0

4 years ago