A 500 kg motorcycle accelerates at a rate of 2 m/s .how much force was applied to the motorcycle?

A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is to be replaced by a high- efficiency moto

daser333 [38]

Convert the shaft ouput from HP to kW

Shaft output = 75HP = 55.93kW

1st: Finding for the power consumption based on 55.93kW output

Power consumption (Old) = 55.93kW / .910 = 61.46kW

Power consumption (New) = 55.93kW / .954 = 58.63kW

2nd: Total power used in kWh:

Power Used = Power consumption * load factor * Hours:

Power (Old) = 61.46kW * .75 * 4368 = 201343 kWh

Power (New) = 58.63kW * .75 * 4368 = 192072 kWh

Energy saved = 201343 kWh - 192072 kWh = 9,271 kWh

3rd: Calculating for the price:

Price = kW-Hr * $/kWh

Price (Old) = 201343kWh * $0.08/kWh = $16107.44

Price (New) = 192072 kWh * $0.08/kWh = $15365.76

Cost saved = $16107.44 - $15365.76 = $741.68/yr

4th: Setting up the cost equation:

Cost over time, F(t) = Motor_Cost + (Price * Number of Years, t)

Cost (Old) = 5449 + 16107.44*t

Cost (New) = 5520 + 15365.76*t

Equate the two to find for t when they cost equally:

5449 + 16107.44*t = 5520 + 15365.76*t

16107.44*t = 15365.76*t +71

16107.44*t - 15365.76*t = 71

741.68*t = 71

t = 71 / 741.68 = .095 years = 35 days

So the payback period is after 35 days.

6 0

2 years ago

A firecracker breaks up into several pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.

MakcuM [25]

Answer:

The magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.

Explanation:

Hi there!

The total momentum is calculated as the sum of the momenta of the pieces.

The momentum of each piece is calculated as follows:

p = m · v

Where:

p = momentum.

m = mass.

v = velocity.

The momentum is a vector. The 200 g-piece flies along the x-axis then, its momentum will be:

p = (m · v, 0)

p = (0.200 kg · 82.0 m/s, 0)

p = (16.4 kg m/s, 0)

The 300 g-piece flies along the y-axis. Its momentum vector will be:

p =(0, m · v)

p = (0, 0.300 kg · 45.0 m/s)

p = (0, 13.5 kg m/s)

The total momentum is the sum of each momentum:

Total momentum = (16.4 kg m/s, 0) + (0, 13.5 kg m/s)

Total momentum = (16.4 kg m/s + 0, 0 + 13.5 kg m/s)

Total momentum = (16.4 kg m/s, 13.5 kg m/s)

The magnitude of the total momentum is calculated as follows:

$|p| = \sqrt{(16.4 kgm/s)^2+(13.5 kg m/s)^2}= 21.2 kg m/s$

The direction of the momentum vector is calculated using trigonometry:

cos θ = px/p

Where px is the horizontal component of the total momentum and p is the magnitude of the total momentum.

cos θ = 16.4 kg m/s / 21.2 kg m/s

θ = 39.3 (39.5° if we do not round the magnitude of the total momentum)

Then, the magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.

6 0

2 years ago

A common carnival ride, called a gravitron, is a large r = 11 m cylinder in which people stand against the outer wall. the cylin

Leya [2.2K]

Answer:

v = 13.19 m / s

Explanation:

This problem must be solved using Newton's second law, we create a reference system where the x-axis is perpendicular to the cylinder and the Y-axis is vertical

X axis

N = m a

Centripetal acceleration is

a = v² / r

Y Axis

fr -W = 0

fr = W

The force of friction is

fr = μ N

Let's calculate

μ (m v² / r) = mg

μ v² / r = g

v² = g r / μ

v = √ (g r /μ)

v = √ (9.8 11 / 0.62)

v = 13.19 m / s

7 0

2 years ago

A clothes dryer in a home has a power of 4,500 watts and runs on a special 220-volt household circuit,

ladessa [460]

1). Power = (voltage)² / (Resistance)

 4,500 = (220)² / Resistance

Multiply each side by (resistance) : 4,500 x resistance = (220)²

Divide each side by 4,500 : Resistance = (220)² / 4,500 = 10.76 ohms

2). Power = (voltage) x (Current)

Divide each side by (voltage): Power / voltage = Current

 4,500 / 220 = 20.45 Amperes

3). 4,500 watts = 4.5 kilowatts

 (4.5 kilowatts) x (4 hours) = 18 kilowatt-hours

3 0

2 years ago

You are called as an expert witness to analyze the following auto accident: Car B, of mass 2000 kg, was stopped at a red light w

OLga [1]

Answer:

a). va=17.23 $\frac{m}{s}$ or 38.54 mph

b). v=38.54 mph and limit is 35 mph

c). Completely inelastic

d). Eka=192.967 kJ

Ekt=76.071 kJ

Explanation:

$m_{a}=1300kg\\m_{b}=2000kg\\x_{f}=7.25m\\u_{k}=0.65$

The motion is an inelastic collision so

$m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}$

The force of the motion is contrarest by the force of friction so

$F-F_{uk} =0\\F=F_{uk}\\F_{uk}=u_{k}*m*g\\F=m*a\\a=\frac{F}{m}\\ a=\frac{F_{uk}}{m}\\a=\frac{u_{k}*m*g}{m}\\a=u_{k}*g\\a=0.65*9.8\frac{m}{s^{2}} \\a=6.39\frac{m}{s^{2}}$

Now with the acceleration can find the time and the velocity final that make the distance 7.25m being united

$x_{f}=x_{o}+v_{o}*t+2*a*t^{2}\\x_{o}=0\\v_{o}=0\\x_{f}=2*a*t^{2}\\t^{2}=\frac{x_{f}}{2*a}\\t=\sqrt{\frac{7.25m}{6.37\frac{m}{s^{2} } } } \\t=1.06s$

So the velocity final can be find using this time

$v_{f}=v_{o}+a*t\\v_{o}=0\\v_{f}=6.37\frac{m}{s^{2} } *1.06s\\v_{f}=6.79 \frac{m}{s}$

a).

Replacing in the first equation the final velocity can find the initial velocity

$m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}$

$v_{b}=0$

$v_{a}= \frac{(m_{a}+m_{b)*v_{f}}}{m_{a}}\\v_{a}= \frac{(1300+2000)*6.37}{1300}\\v_{a}=17.23 \frac{m}{s}$

b).

$35mph*\frac{1m}{0.000621371mi} *\frac{1h}{3600s}=15.646\frac{m}{s}$

Velocity limit in m/s is 15.646 m/s and the initial velocity is 17.23 m/s

so is exceeding the speed limit in about 1.58 m/s

or in miles per hour

3.5 mph

c).

The collision is complete inelastic because any mass can be returned to the original mass, so even they are no the same mass however in the moment they move the distance 7.25m as a same mass the motion is considered completely inelastic

d).

$Ek=\frac{1}{2}*m*(v)^{2}\\ Eka=\frac{1}{2}*1300kg*(17.23\frac{m}{s})^{2}\\Eka=192.967 kJ\\Ekt=\frac{1}{2}*m*(v)^{2}\\Ekt=\frac{1}{2}*3300kg*(6.79\frac{m}{s})^{2}\\Ekt=76.071 kJ$

8 0

2 years ago