Which of the following is NOT an advantage of using a wheel and axle?

What is the speed of a truck traveling 10km in 10 minutes

user100 [1]

Answer: 58.8235 km/h

speed = distance/time

the distance is 10 km

the time is 10 minutes

the unit is not correct, so we first change minute to hour

so 10/60 is 0.166667, rounded to 0.17.

10 km/ 0.17 hours =

8 0

1 year ago

An elevator motor provides 45.0 kW of power while lifting an elevator 35.0 m. If the elevator contains seven passengers each wit

mylen [45]

Find how much work ∆W is done by the motor in lifting the elevator:

P = ∆W / ∆t

where

• P = 45.0 kW = power provided by the motor

• ∆W = work done

• ∆t = 20.0 s = duration of time

Solve for ∆W :

∆W = P ∆t = (45.0 kW) (20.0 s) = 900 kJ

In other words, it requires 900 kJ of energy to lift the elevator and its passengers. The combined mass of the system is M = (m + 490.0) kg, where m is the mass of the elevator alone. Then

∆W = M g h

where

• g = 9.80 m/s² = acceleration due to gravity

• h = 35.0 m = distance covered by the elevator

Solve for M, then for m :

M = ∆W / (g h) = (900 kJ) / ((9.80 m/s²) (35.0 m)) ≈ 2623.91 kg

m = M - 490.0 kg ≈ 2133.91 kg ≈ 2130 kg

4 0

3 years ago

a cyclist coasting down a 5.0 ◦ incline at a constant speed of 6.0 km/h because of air resistance. If the total mass of the bicy

Dvinal [7]

Answer:

$F_{net}= 85.41\ N$

Explanation:

mass of the bicycle + cyclist = 50 kg

constant speed = 6 km/h

a cyclist coasting down a 5.0° incline

the downward velocity is constant, so net acceleration must be zero

the air drag must be equal to gravitational force downward along the ramp

$F_a = mg sin \theta$

now for upward motion

$F_{net} = mg sin \theta + air\ drag$

$F_{net} = mg sin \theta + mg sin \theta$

$F_{net} = 2 mg sin \theta$

$F_{net} = 2\times 50 \times 9.8 sin 5^0$

$F_{net}= 85.41\ N$

3 0

3 years ago

I need help with questions 4 and 5 because my teacher is not good with explaining them to a better understanding (Geometry)

Helga [31]

Hello! I can help you with this!

4. For this problem, we have to write and solve a proportion. We would set this proportion up as 12/15 = 8/x. This is because we're looking for the length of the shadow and we know the height of the items, so we line them up horizontally and x goes with 8, because we're looking for the shadow length. Let's cross multiply the values. 15 * 8 = 120. 12 * x = 12. You get 120 = 12x. Now, we must divide each side by 12 to isolate the "x". 120/12 is 10. x = 10. There. The cardboard box casts a shadow that is 10 ft long.

5. For this question, you do the same thing. This time, you're finding the height of the tower, so you would do 1.2/0.6 = x/7. Cross multiply the values in order to get 8.4 = 0.6x. Now, divide each side by 0.6x to isolate the "x". 8.4/0.6 is 14. x = 14. There. The tower is 14 m tall.

If you need more help on proportions and using proportions in real life situations, feel free to search on the internet to find more information about how you solve them.

4 0

3 years ago

Two vehicles A and B accelerate uniformly from rest.

spayn [35]

Answer:

(i) Please find attached the required velocity time graphs plotted with MS Excel

(ii) The velocity of vehicle A at the 18th second = 20 m/s

The velocity of vehicle B at the 18th second = 0 m/s

(iii) The distance between the two vehicles at the moment in (ii) above is 60 meters

Explanation:

The given parameters of the motion of vehicles A and B are;

The acceleration of vehicles A and B = Uniform acceleration starting from rest

The maximum velocity attained by vehicle A = 30 m/s

The time it takes vehicle A to attain maximum velocity = 10 s

The maximum velocity attained by vehicle B = 30 m/s

The time it takes vehicle B to attain maximum velocity = The time it takes vehicle A to attain maximum velocity = 10 s

The time duration vehicle A maintains its maximum velocity = 6 s

The time duration vehicle B maintains its maximum velocity = 4 s

(i) From the question, we get the following table;

$\begin{array}{ccc}Time &V_A&V_B\\0&0&0\\10&30&40\\14&30&40\\16&30&20\\18&20&0\\22&0&\end{array}$

From the above table the velocity time graphs of vehicles A and B is created with MS Excel and can included here

(ii) The velocity of vehicle A at the start = 0 m/s

After accelerating for 10 seconds, the velocity of vehicle A = The maximum velocity of vehicle A = 30 m/s

The maximum velocity is maintained for 6 seconds which gives;

At 10 s + 6 s = 16 s, the velocity of vehicle A = 30 m/s

The time it takes vehicle A to decelerate to rest = 6 s

The deceleration of vehicle A, $a_A$ = (30 m/s - 0 m/s)/(6 s) = 5 m/s²

Therefore, we get;

v = u - $a_A$ ·t

At the 18th second, the deceleration time, t = 18 s - 16 s = 2 s

u = 30 m/s

∴ v₁₈ = 30 - 5 × 2 = 20

The velocity of vehicle A at the 18th second, $V_{18A}$ = 20 m/s

For vehicle B, we have;

At the 14th second, the velocity of vehicle B = 40 m/s

Vehicle B decelerates to rest in, t = 4 s

The deceleration of vehicle B, $a_B$ = (40 m/s - 0 m/s)/(4 s) = 10 m/s²

For vehicle B, at the 18th second, t = 18 s - 14 s = 4 s

∴ $v_{18B}$ = 40 m/s - 10 m/s² × 4 s = 0 m/s

The velocity of the vehicle B at 18th second, $v_{18B}$ = 0 m/s

(iii) The distance covered by vehicle A up to the 18th second is given by the area under the velocity-time graph as follows;

The area triangle A₁ = (1/2) × 10 × 30 = 150

Area of rectangle, A₂ = 6 × 30 = 180

Area of trapezoid, A₃ = (1/2) × (30 + 20) × 2 = 50

The distance covered in the 18th second by vehicle $S_A$ = A₁ + A₂ + A₃

∴ $S_A$ = 150 + 180 + 50 = 380

The distance covered in the 18th second by vehicle $S_A$ = 380 m

The distance covered by the vehicle B in the 18th second is given by the area under the velocity time graph of vehicle B as follows;

Area of trapezoid, A₅ = (1/2) × (18 + 4) × 40 = 440

The distance covered by the trapezoid, $S_B$ = 440 m

The distance of the two vehicles apart at the 18t second, $S_{AB}$ = $S_B$ - $S_A$

∴ $S_{AB}$ = 440 m - 380 m = 60 m

The distance of the two vehicles from one another at the 18th second, $S_{AB}$ = 60 m.

5 0

3 years ago