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Naya [18.7K]
2 years ago
10

A force of only 150 N can lift a 600 N sack of flour to a height of 0.50 m when using a lever as shown in the diagram below. a.

Find the work done on the sack of flour (in J). b. Find the distance you must push with the 150 N force-on the left side (in m). c. Briefly explain the benefit of using a lever to lift a heavy object.

Physics
1 answer:
Snowcat [4.5K]2 years ago
8 0
Actually the question is not clear. The Benefit of lifting using lever is that you can apply force in a convenient direction and you can use very less force to lift object by balancing the torque due to object.For example you want to lift a mass of 4N with a force 2N you can use a class 2 lever and maintain the ratio between the distances of the body and the point of application of force from the fulcrum to be 1:2.In any case balance the torque to get the required force.

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In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water d
Alexxandr [17]

Answer:

a) a = g / 3

b) x (3.0) = 14.7 m

c) m (3.0) = 29.4 g

Explanation:

Given:-

- The following differential equation for (x) the distance a rain drop has fallen has the form:

                             x*g = x * \frac{dv}{dt} + v^2

- Where,                v = Speed of the raindrop

- Proposed solution to given ODE:

                             v = a*t

Where,                  a = acceleration of raindrop

Find:-

(a) Using the proposed solution for v find the acceleration a.

(b) Find the distance the raindrop has fallen in t = 3.00 s.

(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.

Solution:-

- We know that acceleration (a) is the first derivative of velocity (v):

                             a = dv / dt   ... Eq 1

- Similarly, we know that velocity (v) is the first derivative of displacement (x):

                            v = dx / dt  , v = a*t ... proposed solution (Eq 2)

                             v .dt = dx = a*t . dt

- integrate both sides:

                             ∫a*t . dt = ∫dt

                             x = 0.5*a*t^2  ... Eq 3

- Substitute Eq1 , 2 , 3 into the given ODE:

                            0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2

                                                = 1.5 a^2 t^2

                            a = g / 3

- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:

                           x (t) = 0.5*a*t^2

                           x (t = 3.0) = 0.5*9.81*3^2 / 3

                           x (3.0) = 14.7 m  

- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:

                           m (t) = k*x (t)

                           m (3.0) = 2.00*x(3.0)

                           m (3.0) = 2.00*14.7

                           m (3.0) = 29.4 g

6 0
2 years ago
A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.
Sedaia [141]

Answer:

335°C

Explanation:

Heat gained or lost is:

q = m C ΔT

where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Heat gained by the water = heat lost by the copper

mw Cw ΔTw = mc Cc ΔTc

The water and copper reach the same final temperature, so:

mw Cw (T - Tw) = mc Cc (Tc - T)

Given:

mw = 390 g

Cw = 4.186 J/g/°C

Tw = 22.6°C

mc = 248 g

Cc = 0.386 J/g/°C

T = 39.9°C

Find: Tc

(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)

Tc = 335

7 0
2 years ago
A 5.00-kg ball is hanging from a long but very light flexible wire when it is struck by a 1.50-kg stone traveling horizontally t
devlian [24]

consider the right direction as positive and left direction as negative.

M = mass of the ball = 5 kg

m = mass of stone = 1.50 kg

V_{bi} = initial velocity of the ball before collision = 0 m/s

V_{si} = initial velocity of the stone before collision = 12 m/s

V_{bf} = final velocity of the ball after collision = ?

V_{sf} = final velocity of the stone after collision = - 8.50 m/s

using conservation of momentum

MV_{bi} + mV_{si} = MV_{bf} + mV_{sf}

(5) (0) + (1.5) (12) = 5 V_{bf} + (1.50) (- 8.50)

V_{bf} = 6.15 m/s

h = height gained by the ball

using conservation of energy

Potential energy gained by ball at Top = kinetic energy at the bottom

Mgh = (0.5) MV_{bf}^{2}

(9.8) h = (0.5) (6.15)²

h = 1.93 m

8 0
2 years ago
Read 2 more answers
A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force
VikaD [51]

Answer:

a) the values of the angle α is 45.5°

b) the required magnitude of the vertical force, F is 41 lb

Explanation:

Applying the free equilibrium equation along x-direction

from the diagram

we say

∑Fₓ = 0

Pcosα - 425cos30° = 0

525cosα - 368.06 = 0

cosα = 368.06/525

cosα = 0.701

α = cos⁻¹ (0.701)

α = 45.5°

Also Applying the force equation of motion along y-direction

∑Fₓ = ma

Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)

525sin45.5° + F + 212.5 - 600 = 27.95

374.46 + F + 212.5 - 600 = 27.95

F - 13.04 = 27.95

F = 27.95 + 13.04

F = 40.99 ≈ 41 lb

8 0
2 years ago
A bicyclist travels the first 800 m of a trip 1.4 minutes, the next 500 m in 1.6 minutes, and finishes up the final 1200 m in 2
soldier1979 [14.2K]

Answer:

 v_average = 500 m / min

Explanation:

Average speed is defined

         v = (x_{f} -x₀) / Δt

let's look in each section

section 1

the variation of the distance is 800 in a time of 1.4 min

         v₁ = 800 / 1.4

         v₁ = 571.4 m / min

section 2

distance interval 500 in a 1.6 min time interval

         v₂ = 500 / 1.6

         v₂ = 312.5 m / min

section 3

distance interval 1200 m in a time 2 min

         v₃ = 1200/2

         v₃ = 600 m / min

taking the speed of each section we can calculate the average speed

         

the distance traveled

        Δx = 800 + 500 + 1200

        Δx = 2500 m

the time spent

        Δt = 1.4 + 1.6+ 2

        Δt = 5 min

         v_average = Δx / Δt

         v_average = 2500/5

         v_average = 500 m / min

7 0
2 years ago
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