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2 years ago

12

what is the mechanical advantage of a crowbar when a worker uses 10N of force to pry open a window that has a resistance of 500N

?

2 answers:

Brrunno [24]2 years ago

5 0

Answer:50

Explanation:

Oksana_A [137]2 years ago

3 0

Answer:

50

Explanation:

The mechanical advantage of a machine is given by

$MA=\frac{F_{out}}{F_{in}}$

where

$F_{out}$ is the output force

$F_{in}$ is the input force

For the crowbar in this problem,

$F_{in}=10 N$ is the force in input applied by the worker

$F_{out}=500 N$ is the force that the machine must apply in output to overcome the resistance of the window and to open it

Substituting into the equation, we find

$MA=\frac{500}{10}=50$

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Explanation:

it’s something that’s in intentional out of body experience

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2 years ago

Read 2 more answers

. A person weighing 750 N gets on an elevator.

Kobotan [32]

F = 750 N (Force)

d = 10 m (displacement )

t = 25 s (time)

L = ? (Mechanical work ) = (Energy)

P = ? (Power)

Solve:

L = F × d = 750 × 10 = 7500 Joules

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2 years ago

An object with velocity 141 ft/s has a kinetic energy of 1558.71 ftâˆ™lbf, on a planet whose gravity is 31.5 ft/s2. What is its

Sidana [21]

Answer:

The mass of the object is 5.045 lbm.

Explanation:

Given;

kinetic energy of the object, K.E = 1558.71 ft.lbf

velocity of the object, V = 141 ft/s

The kinetic energy of the object is calculated as;

$K.E = \frac{1}{2} mV^2\\\\mV^2 = 2K.E\\\\m = \frac{2K.E}{V^2} \\\\1 \ lbf = 32.174 \ lbm.ft/s^2\\\\m = \frac{2 \ \times \ 1558.71 \ ft.lbf \ \times \ 32.174 \ lbm.ft/s^2 }{(141 \ ft/s)2 \ \ \times \ \ \ \ 1 \ lbf\ }$

$m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm.ft^2/s^2 }{(141 )^2\ ft^2/s^2 }\\\\m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm }{(141 )^2 }\\\\m = 5.045 \ lbm$

Therefore, the mass of the object is 5.045 lbm.

6 0

2 years ago

An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and

Natasha_Volkova [10]

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

$\Delta P.E=mg(h_{2}-h_{1})$

$\Delta P.E=10000\times9.8\times(10000-0)$

$\Delta P.E=10000\times9.8\times10000$

$\Delta P.E=980000000\ J$

$\Delta P.E=980\ MJ$

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

$\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)$

$\Delta K.E=148298642\ J$

$\Delta K.E=148.3\ MJ$

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

7 0

2 years ago

A ball is thrown with an initial velocity of u=(10i +15j) m/s. Whan it reaches the top of it trajectory neglecting air resistanc

liraira [26]

Answer:

v = (10 i ^ + 0j ^) m / s, a = (0i ^ - 9.8 j ^) m / s²

Explanation:

This is a missile throwing exercise.

On the x axis there is no acceleration so the velocity on the x axis is constant

v₀ₓ = 10 m / s

On the y-axis velocity is affected by the acceleration of gravity, let's use the equation

v_y = $v_{oy}$ - g t

$v_{y}^2 = v_{oy}^2 - 2 g (y - y_o)$

at the highest point of the trajectory the vertical speed must be zero

v_y = 0

therefore the velocity of the body is

v = (10 i ^ + 0j ^) m / s

the acceleration is

a = (0 i ^ - g j⁾

a = (0i ^ - 9.8 j ^) m / s²

5 0

1 year ago