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Degger [83]
4 years ago
15

The ideal mechanical advantage of a machine reflects the increase or decrease in force there world be without friction, it is al

ways greater than the actual mechanical advantage because all machines must overcome friction.
options:
True
False
Physics
1 answer:
brilliants [131]4 years ago
8 0

True: the ideal mechanical advantage of a machine is always greater than the actual mechanical advantage because all machines must overcome friction.

Explanation:

For a simple machine, it is possible to calculate two types of mechanical advantage:

1) The Ideal Mechanical Advantage (IMA) is given by

IMA=\frac{d_e}{d_r}

where

d_r is the resistance arm

d_e is the effort arm

The IMA gives the mechanical advantage of the machine if there are no friction forces acting on it, and if all the work in input is converted into work in output with no loss of energy

2) The Actual Mechanical Advantage (AMA) is given by

AMA=\frac{L}{E}

where

L is the load (the force in output)

E is the effort (the force in input)

The AMA gives the real mechanical advantage of the machine. For an ideal machine,

AMA=IMA

Because there is no loss of energy due to friction.

For a real machine instead,

AMA

because part of the input energy is converted into thermal energy and other forms of energy due to the presence of friction, so it is "wasted" energy.

Learn more about levers and machines:

brainly.com/question/5352966

#LearnwithBrainly

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<em>A clamp-type measuring instrument operates on the principle of; </em>

A. induction

8 0
3 years ago
A vertical spring with a spring constant of 420 N/m is mounted on the floor. From directly above the spring, which is unstrained
mixas84 [53]

Answer:

19.53 cm

Explanation:

The computation of the height is as follows:

Here we applied the conservation of the energy formula

As we know that

P.E of the block = P.E of the spring

 m g h = ( 1 ÷ 2) k x^2

where

m = 0.15

g = 9.81

k = 420

x = 0.037

So now put the values to the above formula

(0.15) (9.81) (h) = 1 ÷2 × 420 × (0.037)^2

1.4715 (h) = 0.28749

h = 0.19537 m

= 19.53 cm

3 0
3 years ago
A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve
Gemiola [76]

Answer:

(a)   v = 5.42m/s

(b)   vo = 4.64m/s

(c)   a = 2874.28m/s^2

(d)   Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

v=\sqrt{2gh}        (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

h_{max}=\frac{v_o^2}{2g}       (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m

THe compression of the ball when it strikes the floor is 5.11*10^-3m

4 0
4 years ago
Label the following descriptions with (M) for Mass or (W) for Weight. ____measured with a spring scale ____measured in Newtons _
Valentin [98]
1. mass
2. weight
3. weight
4. weight
5. mass
6. mass
8 0
3 years ago
Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27
densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

\rho=n\times\dfrac{m}{V_{c}\times N_{A}}

n=\dfrac{\rho\timesV_{c}\times N}{m}....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100

\rho=5.98

We calculate the mass of the alloy

m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100

m=111.15

Put the value into the equation (I)

n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}

n=1.99\approx 2\ atoms/cell

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

5 0
4 years ago
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