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3 years ago

Two very quick questions!!

2 answers:

8 0

<span>The weight lifted by a machine to the applied force on a machine is called mechanical advantage. This is written as Mechanical advantage, M. A, = load(weight)/effort. So for 1) M.A = 2 and load = 2, 000lb = 8896.446N. So 2 = 8896.446/ effort Effort = 8896.446/2 = 4448.48 Similarly for M.A of 2, 000, 000 we have Effort = 8896.446/ 2, 000, 000 = 0.004448</span>

lana [24]3 years ago

5 0

Explanation:

1. Mechanical advantage of a pulley system, m = 2

Mechanical advantage of a machine is defined as the ratio of load to the effort force i.e.

$m=\dfrac{F_L}{F_E}$

Here, $F_L=2000\ lb$

$F_E=\dfrac{F_L}{m}$

$F_E=\dfrac{2000\ lb}{2}$

$F_E=1000\ lb$

Hence, 1000 lb effort will be needed to move the car.

2. Mechanical advantage of a pulley system, m = 2,000,000

Value of load, $F_L=2000\ lb$

Mechanical advantage, $m=\dfrac{F_L}{F_E}$

$F_E=\dfrac{F_L}{m}$

$F_E=\dfrac{2000\ lb}{2000000}$

$F_E=0.001\ lb$

Hence, 0.001 lb effort would be needed to move the car.

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Jim began a 153-mile bicycle trip to build up stamina for a triathlon competition. Unfortunately, his bicycle chain broke, so he

NARA [144]

consider the motion when Jim travels by bicycle

D = total distance to be traveled for the trip = 153 mile

v = speed of riding = 42 mph

t = time taken to complete the trip by bicycle = ?

using the equation

t = D/ v

inserting the values

t = 153/42

t = 3.64 h

while walking and riding :

T = total time taken to complete the trip by walking and riding = 5 hours

t ' = time of walking = T - t = 5 - 3.64 = 1.36 hours

v' = speed of walking = 4 mph

d' = distance traveled by walking = ?

distance traveled by walking is given as

d' = v' t'

d' = 4 x 1.36

d' = 5.44 miles

d = distance traveled by bicycle = D - d' = 153 - 5.44 = 147.56 miles

v = speed of riding = 42 mph

t = time spent on the bicycle = ?

time spent on the bicycle is given as

t = d/v

t = 147.56/42

t = 3.51 hours

5 0

3 years ago

Is diet Pepsi heterogeneous or is it homogeneous

Lana71 [14]

Heterogeneous because it has less amount of sugar than the regular Pepsi has

6 0

3 years ago

Read 2 more answers

A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency. In general

valentina_108 [34]

Answer:

Explanation:

(A)

The string has set of normal modes and the string is oscillating in one of its modes.

The resonant frequencies of a physical object depend on its material, structure and boundary conditions.

The free motion described by the normal modes take place at the fixed frequencies and these frequencies is called resonant frequencies.

Given below are the incorrect options about the wave in the string.

• The wave is travelling in the +x direction

• The wave is travelling in the -x direction

• The wave will satisfy the given boundary conditions for any arbitrary wavelength $\lambda_i$

• The wave does not satisfy the boundary conditions $y_i(0;t)=0
$

Here, the string of length L held fixed at both ends, located at x=0 and x=L

The key constraint with normal modes is that there are two spatial boundary conditions, $y(0,1)=0
$

and $y(L,t)=0$

.The spring is fixed at its two ends.

The correct options about the wave in the string is

• The wavelength $\lambda_i$ can have only certain specific values if the boundary conditions are to be satisfied.

(B)

The key factors producing the normal mode is that there are two spatial boundary conditions, $y_i(0;t)=0$ and $y_i(L;t)=0$ , that are satisfied only for particular value of $\lambda_i$ .

Given below are the incorrect options about the wave in the string.

• $A_i$ must be chosen so that the wave fits exactly o the string.

• Any one of $A_i$ or $\lambda_i$ or $f_i$ can be chosen to make the solution a normal mode.

Hence, the correct option is that the system can resonate at only certain resonance frequencies $f_i$ and the wavelength $\lambda_i$ must be such that $y_i(0;t) = y_i(L;t)=0
$

(C)

Expression for the wavelength of the various normal modes for a string is,

$\lambda_n=\frac{2L}{n}$ (1)

When $n=1$ , this is the longest wavelength mode.

Substitute 1 for n in equation (1).

$\lambda_n=\frac{2L}{1}\\\\2L$

When $n=2$ , this is the second longest wavelength mode.

Substitute 2 for n in equation (1).

$\lambda_n=\frac{2L}{2}\\\\L$

When $n=3$ , this is the third longest wavelength mode.

Substitute 3 for n in equation (1).

$\lambda_n=\frac{2L}{3}$

Therefore, the three longest wavelengths are $2L$ , $L$ and $\frac{2L}{3}$ .

(D)

Expression for the frequency of the various normal modes for a string is,

$f_n=\frac{v}{\lambda_n}$

For the case of frequency of the $i^{th}$ normal mode the above equation becomes.

$f_i=\frac{v}{\lambda_i}$

Here, $f_i$ is the frequency of the $i^{th}$ normal mode, v is wave speed, and $\lambda_i$ is the wavelength of $i^{th}$ normal mode.

Therefore, the frequency of $i^{th}$ normal mode is $f_i=\frac{v}{\lambda_i}$

6 0

3 years ago

Many Amtrak trains can travel at a top speed of 42.0 m/s. Assuming a train maintains that speed for several hours, how many kilo

777dan777 [17]

Answer:

605 km

Explanation:

Hello

the same units of measure should be used, then

Step 1

convert 42 m/s ⇒ km/h

1 km =1000 m

1 h = 36000 sec

$42 \frac{m}{s}*\frac{1\ km}{1000\ m}=0.042\ \frac{km}{s}\\ 0.042\ \frac{km}{s}\\$

$0.042\ \frac{km}{s}*\frac{3600\ s}{1\ h} =151.2 \frac{km}{h}\\ \\Velocity =151.2\ \frac{km}{h}$

Step 2

find kilometers traveled after 4 hours

$V=\frac{s}{t}\\ \\$

V,velocity

s, distance traveled

t. time

now, isolating s

$V=\frac{s}{t} \\s=V * t\\$

and replacing

$s=V * t\\s=151.2\frac{km}{h}*4 hours\\ s=604.8 km\\$

S=604.8 Km

Have a great day

4 0

3 years ago

How deep is the event horizon

Basile [38]

Explanation:

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7 0

3 years ago