All categories

3 years ago

A student is using a 12.9 ft ramp to raise an object 6 ft above the ground.

Engineering

1 answer:

Hatshy [7]3 years ago

8 0

Answer:

2.150

Explanation:

The ideal mechanical advantage of the ramp is given as a ratio between the height of the ramp and the distance from the ground ie

Ideal mechanical advantage of the ramp= $\frac {12.9}{6}=2.150$

Therefore, the answer is 2.150 to the nearest thousandth

You might be interested in

If you add 10 J of heat to a system so that the final temperature of the system is 200K, what is the change in entropy of the sy

Elden [556K]

Answer:

0.05 J/K

Explanation:

Given data in question

heat (Q) = 10 J

temperature (T) = 200 K

to find out

the change in entropy of the system

Solution

we will solve this by the entropy change equation

i.e ΔS = ΔQ/T ...................1

put the value of heat Q and Temperature T in equation 1

ΔS is the enthalpy change and T is the temperature

so ΔS = 10/200

ΔS = 0.05 J/K

8 0

3 years ago

What are flip flops and what do they look like

Vinil7 [7]

flip flops are shoes and they look like shoes

4 0

2 years ago

Read 2 more answers

(20pts) Air T[infinity] = 10 °C and u[infinity] = 100 m/s flows over a flat plate. Assume that the density of air is 1.0 kg/m3 a

miskamm [114]

Answer:

attached below

Explanation:

8 0

3 years ago

A PMMA plate with a 25 mm (width) x 6.5 mm (thickness) cross-section has a contained crack of length 2c = 0.5 mm in the center o

victus00 [196]

Answer:

LAOD = 6669.86 N

Explanation:

Given data:

width $= 25 mm = 25\times 10^{-3} m$

thickness $= 6.5 mm = 6.5\times 10^{-3} m$

crack length 2c = 0.5 mm at centre of specimen

$\sigma _{applied} = 1000 N/cross sectional area$

stress intensity factor = k will be

$\sigma_{applied} = \frac{1000}{25\times 10^{-3}\times 6.5\times 10^{-3}}$

$= 6.154\times 10^{6} Pa$

we know that

$k =\sigma_{applied} (\sqrt{\pi C})$

$=6.154\sqrt{\pi (2.5\times 10^{-04})}$ [c =0.5/2 = 2.5*10^{-4}]

K = 0.1724 Mpa m^{1/2} for 1000 load

if $K_C = 1.15 Mpa m^{1/2}$ then load will be

$Kc = \sigma _{frac}(\sqrt{\pi C})$

$1.15 MPa = \sigma _{frac}\times \sqrt{\pi (2.5\times 10^{-04})}$

$\sigma _{frac} = 41.04 MPa$

$load = \sigma _{frac}\times Area$

$load = 41.04 \times 10^6 \times 25\times 10^{-3}\times 6.5\times 10^{-3} N$

LAOD = 6669.86 N

3 0

2 years ago

A ball A is thrown vertically upward from the top of a 30-m-high building with an initial velocity of 5 m>s. At the same inst

expeople1 [14]

Answer:

s= 20.4 m

Explanation:

First lets write down equations for each ball:

s=so+vo*t+1/2a_c*t^2

for ball A:

s_a=30+5*t+1/2*9.81*t^2

for ball B:

s_b=20*t-1/2*9.81*t^2

to find time deeded to pass we just put that

s_a = s_b

30+5*t-4.91*t^2=20*t-4.9*t^2

t=2 s

now we just have to put that time in any of those equations an get distance from the ground:

s = 30 + 5*2 -1/2*9.81 *2^2

s= 20.4 m

6 0

3 years ago