He user of a machine applies force to the machine over the ____

A river has a steady speed of 0.566 m/s. A student swims upstream a distance of 1.57 km and returns (still swimming) to the star

MArishka [77]

Answer:

3503.72 seconds

819.96 seconds

Explanation:

$V_r$ =Velocity of river = 0.566 m/s

$V_s$ =Velocity of student = 1.17 m/s

Distance to travel = 1.57 km = 1570 m

So,

Time = Distance / Speed

$\frac{1570}{V_s-V_r}+\frac{1570}{V_s+V_r}=t\\\Rightarrow t=\frac{1570}{1.17-0.566}+\frac{1570}{1.17+0.566}\\\Rightarrow t=3503.72\ s$

Time taken by the student to complete the trip is 3503.72 seconds

In still water

$\frac{1570}{V_s}+\frac{1570}{V_s}=t\\\Rightarrow t=\frac{1570}{1.17}+\frac{1570}{1.17}\\\Rightarrow t=2683.76\ s$

The difference in time between moving water and still water is 3503.72-2683.76 = 819.96 seconds

5 0

2 years ago

Which of the following is an example of an allocation decision?|

Oksana_A [137]

Answer:

The answer is D- Coal is set aside to burn as heating fuel

Explanation:

Allocation decisions set aside resources for a specific purpose.

7 0

3 years ago

Read 2 more answers

A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f

Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

$T=\dfrac{2\pi}{\omega}$

${\omega}=\dfrac{2\pi }{0.2}\ rad/s$

${\omega}=31.41\ rad/s$

We know that

${\omega}^2=m\ K$

K=Spring constant

$K=\dfrac{\omega^2}{m}$

$K=\dfrac{31.41^2}{0.12}\ N/m$

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

3 0

3 years ago

A fire helicopter carries a 700-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant spee

rodikova [14]

Answer:

F = 50636.873 N

Explanation:

given,

bucket of water = 700-kg

length of cable = 20 m

Speed = 40 m/s

angle of the cable = 38.0°

let air resistance be = F

tension in rope be = T

T cos 38° = m×g..................(1)

$T sin 38^0= \dfrac{mv^2}{l} + F$ ..........(2)

equation (1)/(2)

$tan 38^0 =\dfrac{\dfrac{mv^2}{l} + F}{mg}$

$0.781=\dfrac{\dfrac{700\times 40^2}{20} + F}{700\times 9.8}$

F = 50636.873 N

Hence the force exerted on the bucket is equal to F = 50636.873 N

5 0

2 years ago

A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten

Lyrx [107]

Answer:

a) the maximum shear stress τ $_{max}$ the bar is 16T $_{max}$ /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ $_{max}$ = 16T $_{max}$ /πd³

Therefore, the maximum shear stress τ $_{max}$ the bar is 16T $_{max}$ /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta ( $\theta$ ) be the angle of twist

polar moment of inertia $I_p}$ = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d $\theta$ = T(x)dx / G $I_{p}$

so

d $\theta$ = tx.dx / G(πd⁴/32)

d $\theta$ = 32tx.dx / πGd⁴

so total angle of twist $\theta$ will be;

$\theta$ = $\int\limits^L_0 \, d\theta$

$\theta$ = $\int\limits^L_0 \,$ 32tx.dx / πGd⁴

$\theta$ = 32t / πGd⁴ $\int\limits^L_0 \, xdx$

$\theta$ = 32t / πGd⁴ [ L²/2]

$\theta$ = 16tL² / πGd⁴

Therefore, the angle of twist between the ends of the bar is 16tL² / πGd⁴

7 0

2 years ago

He user of a machine applies force to the machine over the _____ distance.