All categories

3 years ago

What do simple machines accomplish

2 answers:

8 0

A simple machine is an unpowered mechanical device that accomplished a task with only one movement. Simple machines work with forces. The input force is the force you apply to the machine. The output force is the force the machine applies to what you are trying to move

xz_007 [3.2K]3 years ago

3 0

Answer:

not many things maybe they accomplish one job the machine has been programmed to do.

Explanation:

You might be interested in

n the Bohr model of the hydrogen atom (see Section 39.3), in the lowest energy state the electron orbits the proton at a speed o

Ivahew [28]

Answer:

(a) $T=1.5*10^{-6}s$

(b) $I=1.1*10^{-3}A$

Explanation:

(a) The orbital period is the time that the electron spend to travel the orbit of the atom. Thus, it is given by the length of the circular orbit divided by its velocity:

$T=\frac{2\pi r}{v}\\T=\frac{2\pi(5.3*10^{-11}m)}{2.2*10^{6}\frac{m}{s}}\\T=1.5*10^{-6}s$

(b) Current means charge over time, So, in this case is charge over period:

$I=\frac{q}{t}\\I=\frac{e}{T}\\I=\frac{1.6*10^{-19}C}{1.5*10^{-6}s}\\\\I=1.1*10^{-3}A$

$\mu=IA$

Here A is the area of the orbit.

$\mu=I\pi r^2\\\mu=(1.1*10^{-3}A)\pi(5.3*10^{-11}m)^2\\\mu=9.71*10^{-24}A\cdot m^2$

4 0

3 years ago

Question 10 of 10

inessss [21]

Answer:

Super idoo di shaw lung domini di shaw

6 0

2 years ago

ASTM A229 oil-tempered carbon steel is used for a helical coil spring. The spring is wound with D = 50 mm d = 10.0 mm, and a pit

horrorfan [7]

Answer

given,

D = 50 mm = 0.05 m

d = 10 mm = 0.01 m

Force to compress the spring

$F = \dfrac{d^4G\delta}{8D^3N}$

$\dfrac{\delta}{N} = p - d = 14 - 10 = 4 mm$

$F = \dfrac{d^4G}{8D^3}\times 0.004$

$F = \dfrac{0.1^4\times 79\times 10^9}{8\times 0.05^3}\times 0.004$

F = 3160 N

stress correction factor from stress correction curve is equal to 1.1

now, calculation of corrected stress

$\tau = \dfrac{8FDk_s}{\pi d^3}$

$\tau = \dfrac{8\times 3160 \times 0.05 \times 1.1}{\pi \times 0.01^3}$

= 442.6 Mpa

The tensile strength of the steel material of ASTM A229 is equal to 1300 Mpa

now,

$\tau_s \leq 0.45 S_u$

$\tau_s \leq 0.45 \times 1300$

$\tau_s \leq 585\ Mpa$

since corrected stress is less than the $\tau_s$

hence, spring will return to its original shape.

6 0

3 years ago

Read 2 more answers

Humberto builds two circuits using identical components.

vlada-n [284]

Humberto should expect to see that all bulbs in circuit 1 will shine more dimly than the original bulbs, while all bulbs in circuit 2 will have the same brightness as the original bulbs.

B) All bulbs in circuit 1 will shine more dimly than the original bulbs, while all bulbs in circuit 2 will have the same brightness as the original bulbs.

6 0

3 years ago

Read 2 more answers

The Kapandu boys dormitory in the UPNG is going up in flames. Rachel, a former student of Physics, who

Tom [10]

Answer:

1108.464 N of force

Explanation:

diameter of water hose = 70 cm = 0.7 m

radius = 0.7/2 = 0.35 m

volumetric flow rate Q = 420 L/min

1 L = 0.001 m^3

1 min = 60 s

therefore,

Q = 420 L/min = (420 x 0.001)/60 = 0.007 m^3/s

Area A of fire hose = π $r^{2}$ = 3.142 x $0.35^{2}$ = 0.38 m^2

From continuity equation, Q = AV

where V1 is the velocity of the water through the pipe, and A1 is the area of the pipe.

Q = A1V1

0.007 = 0.38V1

V1 = 0.007/0.38 = 0.018 m/s.

Nozzle diameter = 0.75 cm = 0.0075 m

radius = 0.00375

Area = π $r^{2}$ = 3.142 x $0.00375^{2}$ = 4.42 x $10^{-5}$ m^2

velocity of water through the nozzle will be

V2 = Q/A2 = 0.007 ÷ (4.42 x $10^{-5}$ ) = 158.37 m/s

From

F = ρQ(v2 - v1)

Where,

F = force exerted

p = density of water = 1000 kg/m^3

F = 1000 x 0.007 x (158.37 - 0.018) = 1108.464 N of force

4 0

3 years ago