All categories

2 years ago

What us meant by Velocity ratio of simple machine is 4?

1 answer:

8 0

The force ratio of a machine is 4 and it velocity ratio is 4 means that the load moved is four times the effort applied and the distance moved by the effort is five times the distance moved by the load at the same time interval.

Please mark as brainliest

You might be interested in

The flywheel of a steam engine runs with a constant angular velocity of 150 rev/min. When steam is shut off, the friction of the

xz_007 [3.2K]

Answer:

a) -1.14 rev/min²

b) 9900 rev

c) -9.92×10⁻⁴ m/s²

d) 30.8 m/s²

Explanation:

First, convert hours to minutes:

2.2 h × 60 min/h = 132 min

a) Angular acceleration is change in angular velocity over change in time.

α = (ω − ω₀) / t

α = (0 rev/min − 150 rev/min) / 132 min

α = -1.14 rev/min²

b) θ = θ₀ + ω₀ t + ½ αt²

θ = 0 rev + (150 rev/min) (132 min) + ½ (-1.14 rev/min²) (132 min)²

θ = 9900 rev

c) The tangential component of linear acceleration is:

a_t = αr

First, convert α from rev/min² to rad/s²:

-1.14 rev/min² × (2π rad/rev) × (1 min / 60 s)² = -1.98×10⁻³ rad/s²

Therefore:

a_t = (-1.98×10⁻³ rad/s²) (0.50 m)

a_t = -9.92×10⁻⁴ m/s²

d) The magnitude of the net linear acceleration can be found from the tangential component and the radial component:

a² = (a_t)² + (a_r)²

The radial component is the centripetal acceleration:

a_r = v² / r

a_r = ω² r

First, convert 75 rev/min to rad/s:

75 rev/min × (2π rad/rev) × (1 min / 60 s) = 7.85 rad/s

Find the radial component:

a_r = (7.85 rad/s)² (0.50 m)

a_r = 30.8 m/s²

Now find the net linear acceleration:

a² = (-9.92×10⁻⁴ m/s²² + (30.8 m/s²)²

a = 30.8 m/s²

5 0

2 years ago

A puck moves 2.35 m/s in a -22 degree direction. A hockey stick pushes it for 0.215 s, changing its velocity to 6.42 m/s in a 50

Whitepunk [10]

Answer:

Displacement in Y direction is 0.434 m

Explanation:

initial velocity of the puck is 2.35 m/s at -22 degree

so here it is given as

$v_i = 2.35 cos22 \hat i - 2.35 sin22 \hat j$

$v_i = 2.18 \hat i - 0.88 \hat j$

final velocity is given as 6.42 m/s at 50 degree

so we have

$v_f = 6.42 cos50 \hat i + 6.42 sin50\hat j$

$v_f = 4.13 \hat i + 4.92\hat j$

now displacement in Y direction is given as

$\Delta y = \frac{v_f + v_i}{2}t$

$\Delta y = \frac{-0.88 + 4.92}{2} (0.215)$

$\Delta y = 0.434 m$

5 0

3 years ago

A young child cannot understand<br>another person's perspective because he is

natima [27]

Answer:

egocentric

Explanation:

6 0

3 years ago

A 53.3 kg woman slides down a 35.0° hill with an acceleration of 4.10 m/s. What is the friction force acting on the woman?

lorasvet [3.4K]

Answer:

I attached an image that should help.

Explanation:

Check it out.

5 0

2 years ago

A rectangular barge, 5.2 m long and 2.4 m wide, floats in fresh water. Suppose that a 410-kg crate of auto parts is loaded onto

sesenic [268]

<h2>Answer:</h2><h2>The depth of barge float=3 cm</h2><h2>Explanation:</h2>

Length of rectangular barge=5.2 m

Width of rectangular barge=2.4m

Mass of crate=410 kg

Let h be the height of barge float

Volume of barge float= $l\times b\times h=5.2\times 2.4\times h=12.48h$

Density of water= $10^3kg/m^3$

Weight of water displaced by barge=Buoyant force=-Weight of horse

$Volume\;of\;water\times density\;of\;water\times g=410\times g$

$12.48h\times 1000=410$

$h=\frac{410}{12.48\times 1000}=0.03 m$

1 m=100 cm

$0.03 m=0.03\times 100=3$ cm

Hence, the depth of barge float=3 cm

<h2 />

4 0

3 years ago

Read 2 more answers

What us meant by Velocity ratio of simple machine is 4?​

What us meant by Velocity ratio of simple machine is 4?