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2 years ago

Which is a result of using a machine?

Physics

1 answer:

kakasveta [241]2 years ago

4 0

Answer:

Yes labor is required therefor jobs like cashier, and other low skill labor jobs I'll not be required

Explanation:

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6.6 kg block initially at rest is pulled to theright along a horizontal, frictionless surfaceby a constant, horizontal force of

neonofarm [45]

Answer:

v = 3.04 m/s

Explanation:

given,

mass of the block, M = 6.6 Kg

horizontal force, F = 12.2 N

distance, L = 2.5 m

initial speed = 0 m/s

speed of the block,v = ?

we now

Work done is equal to change in Kinetic energy.

Work done = Force x displacement

W = Δ K E

Δ K E = Force x displacement

$\dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2= F .s$

$\dfrac{1}{2}\times 6.6 \times v^2 - 0= 12.2\times 2.5$

3.3 v² = 30.5

v² = 9.242

v = 3.04 m/s

speed of the block is equal to 3.04 m/s

4 0

2 years ago

A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, $V_{Hall} = 20\ mV = 0.02\ V$

Magnetic Field, B = 0.10 T

Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

Drift velocity, $v_{d} = \frac{V_{Hall}}{B}$

$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

$E_{Hall} = Bv_{d}sin\theta$ (1)

And

$E_{Hall} = V_{Hall}d$

Thus eqn (1) becomes

$V_{Hall}d = dBv_{d}sin\theta$

where

d = distance

$B = \frac{V_{Hall}}{v_{d}sin\theta}$ (2)

(a) When $\theta = 90^{\circ}$

$B = \frac{0.069}{0.2\times sin90} = 0.345\ T$

(b) When $\theta = 60^{\circ}$

$B = \frac{0.069}{0.2\times sin60} = 0.398\ T$

5 0

2 years ago

2. Is this chemical equation balanced?<br> 2C4H10 (g) + 13O2 (g) = 8CO2 (g) → 10H2O (l)

ehidna [41]

Yes that is a balaned equation

3 0

2 years ago

Please help i’ll mark you

suter [353]

Answer:

What should I do. reply quickly for a quick answer

8 0

1 year ago

Read 2 more answers

A: hydrogen and hydrogen b: copper and copper c: copper and oxygen Rank the above bonds in terms of increasing bond strength. Ch

pashok25 [27]

Answer: Copper and oxygen

Explanation:

Copper and oxygen shares the ionic bond. As we know that ionic bond is the most strongest bond. Here is the order:

Hydrogen bond< Metallic bond< Ionic bond.

That means order in terms of increasing bond strength is :

Hydrogen and hydrogen< Copper and copper< Copper and oxygen.

5 0

2 years ago