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3 years ago

Which of the following tools is used to mark the spot where a hole is to be drilled in a piece of metal?

Physics

2 answers:

Oxana [17]3 years ago

3 0

That's a job for a center-punch.

Snowcat [4.5K]3 years ago

3 0

Answer: Option (b) is the correct answer.

Explanation:

Center punch is a tool which helps to mark a point where we intend to punch a hole in the material without slipping in any direction.

With the help of this mark, it provides a resting place to the drill so that while drilling the hole will be pointed in the right direction.

Thus, we can conclude that out of the given options center punch is the tool used to mark the spot where a hole is to be drilled in a piece of metal.

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Physics - Electricity and Magnetism

Orlov [11]

Answer:

Explanation:

because the unit for resistance is in ohms

4 0

2 years ago

A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th

melomori [17]

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore, the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F = 624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

5 0

3 years ago

Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce

lutik1710 [3]

Answer:

speed of electrons = 3.25 × $10^{7}$ m/s

acceleration in term g is 3.9 × $10^{17}$ g.

radius of circular orbit is 2.76 × $10^{-4}$ m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = $\sqrt{\frac{2qV}{m}}$

v = $\sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}$

v = 3.25 × $10^{7}$ m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = $\frac{Bqv}{m}$

a = $\frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}$

a = 3.82 × $10^{18}$ m/s²

and acceleration in term g

a = $\frac{3.82\times 10^{18}}{9.81}$

a = 3.9 × $10^{17}$ g

acceleration in term g is 3.9 × $10^{17}$ g.

and

electron moving in circular orbit has centripetal force

F = $\frac{mv^2}{r}$

Bqv = $\frac{mv^2}{r}$

r = $\frac{mv}{Bq}$

r = $\frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}$

r = 2.76 × $10^{-4}$ m

radius of circular orbit is 2.76 × $10^{-4}$ m

8 0

3 years ago

The latent heat of fusion for Aluminium is 3.97 x 105. How much energy would be required to melt 0.75 kg of it?

RoseWind [281]

Answer:

$E = 2.9775\times10^5 J$

Explanation:

Given: The latent heat of fusion for Aluminum is $L = 3.97\times10^5 J/Kg$

mass to be malted m = 0.75 Kg

Energy require to melt E = mL

$E = 3.97\times10^5\times0.75 = 2.9775\times10^5 J$

Therefore, energy required to melt 0.75 Kg aluminum

$E = 2.9775\times10^5 J$

5 0

3 years ago

For a ∆x of 0.1mm, what is ∆px, the uncertainty in the transverse momentum of a photon passing through a slit (where uncertainty

Colt1911 [192]

Answer:

$0.53\times 10^{-30}kgms^{-1}$

Explanation:

Uncertainty principle say that the position and momentum can not be measured simultaneously except one relation which is described below,

$\Delta x\Delta p=\frac{h}{4\pi }$

Given that the uncertainty in x is 0.1 mm.

Therefore,

$\Delta p=\frac{6.626\times 10^{-34} }{4\times 3.14\times 1\times 10^{-4}m }\\\Delta p=0.53\times 10^{-30}kgms^{-1}$

Therefore, uncertainty in the transverse momentum of photon is $0.53\times 10^{-30}kgms^{-1}$

6 0

3 years ago