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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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Compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for spee

JulsSmile [24]

Answer:

The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

Explanation:

Given that,

Mass of proton $m=1.67\times10^{-27}\ kg$

Speed $v= 9.00\times10^{7}\ m/s$

We need to calculate the kinetic energy for non relativistic

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times1.67\times10^{-27}\times(9.00\times10^{7})^2$

$K.E=6.76\times10^{-12}\ J$

We need to calculate the kinetic energy for relativistic

Using formula of kinetic energy

$K.E=mc^2(\sqrt{(\dfrac{1}{1-\dfrac{v^2}{c^2}})}-1)$

$K.E=1.67\times10^{-27}\times(3\times10^{8})^{2}\cdot\left(\sqrt{\frac{1}{1-\frac{\left(9.00\times10^{7}\right)^{2}}{(3\times10^{8})^{2}}}}-1\right)$

$K.E=7.25\times10^{-12}\ m/s$

Hence, The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

7 0

3 years ago

You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s

densk [106]

Here when car in front of us applied brakes then it is slowing down due to frictional force on it

So here we can say that friction force on the car front of our car is given as

$F_f = \mu m g$

So the acceleration of car due to friction is given as

$F_{net} = - \mu mg$

$a = \frac{F_{net}}{m}$

$a = -\mu g$

now it is given that

$\mu = 0.868$

$g = 9.81 m/s^2$

so here we have

$a = -0.868 * 9.81$

$a = -8.52 m/s^2$

so the car will accelerate due to brakes by a = - 8.52 m/s^2

4 0

3 years ago

Except when necessary for takeoff and landing, what is the minimum safe altitude required for a pilot to operate an aircraft ove

MAVERICK [17]

Except when necessary for takeoff and landing, <span>the minimum safe altitude required for a pilot to operate an aircraft over other than congested area is an altitude of 1000 ft above the highest obstacle within a 2000 ft horizontal radius of the aircraft.

It is also good to know that apart from taking off and landing, the aircraft must not operate at a distance less than 500 ft from any person, vessel, structure or vehicle.</span>

5 0

3 years ago

Your brother bikes 350 miles in 27 seconds how fast did he go

Harrizon [31]

His speed is exactly (350/27) miles per second ... about 46,667 mph. Wotta guy !

6 0

3 years ago

Which of the following BEST describes all planets in the universe?

Harman [31]

Answer:

c large, spherical body that orbits in a clear path around a star

Explanation:

you can not say b because the sun is a star and you cant say a and d because all planets are not made of rock and all planets are not made of gas

7 0

2 years ago