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3 years ago

What type of clutch is used in a hammer drill to enable or disable the spindle drive's final reduction gearing?

1 answer:

3 0

The clutch that is used in a hammer drill to enable or disable the spindle drives final reduction gearing will be, D. Tapered jaw

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How does atomic bombs work?

yarga [219]

Answer:

Nuclear fission produces the atomic bomb, a weapon of mass destruction that uses power released by the splitting of atomic nuclei. When a single free neutron strikes the nucleus of an atom of radioactive material like uranium or plutonium, it knocks two or three more neutrons free.

Explanation:

this chain reaction releases tremendous amount of energy

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3 years ago

Read 2 more answers

A racecar on a straight track, starting from rest*, steps on the

kirill115 [55]

I think the answer is C

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3 years ago

A student drops a rock from a bridge to the water 12.7 m below. with what speed does the rock strike the water? the acceleration

kolezko [41]

The rock strike the water with the speed of 15.78 m/sec.

The speed by which rock hit the water is calculated by the formula

v= $\sqrt{2gh}$

v= $\sqrt{2*9.8*12.7}$

v=15.78 m/sec

Hence, the rock strike the water with the speed of 15.78 m/sec.

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3 years ago

A patient comes in and tells you that he has arthritis and needs Cerebyx. What do you do? A. You know that Cerebyx and Celebrex

Flura [38]

B. truthfully it's the only one that makes sense

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3 years ago

A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o

Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

$f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz$

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

$\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m$

The fundamental frequency of the wire is given by

$f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N$

The tension in the wire is 67.9064 N

7 0

3 years ago