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3 years ago

An iron nail having threads along its cylindrical surface is

Physics

1 answer:

bulgar [2K]3 years ago

7 0

Answer:

A screw is a mechanism that converts rotational motion to linear motion, and a torque (rotational force) to a linear force.[1] It is one of the six classical simple machines. The most common form consists of a cylindrical shaft with helical grooves or ridges called threads around the outside.[2][3] The screw passes through a hole in another object or medium, with threads on the inside of the hole that mesh with the screw's threads. When the shaft of the screw is rotated relative to the stationary threads, the screw moves along its axis relative to the medium surrounding it; for example rotating a wood screw forces it into wood. In screw mechanisms, either the screw shaft can rotate through a threaded hole in a stationary object, or a threaded collar such as a nut can rotate around a stationary screw shaft.[4][5] Geometrically, a screw can be viewed as a narrow inclined plane wrapped around a cylinder.

Explanation:

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A professional racecar driver buys a car that can accelerate at 5.9 m/s2. The racer decides to race against another driver in a

3241004551 [841]

Answer:

(a) Time will be t = 3.56 sec

(b) Distance traveled by car when they are side by side is 37.38712 m

(b) Velocity of race car = 21.004 m/sec

velocity of stock car = 12.816 m/sec

Explanation:

We have given acceleration of the car $a_1=5.9m/sec^2$

Acceleration of the stock car $a_2=3.6m/sec^2$

When 1st car overtakes the second car then distance traveled by both the car will be same

(a) So $s_1=s_2$

As both car starts from rest so initial velocity of both car will be 0 m/sec

It is given that stock car leaves 1 sec before

So $\frac{1}{2}\times 5.9\times t^2=\frac{1}{2}\times (t+1)^2\times 3.6$

After solving t = 3.56 sec

(b) From second equation of motion $s=ut+\frac{1}{2}at^2=0\times 3.56+\frac{1}{2}\times 5.9\times 3.56^2=37.38712m$

So velocity of race car v = 0+5.9×3.56 = 21.004 m/sec

Velocity of stock car v = 0+ 3.6×3.56 = 12.816 m/sec

3 0

3 years ago

True or False—An external force is defined as a force generated outside the system of interest that acts on an object inside the

inn [45]

Answer:

An external force is a force that acts on an object within the system from outside the system. This type of force is different than an internal force, which acts between two objects that are both within the system. The net external force combines these two definitions; it is the total combined external force

Explanation:

ig the answer is true

5 0

3 years ago

A motorcycle is stopped at a stop light. When the light turns green it

ss7ja [257]

Answer: 18.9 m

Explanation:

i did the kinematic equation & found the answer.

8 0

3 years ago

Calculate the amount of heat (in kJ) required to convert 97.6 g of water to steam at 100° C. (The molar heat of vaporization of

Ratling [72]

Answer:

221.17 kJ

Explanation: Note the heat of vaporization is in kJ/mol,then to determine the number of moles of water: divide the mass by 18. Then multiply the number of moles by the molar heat of vaporization of water.

N = 97.6 ÷ 18

Q=molar heat *moles

Q = (40.79) * (97.6 ÷ 18)

This is approximately 221.17 kJ

4 0

3 years ago

An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric

8090 [49]

Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

Explanation:

Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

Electron q = 1.6×10^-19 C

Electron mass = 9.11×10^-31 Kg

(a) What is the speed of the electron 1.3 ns after entering this region?

E = F/q

F = Eq

Ma = Eq

M × V/t = Eq

Substitute all the parameters into the formula

9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19

V = 7.68×10^-18 /7.0×10^-22

V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres

3 0

3 years ago

An iron nail having threads along its cylindrical surface is​

An iron nail having threads along its cylindrical surface is