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Vilka [71]
3 years ago
7

The material removal processes are a family of shaping operations in which excess material is removed from a starting work part

so that what remains is the desired final geometry:
A. True
B. False
Physics
1 answer:
HACTEHA [7]3 years ago
6 0

Answer:

True.

Explanation:

There are many material removal processes that are used in the manufacturing of gears, shafts etc. These processes removes the extra excess material to make the product of desired symmetry or shape.

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What is the transfer of energy to a system by the application of a force called?
pav-90 [236]
It is called Work done
6 0
3 years ago
The key to making a concise mathematical definition of escape velocity is to consider the energy. If an object is launched at it
Keith_Richards [23]

Answer:

Zero

Explanation:

Total energy of the EARTH  is the sum total of the all the kinetic and the gravitation potential energy of the earth.

Which can be written as

E_{total} = KE+U

= \frac{1}{2}mv^2+ \frac{GMm}{R}

At very large distances the kinetic as well as the gravitational potential energies become zero.

therefore, E_total= 0

5 0
3 years ago
A tennis player tosses a tennis ball straight up and then catches it after 1.77 s at the same height as the point of release. (a
Bogdan [553]

(a) 9.8 m/s^2, downward

There is only one force acting on the ball while it is in flight: the force of gravity, which is

F = mg

where

m is the mass of the ball

g is the gravitational acceleration

According to Newton's second law, the force acting on the ball is equal to the product between the mass of the ball and its acceleration, so

F = mg = ma

which means

a = g

So, the acceleration of the ball during the whole flight is equal to the acceleration of gravity:

g = -9.8 m/s^2

where the negative sign means the direction is downward.

(b) v = 0

Any object thrown upward reaches its maximum height when its velocity is zero:

v = 0

In fact, at that moment, the object's velocity is turning from upward to downward: that means that at that instant, the velocity must be zero.

(c) 8.72 m/s, upward

The initial velocity of the ball can be found by using the equation:

v = u + at

Where

v = 0 is the velocity at the maximum height

u is the initial velocity

a = g = -9.8 m/s^2 is the acceleration

t is the time at which the ball reaches the maximum height: this is half of the time it takes for the ball to reach again the starting point of the motion, so

t=\frac{1.77 s}{2}=0.89 s

So we can now solve the equation for u, and we find:

u=v-at=0-(-9.8 m/s^2)(0.89 s)=8.72 m/s

(d) 3.88 m

The maximum height reached by the ball can be found by using the equation:

v^2 - u^2 = 2ad

where

v = 0 is the velocity at the maximum height

u = 8.72 m/s is the initial velocity

a = g = -9.8 m/s^2 is the gravitational acceleration

d is the maximum height reached

Solving the equation for d, we find

d=\frac{v^2-u^2}{2a}=\frac{0^2-(8.72 m/s)^2}{2(-9.8 m/s^2)}=3.88 m

7 0
3 years ago
How are carbon tetrachloride and sodium chloride different from each other?
Tasya [4]

Answer:

Carbon tetrachloride is insoluble in water and behaves as a bad conductor of electricity while sodium chloride is soluble in water and behave as a good conductor of electricity in their aqueous state.

4 0
4 years ago
Read 2 more answers
Perform an experiment in which you flip a switch 2,300,000 times. Write this
ryzh [129]

Answer:

B. 2.3 x 10⁶

Explanation:

To calculate scientific notation for the number 2,300,000, we have to follow this notation

  • a × 10^b

<u>Step 1 :</u>

To find a we have to write the non-zero digits placing a decimal after the first non-zero digit.

  • 2,300,000 to 2.300000

<u>Step 2 :</u>

Now, to find b count how many numbers of digits are there to the right of the decimal.

  • 2.<u>300000</u>

Hence, there are 6 digits to the right of the decimal.

<u>Step 3 :</u>

Since, we had found the value of a and b, we can now reconstruct the number into scientific notation.

a = 2.3

b = 6

  • 2.3 × 10⁶

<u>Therefore</u><u>,</u><u> </u><u>option </u><u>b </u><u>is </u><u>correct</u><u>.</u>

4 0
2 years ago
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