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3 years ago

Grinding with the portable disc grinder should not be done in an area which

Engineering

1 answer:

emmainna [20.7K]3 years ago

7 0

Nothing flammable of explosive type of material is around

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Crank OA rotates with uniform angular velocity 0  4 rad/s along counterclockwise. Take OA= r= 0.5

mafiozo [28]

CRABK DAT SOUIJA BOI LIKE OOOOO WATCH ME WATCH ME OOOO LASAGNA Can you lend me 700$ because I used my toaster as a bath heater and now my legs are gone plz I need money for bandaids

3 0

3 years ago

A piston–cylinder assembly contains 5 kg of air, initially at 2.0 bar, 30 C. The air undergoes a process to a state where the pr

Ainat [17]

Answer: wor done is 145. 06kJ

Heat transfer is 135.53kJ

Explanation:

No of moles of air = mass/molar mass = 5000g/28gmol^-1 = 172.65mol

P1 = 2bar =2*101300 =202600pa

T1 = 30° +273k = 303k

P2 =p1 = 202600pa

V2 =? T2 =?

Using pV = nRT

R = 8.314 PA m^3 mol^-1 k^-1

V1 = (172.65*8.314*303)/202600

V1 = 2.146m^3

For second state, 1.5pv = const = P1V1

V2 = (202600*2.146)/(1.5*202600)

V2 = 1.43m^3

Volume change = 2.146 - 1.43 =0.715m^3

Word done = pressure* volume change

W = 202600*0.716 = 145061.6J

= 145.061kJ

Using V1/T1 = V2/T2

T2 = V2T1/V1

=(1.43*303)/2.146 = 201.9k

For internal energy U

U = nCv*(T2 - T1)

*CV is the heat capacity at const. vol approximately 0.718J mol^-1 k^-1

U = 172.65*0.718*(201.9-303)

U = -12532.6J = -12.532kJ

The -ve means the system lost internal energy.

Q = U+W = total heat energy of system

Q = - 12.532+145.061 = 132.52 kJ

7 0

4 years ago

What is the tolerance of number 4?

Kamila [148]

Answer:

Answer: ±0.02 units or 20±0.02 units or 19.98-20.02 units depending on how they prefer its written (typically the first or second one)

Explanation:

says on the sheet. Unless otherwise stated 0.XX = ±0.02 tolerance

(based on image sent in other post)

5 0

3 years ago

Coulomb's Law Two point charges experience an attractive force of 10.8 N when they are separated by 2.4 m. VWhat force in newton

SashulF [63]

Answer:

The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons

Explanation:

We know that for two point charges of magnitude $q_{1},q_{2}$ the magnitude of force between them is given by

$F=\frac{k_{e}q_{1}\cdot q_{2}}{r^{2}}$

where

$k_{e}$ is constant

$r$ is the separation between the charges

Initially when the charges are separated by 2.4 meters the force can be calculated as

$F_{1}=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\10.8=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\\therefore k_{e}\cdot q_{1}q_{2}=10.8\times 2.4^{2}=62.208$

Now when the separation is reduced to 0.7 meters the force is similarly calculated as

$F_{2}=\frac{k_{e}\cdot q_{1}q_{2}}{0.7^{2}}$

Applying value of the constant we get

$F_{1}=\frac{62.208}{0.7^{2}}$

Thus $F_{2}=126.955Newtons$

5 0

3 years ago

What kind of microscope would be used to study a whole or opaque object?

djyliett [7]

Answer:

the compound light microscope

Explanation:

The stereomicroscope is to study section to study the entire objects in three dimensions at low magnification. A Compound light microscope is used for small or thinly sliced objects under higher magnification than stereomicroscope.

4 0

4 years ago

Grinding with the portable disc grinder should not be done in an area which​

Grinding with the portable disc grinder should not be done in an area which