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Nadusha1986 [10]
3 years ago
11

Find the mass of a 52.2N bucket.​

Physics
1 answer:
Goshia [24]3 years ago
3 0

Answer:

m = 5.22 kg

Explanation:

The force acting on the bucket is 52.2 N.

We need to find the mass of the bucket.

The force acting on the bucket is given by :

F = mg

g is acceleration due to gravity

m is mass

m=\dfrac{F}{g}\\\\m=\dfrac{52.2}{10}\\\\=5.22\ kg

So, the mass of the bucket is 5.22 kg.

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The primary purpose of a mirror is to _________ light rays.
Effectus [21]
a. reflect (I found this using prior knowledge and process of illumination it can't be absorb cuz you wouldn't see anything it can't be refract because it doesn't reverse the image and it isn't transmit )
8 0
3 years ago
The Assignment: A fixed quantity of an ideal gas (R 0.28 kJ/kgK; Cv-0.71kJ/kgK) is expanded from an initial condition of 35 bar,
Nikolay [14]

Answer:

Index of expansion: 4.93

Δu = -340.8 kJ/kg

q = 232.2 kJ/kg

Explanation:

The index of expansion is the relationship of pressures:

pi/pf

The ideal gas equation:

p1*v1/T1 = p2*v2/T2

p2 = p1*v1*T2/(T2*v2)

500 C = 773 K

20 C = 293 K

p2 = 35*0.1*773/(293*1.3) = 7.1 bar

The index of expansion then is 35/7.1 = 4.93

The variation of specific internal energy is:

Δu = Cv * Δt

Δu = 0.71 * (20 - 500) = -340.8 kJ/kg

The first law of thermodynamics

q = l + Δu

The work will be the expansion work

l = p2*v2 - p1*v1

35 bar = 3500000 Pa

7.1 bar = 710000 Pa

q = p2*v2 - p1*v1 + Δu

q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg

7 0
3 years ago
Vector ????⃗ has a magnitude of 16.6 and is at an angle of 50.5∘ counterclockwise from the +x‑axis. Vector ????⃗ has a magnitude
natka813 [3]

Answer:

For vector u, x component = 10.558 and  y component =12.808

unit vector = 0.636 i+ 0.7716 j

For vector v, x component = 23.6316 and y component = -6.464

unit vector = 0.9645 i-0.2638 j

Explanation:

Let the vector u has magnitude 16.6

u makes an angle of 50.5° from x axis

So u_x=ucos\Theta =16.6\times cos50.5=10.558

Vertical component u_y=usin\Theta =16.6\times sin50.5=12.808

So vector u will be u = 10.558 i+12.808 j

Unit vector u=\frac{10.558i+12.808j}{\sqrt{10.558^2+12.808^2}}=0.636i+0.7716j

Now in second case let vector v has a magnitude of 24.5

Making an angle with -15.3° from x axis

So horizontal component v_x=vcos\Theta =24.5\times cos(-15.3)=23.6316

Vertical component v_y=vsin\Theta =24.5\times sin(-15.3)=-6.464

So vector v will be 23.6316 i - 6.464 j

Unit vector of v =\frac{23.6316i-6.464}{\sqrt{23.6316^2+6.464^2}}=0.9645i-0.2638j

8 0
3 years ago
The black lines that wrap themselves around a basketball represent an example of what type of geometry?
Dmitry [639]
In Euclidean geometry parallel lines never intersect. But in non-Euclidean geometry parallel lines can either curve away from each other, or curve towards each other. Example : the black lines that wrap themselves around the basketball.
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5 0
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NemiM [27]

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for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as-E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero  between the two charges,but we can find the points situated on the axial  line but  outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining  two charges  where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

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