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2 years ago

Can we write names while writing conversation in board exam

Physics

1 answer:

lora16 [44]2 years ago

5 0

Answer:

ya we can write the imaginary character's name .

So that we can identify these imaginary people, as we cannot simply write the conversation and leave it .

Or maybe sometimes the reader will get confused as there is no name for the two people .

So, i suggest that you should write the names

Explanation:

You can even ask to your class teacher for further clarification

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A 43-N crate is suspended from the left end of a plank. The plank weighs 21 N, but it is not uniform, so its center of gravity d

lesya [120]

Answer:

$d_{2}=0.585m$

Explanation:

The torque is the force by the distance so to determinate that both torque are the same magnitude so

$T_{1}-T_{2}=0$

$T_{1}=T_{2}$

$F_{1}*d_{1}=F_{2}*d_{2}$

$43N*0.30m=21N*d_{2}$

Solve to d2

$d_{2}=\frac{41N*0.30m}{21N}$

$d_{2}=0.585m$

5 0

2 years ago

Explain how tangential speed depends on distance.

Alinara [238K]

The equation v=rω says that the tangential speed v is proportional to the distance r from the center of rotation. ... This makes sense because a point farther out from the center has to cover a longer arc length in the same amount of time as a point closer to the center.

6 0

2 years ago

Consider a system two point charges. One has charge +q at (x, y,z) -(a,0,0) and another of charge-q at (x, y, z) = (-a, 0,0). 5.

olga2289 [7]

Answer:

electricfield at (0,0,0) is Et = 2 k q / a²

Explanation:

For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity

For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation

E = k q / r²

Point (0,0,0)

We calculate for the charge -q which is at a distance R = a

E1 = k (-q) / a²

E1 = - kq / a²

As the test charge is positive in the field it goes to the left, attractive force

We calculate for the charge that is also at R = a

E2 = k q / a²

This field goes to the left, repulsive force

We find the total electric field

Et = E1 + E2

Et = kq / a² + kq / a²

Et = 2 k q / a²

Point (0,0, R)

We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a

E1 = k q / (R + a)²

E2 = kq / (R-a)²

Et = kq [1 / (R + a)² + 1 / (R-a)²]

Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}

Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}

If we use the condition that R> a we can despise in the patents "a"

(R² + a²) = R² (1+ a² / R²) ≈ R²

(R + a)² = R² (1 + a / R)² ≈ R²

(R- a)² = R² (1-a / R)² ≈ R²

Substituting in the total electric field

Et = kq {2 R²) / [R²R²]}

Et =kq 2 / R²

7 0

2 years ago

The aqueduct passes under Johnson Road in Lancaster through a siphon. The maximum capacity of the aqueduct is 350 m3/s. The heig

Mariulka [41]

Answer:

D ≈ 8.45 m

L ≈ 100.02 m

Explanation:

Given

Q = 350 m³/s (volumetric water flow rate passing through the stretch of channel, maximum capacity of the aqueduct)

y₁ - y₂ = h = 2.00 m (the height difference from the upper to the lower channels)

x = 100.00 m (distance between the upper and the lower channels)

We assume that:

the upper and the lower channels are at the same pressure (the atmospheric pressure).
the velocity of water in the upper channel is zero (v₁ = 0 m/s).
y₁ = 2.00 m (height of the upper channel)
y₂ = 0.00 m (height of the lower channel)
g = 9.81 m/s²
ρ = 1000 Kg/m³ (density of water)

We apply Bernoulli's equation as follows between the point 1 (the upper channel) and the point 2 (the lower channel):

P₁ + (ρ*v₁²/2) + ρ*g*y₁ = P₂ + (ρ*v₂²/2) + ρ*g*y₂

Plugging the known values into the equation and simplifying we get

Patm + (1000 Kg/m³*(0 m/s)²/2) + (1000 Kg/m³)*(9.81 m/s²)*(2 m) = Patm + (1000 Kg/m³*v₂²/2) + (1000 Kg/m³)*(9.81 m/s²)*(0 m)

⇒ v₂ = 6.264 m/s

then we apply the formula

Q = v*A ⇒ A = Q/v ⇒ A = Q/v₂

⇒ A = (350 m³/s)/(6.264 m/s)

⇒ A = 55.873 m²

then, we get the diameter of the pipe as follows

A = π*D²/4 ⇒ D = 2*√(A/π)

⇒ D = 2*√(55.873 m²/π)

⇒ D = 8.434 m ≈ 8.45 m

Now, the length of the pipe can be obtained as follows

L² = x² + h²

⇒ L² = (100.00 m)² + (2.00 m)²

⇒ L ≈ 100.02 m

6 0

2 years ago

Read 2 more answers

A 1,100 kg car is traveling initially 20 m/s when the brakes are applied. The brakes apply a constant force while bringing the c

Natalka [10]

Answer:

Work done = -220,000 Joules.

Explanation:

<u>Given the following data;</u>

Mass = 1100kg

Initial velocity = 20m/s

To find workdone, we would calculate the kinetic energy possessed by the car.

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

$K.E = \frac{1}{2}MV^{2}$

Where,

K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

Substituting into the equation, we have;

$K.E = \frac{1}{2}*1100*20^{2}$

$K.E = 550*400$

K.E = 220,000J

Therefore, the workdone to bring the car to rest would be -220,000 Joules because the braking force is working to oppose the motion of the car.

4 0

1 year ago

Can we write names while writing conversation in board exam​

Can we write names while writing conversation in board exam