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2 years ago

Question 2 of 10 Which illustration represents low accuracy and low precision?

Physics

1 answer:

Zina [86]2 years ago

6 0

Answer:

D is the correct answer

Explanation:

A) shows low accuracy and high precision since they missed the bull's-eye but are all grouped together.

B) Shows high accuracy and high precision

C) Shows high precision

D is the only one that shows both low accuracy and low precision

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What exactly does it mean to say an object is polarized

Anestetic [448]

It means that it has a force to make something go in one directions only
if a metal is polarized it means it has a pith pole and a north pole like a magnet

5 0

3 years ago

Suppose that she pushes on the sphere tangent to its surface with a steady force of F = 60 N and that the pressured water provid

sergey [27]

Answer:

25.06s

Explanation:

Remaining part of the question.

(A large stone sphere has a mass of 8200 kg and a radius of 90 cm and floats with nearly zero friction on a thin layer of pressurized water.)

Solution:

F = 60N

r = 90cm = 0.9m

M = 8200kg

Moment of inertia for a sphere (I) = ⅖mr²

I = ⅖ * m * r²

I = ⅖ * 8200 * (0.9)²

I = 0.4 * 8200 * 0.81

I = 2656.8 kgm²

Torque (T) = Iα

but T = Fr

Equating both equations,

Iα = Fr

α = Fr / I

α = (60 * 0.9) / 2656.8

α = 0.020rad/s²

The time it will take her to rotate the sphere,

Θ = w₀t + ½αt²

Angular displacement for one revolution is 2Π rads..

θ = 2π rads

2π = 0 + ½ * 0.02 * t²

(w₀ is equal to zero since sphere is at rest)

2π = ½ * 0.02 * t²

6.284 = 0.01 t²

t² =6.284 / 0.01

t² = 628.4

t = √(628.4)

t = 25.06s

8 0

3 years ago

A mirror forms an erect image 40cm from the object and one third its height where must the mirror be situated

iragen [17]

The mirror is located 30 cm from the object

Explanation:

To solve the problem, we can use the magnification equation, which states that:

$M=\frac{y'}{y}=-\frac{q}{p}$

where

M is the magnification

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

In this problem, we notice that the image formed by the mirror is erect and diminished: this means that this is a convex mirror, so the image is virtual, and this means that the image and the object are located on opposite sides of the mirror. Therefore,

$p-q=40 cm$ (distance of the image from the object is 40 cm, but since the image is virtual, q is the negative)

The size of the image is 1/3 that of the object, so

$y'=\frac{1}{3}y$

Solving the equation for p, we find the distance between the object and the mirror:

$\frac{y'}{y}=-\frac{p-40}{p}\\\\p=\frac{40}{1+\frac{y'}{y}}=\frac{40}{4/3}=30 cm$

So, the mirror is 30 cm from the object.

#LearnwithBrainly

5 0

3 years ago

When is thermal equilibrium achieved between two objects

melisa1 [442]

Answer:

when the temperatures of the two objects are equal

Explanation:

Thermal equilibrium is achieved between two objects when the temperatures of the two objects are equal. Heat flows from hot to cold objects. When the two objects attain equal temperatures, we say thermal equilibrium has been achieved

8 0

2 years ago

Four identical masses m are evenly spaced on a frictionless 1D track. The first mass is sent at speed v toward the other three.

SpyIntel [72]

Answer:

The speed decreases 75%.

Explanation:

Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
For the first collission, only mass 1 is moving before it, so we can write the following equation:

$p_{i} = p_{f} = m*v_{o} (1)$

Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:

$p_{f1} = 2*m*v_{1} (2)$

From (1) and (2) we get:
v₁ = v₀/2 (3)
Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:

$p_{1} = 2*m*v_{1} = 2*m*\frac{v_{o} }{2} = m*v (4)$

Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:

$p_{2} = 3*m*v_{2} (5)$

From (4) and (5) we get:
v₂ = v₀/3 (6)

Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:

$p_{2} = 3*m*v_{2} = 3*m*\frac{v_{o} }{3} = m*v (7)$

Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:

$p_{3} = 4*m*v_{3} (8)$

From (7) and (8) we get:
v₃ = v₀/4
This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.

5 0

2 years ago