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Pavel [41]
2 years ago
14

Which statement best describes why some materials are NOT magnetic?

Physics
1 answer:
hammer [34]2 years ago
3 0

B

paired electrons spin in opposite directions cancelling their magnetic fields

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at the extact same time, youj drop a piece of notebook paper and your science out the window. the science book hit first? why?
jeyben [28]
The piece of paper has less mass and will glide down the window, whereas the textbook will go straight to the ground. Since the textbook has more mass and less ways of it being able to 'glide' the textbook will hit the ground first.
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How did mount st. Helen get its name
steposvetlana [31]

The modern name, Mount St. Helen's, was given to the volcanic peak in 1792 by seafarer and explorer Captain George Vancouver of the British Royal Navy. He named it in honor of fellow countryman Alleyne Fitzherbert, who held the title 'Baron St. Helen's.

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3 years ago
Describe how the student should measure the time taken for the toy parachute to
PolarNik [594]

The time must be measured with respect to gravity. As it falls, it has free fall that is the force acting on it will be the gravity.With the distance in account, d = 1/2 gt²

t = √(2d/g)

4 0
2 years ago
A point charge q1=2.0μC is located on the positive y axis at y=0.30m, and an identical charge q2 is at the origin. Find the magn
dalvyx [7]

Answer:

(A) 0.279N at angle 38.02°

(B) 0.701N

(C) 14.19°

Explanation:

(A) The net force on q3 is given as:

F = Fxi + Fyj

Fx is the x component of the force

Fy is the y component of the force

Fx = -F(1, 3)cos(90 - x) + F(2, 3)cos0

Fy = -F(2, 3)cosx - F(2, 3)cos90 = -F(2, 3)cosx

First let us find y and angle x from the diagram.

Using Pythagoras theorem,

y² = 0.3² + 0.4²

y² = 0.25

y = 0.5m

Using SOHCAHTOA to find x,

sinx = 0.4/0.5

x = 53.13°

Electrostatic force, F is given as:

F = kqQ/r²

Where k = Coulumbs constant

F(1,3) = (k*q1*q3) / r²

F(1, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.5²)

F(1, 3) = 0.288N

F(2,3) = (k*q2*q3) / r²

F(2, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)

F(2, 3) = 0.45N

Therefore,

Fx = -0.288cos36.87 + 0.45

Fx = 0.22N

Fy = 0.288cos53.13

Fy = 0.172N

=> F = 0.22i + 0.172j

The magnitude of the force will be

F(mag) = √(0.22² + 0.172²)

F(mag) = 0.279N

The direction of the force makes will be

tanθ = Fy/Fx

tanθ = 0.172/0.22 = 0.781

θ = 38.02° to the x axis.

(B) q2 = - 2.0 * 10^(-6)

This implies that:

F(2,3) = (k*q2*q3) / r²

F(2, 3) = (9 * 10^9 * -2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)

F(2, 3) = -0.45N

Therefore,

Fx = -0.288cos36.87 - 0.45

Fx = -0.68N

Fy = 0.172N

=> F = - 0.68i + 0.172j

The magnitude of the force will be

F(mag) = √((-0.68)² + 0.172²)

F(mag) = 0.701N

(C) The direction of the force makes will be

tanθ = 0.172/0.68

θ = 14.19° to the x axis

4 0
3 years ago
Read 2 more answers
A 30-g bullet is fired with a horizontal velocity of 460 m/s and becomes embedded in block B which has a mass of 3 kg. After the
ICE Princess25 [194]

Answer:

energy loss due to friction and the impacts = 2.97 J; The impact loss due to AB impacting the carrier is =25.72J; The impact loss at first impact is 6,316.64J

Explanation:

First find the velocity of the bullet after the first impact using

M1V1 + 0 = (M1 + M2)v'

Where M1 is the mass of the bullet

M2 is the mass of the block B

M3 is the mass of the carrier

v' is the velocity

v' = M1V1/(M1 + M2)

v'= (30 × 10^-3 kg)(460 m/s) / (30 × 10^-3 kg + 3 kg)

v' = 13.8/3.03

v'= 4.55m/s

Also calculate final velocity of the carrier v2'

v2' = M1V1/(M1 + M2 + M3)

v2'= (30 × 10 kg)(460 m/s) / (30 × 10 kg + 3 kg + 30kg)

v2' =0.42m/s

Now to calculate energy loss due to friction

Normal force

N= W1 + W2 = (m1 + m2)g

Where W1 and W2 is the weight of the bullet and block respectively

g is gravitational acceleration for taken as 9.81m/s

= (0.030 kg + 3 kg)(9.81 m/s) 29.724 N

Friction force = coefficient of kinetic× normal force

Where coefficient of kinetic = 0.2

Ff = (0.2)(29.724)= 5.945 N

Now

Energy loss due to friction = frictional force × distance

Assume distance is 0.5 m.

Energy loss due to friction = 5.945 N × 0.5 m

= 2.97J

Kinetic energy of block with embedded bullet immediately after first impact:

1/2 × (m1 + m2)(v')^2

1/2 × (30 × 10^-3 kg + 3 kg)(4.55m/s)^2

= 1/2 × (3.03kg) × (4.55m/s)^2

= 31.36 J

Final kinetic energy of bullet, Block, and Carrier together

1/2 × (m1 + m2 + m3)(v2')^2

1/2 × (30 × 10^-3 kg + 3 kg + 30kg) (0.42m/s)^2

1/2 × (33.03kg) × (0.42m/s)^2

= 2.91 J

Therefore

Loss due to friction and stopping impact = Kinetic energy of block with embedded bullet immediately after first impact - Final kinetic energy of bullet, Block, and Carrier together

= 31.36 J - 2.91 J

= 28.69 J

Impact loss due to AB impacting the carrier = loss due to friction- energy due to friction

28.69J - 2.97J

=25.72J

Initial kinetic energy of system ABC = 1/2(m1vo)

=1/2(0.030 kg)(460 m/s)^2 = 6,348J

Therefore

Impact loss at first impact = Initial kinetic energy of system ABC - Kinetic energy of block with embedded bullet immediately after first impact:

= 6,348J - 31.36 J

= 6,316.64J

3 0
3 years ago
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