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Damm [24]
4 years ago
9

Argon, neon, and xenon are examples of

Physics
1 answer:
Luda [366]4 years ago
3 0
1. your in highschool you should know this but its A
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Suppose a 96.10 kg hiker has ascended to a height of 1841 m above sea level in the process of climbing Mt. Washington. By what p
lbvjy [14]

Answer:

0.029%

Explanation:

Weight on the surface is given by,

F_W = mg = (96.1Kg)(9.8m/s^2)=941.78N

The acceleration due to gravity at given height is,

g_h=\frac{gR^2}{R+h^2}

The weight in another height is equal to

F_h=mg_h

F_h= m*\frac{gR^2}{R+h^2}

F_h = \frac{F_WR^2}{(R+h)^2}

F_h=\frac{(941.78)(6.371*10^6)}{(6.371*10^6+1841)}

F_h = 941.50N

The change of the Weight is,

\frac{F-F_h}{F}*100=\frac{941.78-941.50}{941.78}*100=0.029\%

6 0
3 years ago
A square loop of wire is carrying current in the counterclockwise direction. There is a horizontal uniform magnetic field pointi
Citrus2011 [14]

1. C) The net force on the loop is zero

Explanation:

The uniform magnetic field points to the right, while the current is counterclockwise. We must analyze the direction of the force on each side of the loop.

However, we notice that:

- the current in the upper and lower side of the loop is parallel/anti-parallel to the magnetic field, so the magnetic force on these sides is zero

- the current in the left side and right side of the loop is perpendicular to the magnetic field, however the direction of the current in the two sides is opposite, so the two forces will be opposite, and since their magnitude is equal (because the field is uniform), then the net force must be zero

2. (B) around the y axis

Explanation:

We already said that the magnetic force on the upper and lower side of the loop is zero. So let's analyze the direction of the force on the left and right side of the loop:

- Left side: the current is flowing downward, while the magnetic field points to the right. Using the right-hand rule:

-- Index finger: current (downward)

-- Middle finger: magnetic field (to the right)

-- Thumb: force (out of the screen, positive z-axis)

- Right side: the current is flowing upward, while the magnetic field points to the right. Using the right-hand rule:

-- Index finger: current (upward)

-- Middle finger: magnetic field (to the right)

-- Thumb: force (into the screen, negative z-axis)

---> this means that the loop of wire will rotate around the y-axis.

7 0
4 years ago
Is the force of friction the sum of all the microscopic contact forces between the bottom of the block and the surface?
kotegsom [21]

The force of friction is that force that tends to oppose the motion of the body with the surface in contact... are you clear with it?

6 0
3 years ago
An example of a solution is
LenaWriter [7]

A solution is a homogeneous mixture such as salt water

5 0
3 years ago
Block 1, of mass m1 = 2.70 kg , moves along a frictionless air track with speed v1 = 27.0 m/s . It collides with block 2, of mas
ki77a [65]

Answer:

a) Block 1 = 72.9kgm/s

Block 2 = 0kgm/s

b) vf = 1.31m/s

c) ∆KE = 936.36Joules

Explanation:

a) Momentum = mass× velocity

For block 1:

Momentum = 2.7×27

= 72.9kgm/s

For block 2:

Momentum = 53(0) (body is initially at rest)

= 0kgm/s

b) Using the law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses of the block

u1 and u2 are their initial velocity

v is the common velocity

Given m1 = 2.7kg, u1 = 27m/s, m2 = 53kg, u2 = 0m/s (body at rest)

2.7(27)+53(0) = (2.7+53)v

72.9 = 55.7v

V = 72.9/55.7

Vf = 1.31m/s

c) kinetic energy = 1/2mv²

Kinetic energy of block 1 = 1/2×2.7(27)²

= 984.15Joules

Kinetic energy of block 2 before collision = 0kgm/s

Total KE before collision = 984.15Joules

Kinetic energy after collision = 1/2(2.7+53)1.31²

= 1/2×55.7×1.31²

= 47.79Joules

∆KE = 984.15-47.79

∆KE = 936.36Joules

7 0
3 years ago
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