The _____ phrase "answers" the antecedent with a similar rhythm and harmony.

Do you think you must have a tree of your?

Novosadov [1.4K]

Answer:

yes I think I must have a tree of me I hope it will help you please follow me

4 0

3 years ago

A current-carrying wire is bent into a circular loop of radius R and lies in an xy plane. A uniform external magnetic field B in

Ilya [14]

Answer:

F = 2π I R B

Explanation:

The magnetic force is described by the equation.

F = q v x B = i L x B

Where i is the current, L is a vector that points in the direction of the current (length) and B is the magnetic field.

This equation can be used in scalar form and the direction of the force found by the right hand ruler, the thumb goes in the direction of L, the fingers extended in the direction of B and the palm of the hand indicates the direction of the force if the load is positive

F = i L B sin θ

In this case the wire is in the xy plane and the z-axis field whereby they are perpendicular, θ = 90º and sin 90 = 1

F = i L B

The loop length is

L = 2π R

F = i 2π R B

F = 2π I R B

The force is in the loop

8 0

3 years ago

The momentum of an electron is 1.75 times larger than the value computed non-relativistically. What is the speed of the electron

FrozenT [24]

Answer:

Speed of the electron is 2.46 x 10^8 m/s

Explanation:

momentum of the electron before relativistic effect = $M_{0} V$

where $M_{0}$ is the rest mass of the electron

V is the velocity of the electron.

under relativistic effect, the mass increases.

under relativistic effect, the new mass M will be

M = $M_{0}/ \sqrt{1 - \beta ^{2} }$

where

$\beta = V/c$

c is the speed of light = 3 x 10^8 m/s

V is the speed with which the electron travels.

The new momentum will therefore be

==> $M_{0}V/ \sqrt{1 - \beta ^{2} }$

It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have

1.75 $M_{0} V$ = $M_{0}V/ \sqrt{1 - \beta ^{2} }$

the equation reduces to

1.75 = $1/ \sqrt{1 - \beta ^{2} }$

square both sides of the equation, we have

3.0625 = 1/ $(1 - \beta ^{2} )$

3.0625 - 3.0625 $\beta ^{2}$ = 1

2.0625 = 3.0625 $\beta ^{2}$

$\beta ^{2}$ = 0.67

β = 0.819

substitute for $\beta = V/c$

V/c = 0.819

V = c x 0.819

V = 3 x 10^8 x 0.819 = 2.46 x 10^8 m/s

6 0

3 years ago

The 45-g arrow is launched so that it hits and embeds in a 1.40 kg block. The block hangs from strings. After the arrow joins th

worty [1.4K]

Question: How fast was the arrow moving before it joined the block?

Answer:

The arrow was moving at 15.9 m/s.

Explanation:

The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

$\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg$

where $m_a$ is the mass of the arrow, $m_b$ is the mass of the block, $\Delta H$ of the change in height of the block after the collision, and $v$ is the velocity of the arrow before it hit the block.