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3 years ago

In which arrangement of magnets will all the magnets attract

Physics

2 answers:

Elodia [21]3 years ago

7 0

when they all are opisite

IDK I just guess

Licemer1 [7]3 years ago

3 0

Answer:

It will attract when there are opposite magnets>>>

Explanation:

Opposites, Attract

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A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Pa

DIA [1.3K]

Given that,

Mass of trackler, m₁ = 100 kg

Speed of trackler, u₁ = 2.6 m/s

Mass of halfback, m₂ = 92 kg

Speed of halfback, u₂ = -5 m/s (direction is opposite)

To find,

Mutual speed immediately after the collision.

Solution,

The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s$

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.

3 0

3 years ago

Alpha particles, each having a charge of +2e and a mass of 6.64 ×10-27 kg, are accelerated in a uniform 0.50 T magnetic field to

sergij07 [2.7K]

Answer:

$KE=1.2036\times 10^{-12}\ J$

Explanation:

Given:

charge on the alpha particle, $q=2e=3.2\times 10^{-19}\ C$

mass of the alpha particle, $m=6.64\times 10^{-27}\ kg$

strength of a uniform magnetic field, $B=0.5\ T$

radius of the final orbit, $r=0.5\ m$

<u>During the motion of a charge the magnetic force and the centripetal forces are balanced:</u>

$q.v.B=m.\frac{v^2}{r}$

$m.v=q.B.r$

where:

v = velocity of the alpha particle

$v=\frac{q.B.r}{m}$

$v=\frac{3.2\times 10^{-19}\times 0.5\times 0.5}{6.64\times 10^{-27}}$

$v=1.2048\times 10^{7}\ m.s^{-1}$

Here we observe that the velocity of the aprticle is close to the velocity of light. So the kinetic energy will be relativistic.

<u>We firstly find the relativistic mass as:</u>

$m'=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } \times m$

$m'=\frac{6.64\times 10^{-27}}{\sqrt{1-\frac{(1.2048\times 10^7)^2}{(3\times 10^8)^2} } }$

$m'=6.6533\times10^{-27}\ kg$

now kinetic energy:

$KE=m'.c-m.c$

$KE=6.6533\times 10^{-27}\times (3\times 10^8)^2-6.64\times 10^{-27}\times (3\times 10^8)^2$

$KE=1.2036\times 10^{-12}\ J$

6 0

3 years ago

Which data set has the largest range?

MrRissso [65]

Answer:

Option C

Explanation:

We have to check range of all options first

For A:

Largest Value: 5

Smallest Value: 1

So range = Largest value - smallest value

5-1 = 4

For B:

Largest Value: 6

Smallest Value: 4

Range = 6-4 = 2

For C:

Largest Value: 9

Smallest Value: 1

Range = 9-1 = 8

For D:

Largest Value = 9

Smallest Value = 3

Range = 9-3=6

So, the data set in option C has the largest range

4 0

3 years ago

Define and describe the following terms: amplitude, wavelength, frequency, period, tension.

maw [93]

Answer:

Amplitude—distance between the resting position and the maximum displacement of the wave

Frequency—number of waves passing by a specific point per second

Period—time it takes for one wave cycle to complete

wavelength λ - the distance between adjacent identical parts of a wave, parallel to the direction of propagation.

Tension - described as the pulling force transmitted axially by the means of a string, a cable, chain, or similar one-dimensional continuous object, or by each end of a rod, truss member, or similar three-dimensional object

7 0

3 years ago

An astronaut on Pluto attaches a small brass ball to a 1.00-m length of string and makes a simple pendulum. She times 10 complet

pickupchik [31]

Answer:

The acceleration due to gravity at Pluto is 0.0597 m/s^2.

Explanation:

Length, L = 1 m

10 oscillations in 257 seconds

Time period, T = 257/10 = 25.7 s

Let the acceleration due to gravity is g.

Use the formula of time period of simple pendulum

$T = 2\pi\sqrt{\frac{L}{g}}\\\\25.7 = 2 \times 31.4\sqrt{\frac{1}{g}}\\\\g = 0.0597 m/s^2$

7 0

3 years ago