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NARA [144]
3 years ago
7

The fact that the magnetic field generates a force perpendicular to the instantaneous velocity of the particle has implications

for the work that the field does on the particle. As a consequence, if only the magnetic field acts on the particle, its kinetic energy will ____________.
__ increase over time
__ decrease over time
__ remain constant
__ oscillate
Physics
1 answer:
Amanda [17]3 years ago
4 0

Answer:

The kinetic energy will remain constant

Explanation:

Charged particles in a magnetic field feel a force perpendicular to their velocity,  since the force is F = qvB, and a charged particle feels a force of constant magnitude always directed perpendicular to its motion which causes circular motion in the magnetic field but the speed and kinetic energy of the particle will remain constant.

Therefore, if only the magnetic field acts on the particle, its kinetic energy will remain constant.

You might be interested in
How many seconds of space should you keep between you if you're driving a 100 foot truck 30 mph?
Marrrta [24]

Answer:

<em> The space in seconds that will be kept = 2.27 seconds</em>

Explanation:

S = d/t..................... Equation 1

making t the subject of formula in the equation above,

t = d/S.................... Equation 2

Where S = speed, d = distance, t = time.

<em>Conversion: (i)if 1 mph = 0.44704 m/s,</em>

<em>                 then, 30 mph = 30×0.44704    </em>

<em>                = 13.41 m/s</em>

<em>               (ii) If 1 foot = 0.3048 m</em>

<em>            then, 100 foot = 30.48 m.</em>

<em>Given: S = 30 mph = 13.41 m/s, d = 100 foot = 30.48 m</em>

<em>Substituting these values into equation 2</em>

<em>t = 30.48/13.41</em>

<em>t = 2.27 seconds.</em>

<em>Therefore the space in seconds that will be kept = 2.27 seconds</em>

7 0
3 years ago
A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.
Veronika [31]

Answer:

A) ω = 6v/19L

B) K2/K1 = 3/19

Explanation:

Mr = Mass of rod

Mb = Mass of bullet = Mr/4

Ir = (1/3)(Mr)L²

Ib = MbRb²

Radius of rotation of bullet Rb = L/2

A) From conservation of angular momentum,

L1 = L2

(Mb)v(L/2) = (Ir+ Ib)ω2

Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.

(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2

(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2

Divide each term by Mr;

vL/8 = (L²/3 + L²/16)ω2

vL/8 = (19L²/48)ω2

Divide both sides by L to obtain;

v/8 = (19L/48)ω2

Thus;

ω2 = 48v/(19x8L) = 6v/19L

B) K1 = K1b + K1r

K1 = (1/2)(Mb)v² + Ir(w1²)

= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)

= (1/8)(Mr)v²

K2 = (1/2)(Isys)(ω2²)

I(sys) is (Ir+ Ib). This gives us;

Isys = (19L²Mr/48)

K2 =(1/2)(19L²Mr/48)(6v/19L)²

= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152

Thus, the ratio, K2/K1 =

[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19

3 0
3 years ago
5,6,1,9,21,37,68 ? find missing​
maria [59]

Answer:

Missing numbers are 3,11,13, 15,17,19, 23,27,29 and wrong are 6,68

4 0
3 years ago
Projectile A is launched horizontally at a speed of 20. Meters per second from the top of a cliff and strikes a level surface be
miv72 [106K]

consider the motion of projectile A in vertical direction :

v₀ = initial velocity of projectile A in vertical direction = 0 m/s         (since the projectile was launched horizontally)

a = acceleration of the projectile = g = acceleration due to gravity = 9.8 m/s²

t = time of travel for projectile A = 3.0 seconds

Y = vertical displacement of projectile A = height of the cliff = h = ?

using the kinematics equation along the vertical direction as

Y = v₀ t + (0.5) a t²

h = (0) (3.0) + (0.5) (9.8) (3.0)²

h = 44.1 m

4 0
3 years ago
Guys please help me ​
Likurg_2 [28]

Answer:

1)t=2.26\: s

2)S=33.9\: m

3)v=26.77\: m/s

4)\alpha=55.92

Explanation:

1)

We can use the following equation:

y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

0=25-0.5*9.81*t^{2}

t=2.26\: s

2)

The equation of the motion in the x-direction is:

v_{ix}=\frac{S}{t}

15=\frac{S}{2.26}

S=33.9\: m

3)

The velocity in the y-direction of the stone will be:

v_{fy}=v_{iy}-gt

v_{fy}=0-(9.81*2.26)

v_{fy}=-22.17\: m/s

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}

v=26.77\: m/s

4)

The angle of this velocity is:

tan(\alpha)=\frac{22.17}{15}

\alpha=tan^{-1}(\frac{22.17}{15})

\alpha=55.92

Then α=55.92° negative from the x-direction.

I hope it helps you!

6 0
3 years ago
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