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snow_lady [41]
3 years ago
13

If a steady magnetic field exerts a force on a moving charge, that force is directed

Physics
1 answer:
lesya [120]3 years ago
4 0

Answer:

C) at right angles to the direction of the motion

Explanation:

The magnetic force exerted on a charged moving particle is given by

F=qvBsin \theta

where

q is the charge of the particle

v is the velocity of the particle

B is the magnetic field intensity

\theta is the angle between the direction of v and B

Moreover, the direction of the force is perpendicular to both v and B. In particular, the direction can be found by using the right hand rule:

- index finger: direction of the velocity

- middle finger: direction of the magnetic field

- thumb: direction of the force (if the charge is positive, otherwise the direction must be reversed if the charge is negative)

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The resistivity of water is increased if salt is added. true or false
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Metals in group 2 on the period table most commonly form which type
LuckyWell [14K]

Answer:

alkaline earth metals

Group 2 metals, the alkaline earth metals, have 2 valence electrons, and thus form M2+ ions. The halogens, Group 17 , reach a full valence shell upon reduction, and thus form X− ions

Explanation:

6 0
3 years ago
A projectile is fired with a velocity of 22 m/s at an angle of 25°. What is the vertical component of the velocity?
7nadin3 [17]

Answer:

Vertical component of velocity is 9.29 m/s

Explanation:

Given that,

Velocity of projection of a projectile, v = 22 m/s

It is fired at an angle of 22°

The horizontal component of velocity is v cosθ

The vertical component of velocity is v sinθ

So, vertical component is given by :

v_y=v\ sin(25)

v_y=22\ m/s\times\ sin(25)

v_y=9.29\ m/s

Hence, the vertical component of the velocity is 9.29 m/s

3 0
3 years ago
A 50 kg bear climbed on the tree branch 10 meters above the ground. If the bear descends to 5 meters above the ground, its poten
Simora [160]
<h3>Option A. 4900 J or joules</h3>
6 0
3 years ago
An electric dipole is formed from two charges, ±q, spaced 0.800 cm apart. The dipole is at the origin, oriented along the y-axis
Simora [160]

Answer: q = 2.781e-9C = 2.781nC

E=200C

Explanation:

E = Qd/(2πEor^3)

Where

E=Electric field intensity

Q=Charge

d=distance between the dipole=0.008m

Eo=permitivitty

400 N/C = Q(0.80e-2 m)/(2πε*(10e-2 m)^3)

Q= (400* 2* 3.142 * 8.85 x 10-12 * 0.1^3)/0.008

q = 2.781e-9C = 2.781nC

b)

Though the dipole are two separate charges. And since the point is on the x-axis, the electric field strengths are equivalent. The magnitude of the vector sum is:

E = kq*2sin θ/r^2

= 2(8.99e9 N*m^2/C^2)(2.781e-9 C)*sin(arctan(.4/10))/(10e-2 m)^2

= 2(8.99e9) * (2.781e-9) * sin(2.290)/(10e-2 m)^2

=200 C

3 0
2 years ago
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