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3 years ago

If the magnetic force is 3.5 × 10–2 N, how fast is the charge moving?

Physics

2 answers:

Sveta_85 [38]3 years ago

6 0

Answer:

The velocity of charge is 1.1×10⁴

Option(D)

Explanation:

F = q(v×B)

3.5×10-²= (8.4×10-⁴)*(v)*(6.7×10-³)*sin35

v = (3.5)/(8.4×6.7×0.57) × 10⁵

v = 3.5/32.08 × 10⁵

v = 0.109 × 10⁵

v ≈ 1.1 × 10⁴ m/s

zaharov [31]3 years ago

3 0

Answer:

Explanation:

Took it on edg

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$\Delta KE = \Delta PE$

$\frac{1}{2}mv_f^2-\frac{1}{2} mv_0^2 = mgh_2-mgh_1$

Where,

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7 0

3 years ago

Two balls with masses of 2.0 kg and 6.0 kg travel toward each other at speeds of 12 m/s and 4.0 m/s, respectively. If the balls

Alina [70]

Answer:

The kinetic energy lost in the collision is 48 J

Explanation:

Given;

mass of the first ball, m₁ = 2.0 kg

mass of the second ball, m₂ = 6.0 kg

initial speed of the first ball, u₁ = 12 m/s

initial speed of the second ball, u₂ = 4 m/s

let v be the final velocity of the two balls after the inelastic collision

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2 x 12 + 6 x 4 = v(2 + 6)

48 = 8v

48 / 8 = v

v = 6 m/s

The initial kinetic energy of the balls is calculated as;

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(2)(12²) + ¹/₂(6)(4)²

K.E₁ = 144 + 48

K.E₁ = 192 J

The final kinetic of the balls is calculated as;

K.E₂ = ¹/₂(m₁ + m₂)(v²)

K.E₂ = ¹/₂(2 + 6)(6²)

K.E₂ = ¹/₂(8)(6²)

K.E₂ = 144 J

The lost in kinetic energy of the balls is K.E₂ - K.E₁ = 144 J - 192 J = -48 J

Therefore, the kinetic energy lost in the collision is 48 J

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3 years ago

Why is random sampling and random assignment in experimental research important?

Hunter-Best [27]

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3 years ago

An assault rifle fires an eight-shot burst in 0.40 s. Each bullet has a mass of 7.5 g and a speed of 300 m/s as it leaves the gu

myrzilka [38]

Answer:

The average recoil force on the gun during that 0.40 s burst is 45 N.

Explanation:

Mass of each bullet, m = 7.5 g = 0.0075 kg

Speed of the bullet, v = 300 m/s

Time, t = 0.4 s

The change in momentum of an object is equal to impulse delivered. So,

$F\times t=mv\\\\F=\dfrac{mv}{t}$

For 8 shot burst, average recoil force on the gun is :

$F=\dfrac{8mv}{t}\\\\F=\dfrac{7.5}{1000}\cdot\dfrac{300}{0.4}\cdot8\\\\F=45\ N$

So, the average recoil force on the gun during that 0.40 s burst is 45 N.

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3 years ago

a rocket engine can accelerate a rocket launched from rest vertically up with an acceleration of 22.9 m/s2. however, after 50.0

xxTIMURxx [149]

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Hence, There is no acceleration at the start. 22. This is the rocket's displacement in 50 seconds, so this is the rocket's altitude when the engine fails.

To know more about Accelerate refer to:

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1 year ago