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oksian1 [2.3K]
2 years ago
8

A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 N/C is a

cting down, how long does it take the proton to return to the horizontal plane?
Physics
1 answer:
GuDViN [60]2 years ago
6 0

Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}

T = 7.83 X10⁻⁷ s

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Sound waves or bounces off the wall and light waves are waves of light
7 0
3 years ago
A 97.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.63 r
sammy [17]

Answer:

the final angular velocity of the platform with its load is 1.0356 rad/s

Explanation:

Given that;

mass of circular platform m = 97.1 kg

Initial angular velocity of platform ω₀ = 1.63 rad/s

mass of banana m_{b} = 8.97 kg

at distance r = 4/5  { radius of platform }

mass of monkey m_{m} = 22.1 kg

at edge = R

R = 1.73 m

now since there is No external Torque

Angular momentum will be conserved, so;

mR²/2 × ω₀ = [ mR²/2 + m_{b} (\frac{4}{5} R)² + m_{m}R² ]w

m/2 × ω₀ = [ m/2 + m_{b} (\frac{4}{5} )² + m_{m} ]w

we substitute

w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1

w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )

w = 48.55 × [ 1.63 / ( 76.3908 ) ]

w = 48.55 × 0.02133

w = 1.0356 rad/s

Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s

8 0
3 years ago
Which is harder, air tight or water tight seals?
Vladimir [108]
Should be an air tight seal
4 0
3 years ago
A car with a mass of 1.50x10^3 kg starts from rest and accelerates to a speed of 18.0m/s in 12.0 s. assume that the force of res
Luden [163]
The first thing you should know for this case is the definition of distance.
 d = v * t
 Where,
 v = speed
 t = time
 We have then:
 d = v * t
 d = 9 * 12 = 108 m
 The kinetic energy is:
 K = ½mv²
 Where,
 m: mass
 v: speed
 K = ½ * 1500 * (18) ² = 2.43 * 10 ^ 5 J
 The work due to friction is
 w = F * d
 Where,
 F = Force
 d = distance:
 w = 400 * 108 = 4.32 * 10 ^ 4
 The power will be:
 P = (K + work) / t
 Where,
 t: time
 P = 2.86 * 10 ^ 5/12 = 23.9 kW
 answer:
 the average power developed by the engine is 23.9 kW
8 0
3 years ago
A scuba diver is sitting on a boat while waiting to go on a dive and sees light reflected from the water's surface. At what angl
LUCKY_DIMON [66]

Answer:

θ_p = 53.0º

Explanation:

For reflection polarization occurs when a beam is reflected at the interface between two means, the polarization in total when the angle between the reflected and the transmitted beam is 90º

Let's write the transmission equation

     n1 sin θ₁ = ne sin θ₂

The angle to normal (vertcal) is

    180 = θ2 + 90 + θ_p

    θ₂ = 90 - θ_p

Where θ₂ is the angle of the transmitted ray θ_p is the angle of the reflected polarized ray

We replace

     n1 sin θ_p = n2 sin (90 - θ_p)

Let's use the trigonometry relationship

    Sin (90- θ_p) = sin 90 cos θ_p - cos 90 sin θ_p = cos θ_p

In the law of reflection  incident angle equals reflected angle,  

    ni sin θ_p = ns cos θ_p

    n₂ / n₁ = sin θ_p / cos θ_p

    n₂ / n₁ = tan θ_p

    θ_p = tan⁻¹ (n₂ / n₁)

Now we can calculate it

The refractive index of air is 1 (n1 = 1) the refractive index of seawater varies between 1.33 and 1.40 depending on the amount of salts dissolved in the water

n₂ = 1.33

      θ_p = tan⁻¹ (1.33 / 1)

      θ_p = 53.0º

n₂ = 1.40

      θ_p = tan⁻¹ (1.40 / 1)

      Tep = 54.5º

4 0
3 years ago
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