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Novosadov [1.4K]
3 years ago
13

Why does it take significantly stronger magnetic and electric field strengths to move the beam of alpha particles compared with

the beam of electrons(beta particles)?Medal will be given to best answer
Physics
1 answer:
wlad13 [49]3 years ago
4 0
It takes significantly stronger magnetic and electric field strengths to move a beam of alpha particles compared with the beam of electrons(betaparticles) because the charge of an alpha particle is twice stronger than a beta particle. Therefore, more energy is needed to move the alpha particle.
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A block starting from rest slides down the length of an 18 plank with an acceleration of 4.0 meters per second. How long does th
Snowcat [4.5K]

Answer:

\boxed{\text{\sf \Large 3.0 s}}

Explanation:

Use distance formula

\displaystyle d=ut+\frac{1}{2} at^2

u= \text{\sf  initial velocity}\\d= \text{\sf  distance}\\a= \text{\sf  acceleration}\\t= \text{\sf  time taken}

\displaystyle 18=0 \times t+\frac{1}{2} \times 4 \times t^2

t=3

4 0
3 years ago
In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st
bekas [8.4K]

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

a)

  • Total time will be the sum of the partial times between stations plus the time stopped at the stations.
  • Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
  • At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
  • In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

       v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s  (1)

  • Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

       t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)

  • Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

       x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m  (3)

  • In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

       t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)

  • We can find the distance traveled while the train was decelerating as follows:

       x_{3} = (v_{1} * t_{3})   + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m  (5)

  • Finally, we need to know the time traveled at constant speed.
  • So, we need to find first the distance traveled at the constant speed of 26.4m/s.
  • This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
  • x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s   (7)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
  • Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
  • tm = t*5 = 131.6 * 5 = 658.2 s (9)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
  • Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

  • Using all the same premises that for a) we know that the only  difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
  • Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
  • We can find the distance traveled at constant speed, rewriting (6) as follows:
  • x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s   (12)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
  • Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
  • tm = t*3 = 207.4 * 3 = 622.2 s (14)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
  • Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)
7 0
3 years ago
A car moving with an initial speed of 25 m/s slows down to a speed of 5 m/s in 10 seconds Calculate a) the acceleration of the c
stealth61 [152]

Answer :

(a) The acceleration  of the car is, -2m/s^2

(b) The distance covered by the car is, 150 m

Explanation :  

By the 1st equation of motion,

v=u+at ...........(1)

where,

v = final velocity = 5 m/s

u = initial velocity  = 25 m/s

t = time = 10 s

a = acceleration  of the car = ?

Now put all the given values in the above equation 1, we get:

5m/s=25m/s+a\times (10s)

a=-2m/s^2

The acceleration  of the car is, -2m/s^2

By the 2nd equation of motion,

s=ut+\frac{1}{2}at^2 ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity  = 25 m/s

t = time = 10 s

a = acceleration  of the car = -2m/s^2

Now put all the given values in the above equation 2, we get:

s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2

By solving the term, we get:

s=150m

The distance covered by the car is, 150 m

8 0
3 years ago
Give at least one fact about subatomic particles
victus00 [196]

Answer:

- Particles smaller than atoms are called subatomic particles .

- There are three famous subatomic particles, proton, neutron and electron .

- The study of sub atomic particles are called particle physics

- These particles can be divided as Brayons and Leptons

- These particles are often held together by one of the four fundamental particles ( Weak force, strong force, electromagnetic force, gravitational force).

6 0
3 years ago
Solve -7= sqrt 2x-9​
Sauron [17]

Answer:

x = 2

Explanation:

if it was -7 = the square root of both 2x-9 together, it would be false.

if it was square root of just 2x in the equation, the answer is:

x = 2

°°°°°°°°°

-7 = √2x - 9

-√2x = -9 + 7

√-2x = -2

√2x = 2

2x = 4

x = 2

4 0
3 years ago
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