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Artist 52 [7]
4 years ago
13

Rutherford hypothesized that, in his experiment,more alpha particles would be deflected if ?

Physics
1 answer:
Alex17521 [72]4 years ago
5 0
This is the CBS's this is a B a bit
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An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the sp
laila [671]

Answer:

The centripetal acceleration changed by a factor of 0.5

Explanation:

Given;

first radius of the horizontal circle, r₁ = 500 m

speed of the airplane, v = 150 m/s

second radius of the airplane, r₂ = 1000 m

Centripetal acceleration is given as;

a = \frac{v^2}{r}

At constant speed, we will have;

v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1

a₂ = 0.5a₁

Therefore, the centripetal acceleration changed by a factor of 0.5

7 0
4 years ago
Part B<br> Do the atoms in salt follow a pattern?
erastovalidia [21]

Answer: Yes.

Explanation: Because of the crystalline structure.

4 0
3 years ago
Relative motion can best be defined as
creativ13 [48]

Relative motion is the motion of an object relative to another object,
or as it would appear to an observer stuck to the other object and
riding on it.

6 0
3 years ago
A roller coaster is at the top of a 102 m hill and a mass of 120 kg. The coaster (at this
dybincka [34]
  • Height=h=102m
  • Mass=m=120g
  • Acceleration due to gravity=g=10m/s^2

\\ \sf\longmapsto P.E=mgh

  • P E means potential energy

\\ \sf\longmapsto P.E=102(120)(10)

\\ \sf\longmapsto P.E=122400J

7 0
3 years ago
A car accelerates uniformly from rest to 20 m/sec in 5.6 sec along a level stretch of road. Ignoring friction, determine the ave
bonufazy [111]

Answer:

(a) P=33000W

(b) P=51000W

Explanation:

The average power is defined as the amount of work done during a time interval:

P=\frac{W}{t}(1)

According to work-energy theorem, the work done is equal to the change in kinetic energy. So, we have:

W=\Delta K\\W=K_f-K_0\\W=\frac{mv_f^2}{2}-\frac{mv_0^2}{2}\\(2)

Recall that the weight is given by:

w=mg\\m=\frac{w}{g}(3)

The car accelerates uniformly from rest (v_0=0). Replacing (3) in (2), we have:

W=\frac{wv_f^2}{2g}

(a) Finally, we replace this in (1):

P=\frac{wv_f^2}{2gt}\\P=\frac{9000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=33000W

(b)

P=\frac{14000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=51000W

3 0
4 years ago
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