Answer:
(1) passed through the foil
Explanation:
Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.
- When the experiment was conducted he found that most of the alpha particles went away without any deflection (due to the empty space) glowing the fluorescent screen right at the point of from where they were emitted.
- While a few were deflected at reflex angle because they were directed towards the center of the nucleus having the net effective charge as positive.
- And some were acutely deflected due to the field effect of the positive charge of the proton inside the nucleus. All these conclusions were made based upon the spot of glow on the fluorescent screen.
Answer:
B.It is a satellite that collects data about rain and snow
C.Its orbit covers 90 percent of Earth’s surface
F.The sensors measure microwaves
Answer:
to the right.
to in the upwards direction.
Explanation:
In order to solve this problem, we must first start by drawing a diagram of the situation. (See attached diagram).
So, remember that a force is determined by multiplying the mass of the parcticle by its acceleration:
F=ma
so in order to find the components of the force, we need to start by finding its acceleration.
Acceleration is found by using the following formula:

so we can subtract the two vectors, like this:

which yields:

or:

so now I can find the components of the force:

which yields:
F=(2.31i+2.1j)N
so the components of the force are:
to the right.
to in the upwards direction.
Answer:
20 rad/s
Explanation:
mass, m = 12 kg
radius, r = 0.250 m
Moment of inertia of cylinder, I = 1/2 mr²
I = 0.5 x 12 x 0.250 x 0.250 = 0.375 kgm^2
Work done = Change in kinetic energy
Initial K = 0
Final K = 1/2 Iω²
W = 1/2 Iω²
ω² = 2W/ I = 2 x 75 / (0.375)
ω = 20 rad/s
Thus, the final angular velocity is 20 rad/s .
Answer:
Explanation:
Electric field due to charge at origin
= k Q / r²
k is a constant , Q is charge and r is distance
= 9 x 10⁹ x 5 x 10⁻⁶ / .5²
= 180 x 10³ N /C
In vector form
E₁ = 180 x 10³ j
Electric field due to q₂ charge
= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²
= 30.33 x 10³ N / C
It will have negative slope θ with x axis
Tan θ = .5 / √.5² + .8²
= .5 / .94
θ = 28°
E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j
= 26.78 x 10³ i - 14.24 x 10³ j
Total electric field
E = E₁ + E₂
= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j
= 26.78 x 10³ i + 165.76 X 10³ j
magnitude
= √(26.78² + 165.76² ) x 10³ N /C
= 167.8 x 10³ N / C .