Answer:
The acceleration due to gravity of that planet is, gₐ = 1.25 m/s²
Explanation:
Given that,
Mass of the planet, m = 1/2M
Radius of the planet, r = 2R
Where M and R is the mass and radius of the Earth respectively.
The acceleration due to gravity of Earth, g = 10 m/s²
The acceleration due to gravity of Earth is given by the relation,
g = GM/R²
Similarly, the acceleration due to gravity of that planet is
gₐ = Gm/r²
where G is the Universal gravitational constant
On substituting the values in the above equation
gₐ = G (1/2 M)/4 R²
= GM/8R²
= 1/8 ( 10 m/s²)
= 1.25 m/s²
Hence, the acceleration due to gravity of that planet is, gₐ = 1.25 m/s²
Hello!
We can use Faraday's Law of Electromagnetic Induction to solve.
ε = Induced emf (4.08 V)
N = Number of loops (?)
= Magnetic Flux (Wb)
t = time (s)
**Note: The negative sign can be disregarded for this situation. The sign simply shows how the induced emf OPPOSES the current.
Now, we know that is analogous to the change in magnetic flux over change in time, or , so:
Rearrange the equation to solve for 'N'.
Plug in the given values to solve.
**Rounding up because we cannot have a part of a loop.
Answer:
a) vₓ = 6,457 m / s
, v_{y} = 0.518 m / s
, b) v = 6.478 m / s, θ = 4.9°
Explanation:
a) This is a kinematic problem, let's use trigonometry to find the components of acceleration
sin 31 = / a
cos 31 = aₓ = a
a_{y} = a sin31
aₓ = a cos 31
Now let's use the kinematic equation for each axis
X axis
vₓ = v₀ₓ + aₓ (t-t₀)
vₓ = v₀ₓ + a cos 31 (t-t₀)
vₓ = 2.6 + 0.45 cos 31 (20-10)
vₓ = 6,457 m / s
Y Axis
v_{y} = v_{oy} + a_{y} t
v_{y} = v_{oy} + a_{y} sin31 (t-to)
v_{y} = -1.8 + 0.45 sin31 (20-10)
v_{y} = 0.518 m / s
b) let's use Pythagoras' theorem to find the magnitude of velocity
v = √ (vₓ² + v_{y}²)
v = √ (6,457² + 0.518²)
v = √ (41.96)
v = 6.478 m / s
We use trigonometry for direction
tan θ = v_{y} / vₓ
θ = tan⁻¹ v_{y} / vₓ
θ = tan⁻¹ 0.518 / 6.457
θ = 4.9°
c) let's look for the vector at the initial time
v₁ = √ (2.6² + 1.8²)
v₁ = 3.16 m / s
θ₁ = tan⁻¹ (-1.8 / 2.6)
θ₁ = -34.7
We see that the two vectors differ in module and direction, and that the acceleration vector is responsible for this change.
a = (v₂ -v₁) / (t₂-t₁)
Answer:
He did not do it very well.
An object is moving along a straight line and the uncertainty in its position is 0.1 m.
(a) Find the minimum uncertainty in the momentum of the object.
kg · m/s
(b) Find the minimum uncertainty in the object's velocity, assuming that the object is a golf ball (mass = 0.045 kg).
m/s
(c) Find the minimum uncertainty in the object's velocity, assuming that the object is an electron.
m/s