Answer:
aₓ = 0
, ay = -6.8125 m / s²
Explanation:
This is an exercise that we can solve with kinematics equations.
Initially the rabbit moves on the x axis with a speed of 1.10 m / s and after seeing the predator acceleration on the y axis, therefore its speed on the x axis remains constant.
x axis
vₓ = v₀ₓ = 1.10 m / s
aₓ = 0
y axis
initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration
= v_{oy} -ay t
ay = (v_{oy} -v_{y}) / t
ay = (0 -10.9) / 1.6
ay = -6.8125 m / s²
the sign indicates that the acceleration goes in the negative direction of the y axis
The gravitational force between two masses m₁ and m₂ is
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d = distance between the masses.
Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg
Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
= 0.0222
d = 0.1492 m = 149.2 mm
Answer: 149.2 mm
Answer:
Given that
T= 0.43 s
Radius of the ball path's , r=2.1 m
a)
We know that
f= 1/T
Here f= frequency
T= Time period
Now by putting the values
f= 1/T
T= 0.43 s
f= 1/0.43
f=2.32 Hz
b)
We know that
V= ω r
ω = 2 π f
ω=Angular speed
V= Linear speed
ω = 2 π f=ω = 2 x π x 2.32 =14.60 rad/s
V= ω r= 14.60 x 2.1 = 30.66 m/s
c)
Acceleration ,a
a =ω ² r
a= 14.6 ² x 2.1 = 447.63 m/s²
We know that g = 10 m/s²
So
a= a/g= 447.63/10 = 44.7 g m/s²
a= 44.7 g m/s²
Answer:
Scalar product is between ║A║ ║ B║ and -║A║ ║ B║
Explanation:
Dot product between vec A and vec B is
A.B = ║A║ ║ B║ cos θ
Here, both ║A║ and ║B║ are positive and value of cos θ depends upon θ and lies between 1 and -1
So, Scalar product is between ║A║ ║ B║ and -║A║ ║ B║
