Answer:
Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m
Explanation:
The attractive force experienced by two mass objects is known as Gravitational force.
The gravitational force is determine by the relation:
....(1)
According to the problem,
Mass of Moon, m₁ = 7.35 x 10²² kg
Mass of Earth, m₂ = 5.97 x 10²⁴ kg
Gravitational force experienced by them, F = 1.98 x 10²⁰ N
Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²
Substitute these values in equation (1).
d = 3.85 x 10⁸ m
Answer:
Mass = 18.0 kg
Explanation:
From Hooke's law,
F = ke
where: F is the force, k is the spring constant and e is the extension.
But, F = mg
So that,
mg = ke
On the Earth, let the gravitational force be 10 m/
.
3.0 x 10 = k x 5.0
30 = 5k
⇒ k = 
................ 1
On the Moon, the gravitational force is 
of that on the Earth.
m x 
= k x 5.0
= 5k
⇒ k = 
............. 2
Equating 1 and 2, we have;
=
m = 
= 18.0
m = 18.0 kg
The mass required to produce the same extension on the Moon is 18 kg.
Answer:
230.26 N
Explanation:
Since the speed is constant, acceleration is zero hence the net force will be given by the product of mass, coefficient of friction and acceleration due to gravity
F=0.72*32.6*9.81=230.26 N
Answer:
its will explode because it's like a soda with baking soda added together and you shake it the small hole can make a huge explosion
Answer:
Δy= 5,075 10⁻⁶ m
Explanation:
The expression that describes the interference phenomenon is
d sin θ = (m + ½) λ
As the observation is on a distant screen
tan θ = y / x
tan θ= sin θ/cos θ
As in ethanes I will experience the separation of the vines is small and the distance to the big screen
tan θ = sin θ
Let's replace
d y / x = (m + ½) λ
The width of a bright stripe at the difference in distance
y₁ = (m + ½) λ x / d
m = 1
y₁ = 3/2 λ x / d
Let's use m = 1, we look for the following interference,
m = 2
y₂ = (2+ ½) λ x / d
The distance to the screen is constant x₁ = x₂ = x₀
The width of the bright stripe is
Δy = λ x / d (5/2 -3/2)
Δy = 630 10⁻⁹ 2.90 /0.360 10⁻³ (1)
Δy= 5,075 10⁻⁶ m
