The gravitational force between the two balls is 
Explanation:
The magnitude of the gravitational force between two objects is given by:
where
:
is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between the objects
For the balls in this problem, we have
r = 0.74 m
Substituting into the equation, we find the gravitational force between the two balls:
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Answer:
157.9 kg
Explanation:
Density: This can be defined as the ratio of the mass of a body and it's volume.
The S.I unit of density is kg/m³.
From the question,
Density = Mass/volume
D = m/v............................ Equation 1
Where D = Density of gold, m = mass of gold, v = volume of gold.
make m the subject of the equation
m = Dv.................... Equation 2
Since the gold is a cube,
v = l³................... Equation 3
Where l = length of the gold cube.
Substitute equation 3 into equation 2
m = Dl³............... Equation 4
Given: D = 19300 kg/m³, l = 0.2015 m
Substitute into equation 4
m = 19300(0.2015)³
m = 157.9 kg.
Answer:
Small sparks can lead to huge explosion if they are left unattended.
Explanation:
Small sparks are not harmful but if these sparks happen near some hazardous material or object then it could lead to heavy explosion. If there is some chemical substance near the spark or there are magnetic lines which can explode the spark then these minor sparks could result in heavy disastrous explosion.
Answer:
Option B is the correct answer.
Explanation:
Shear stress is the ratio of shear force to area.
We have
Shear stress = 3 N/mm² = 3 x 10⁶ N/m²
Area = Area of rectangle = 10 x 10⁻² x d = 0.1d
Shear force = 50000 N
Substituting
Width of beam = 16.67 cm
Option B is the correct answer.
For a circular aperture, the first minima (n=1) as an angular separation from the peak of the central maxima given by
Sinθ = 1.22λ / d
Where,
d is the aperture or pupil diameter
d = 4.69 mm = 4.69 × 10^-3m
λ is the wavelength
λ = 545 nm = 545 × 10^-9 m
Then,
Sinθ = 1.22λ / d
Sinθ = 1.22 × 545 × 10^-9 / 4.69 × 10^-3
Sinθ = 1.418 × 10^-4 rad
Then, the head light sources have the same angular separation θ from the eye as the image have inside the eye.
For the headlight
Sinθ ≈ light separation / distantce for the eye
Light separation is give as x = 0.659 m
And let the distance of the eye be D
Then,
Sinθ = x / D
Make D subject of formula
D = x / Sinθ
D = 0.695 / 1.418 × 10^-4
D = 4902.316m
To km, 1km = 1000m
D ≈ 4.9 km
