The new gravitational attraction will be 1/4 as much
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 them
In this problem, the original force between the two objects is F, when they are separated by a distance r.
Later, the distance between the two objects is doubled, so the new distance is
Therefore, the new force will be
Therefore, the new force will be one-fourth as much.
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Answer:
7800kg/m³
Explanation:
Density of iron in CGS unit is 7.8 g/cm3. Its density is SI is
Given the density of iron = 7.8 g/cm3.
The SI units must be in kg/m³
7.8g = 7.8/1000 kg
7.8g = 0.0078kg
1cm³ = 0.000001m³
7.8g/cm³
= 0.0078/0.000001 kg/m³
= 7800kg/m³
Hence the density in SI unit is 7800kg/m³
Let the angle be Θ (theta)
Let the mass of the crate be m.
a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.
Normal force (N) = mg CosΘ
μ (coefficient of static friction) = 0.29
Static friction = μN = μmg CosΘ
Now, along the ramp, the equation of net force will be:
mg SinΘ - μmg CosΘ = 0
mg SinΘ = μmg CosΘ
tan Θ = μ
tan Θ = 0.29
Θ = 16.17°
b) Let the acceleration be a.
Coefficient of kinetic friction = μ = 0.26
Now, the equation of net force will be:
mg sinΘ - μ mg CosΘ = ma
a = g SinΘ - μg CosΘ
Plugging the values
a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96
a = 2.7244 - 2.44608
a = 0.278 m/s^2
Hence, the acceleration is 0.278 m/s^2
Answer:
0.45 seconds
Explanation:
Letting the value of g = 10 m/s/s
final velocity (v) = 0 m/s (since the egg will come to rest at the maximum height)
initial velocity(u) = 4.5 m/s
acceleration = -10 m/s/s (since the gravity is acting against the egg)
time = t seconds
From the first equation of motion:
<em>v = u + at</em>
<em>0 = 4.5 + (-10)t</em>
<em>t = -4.5 / -10</em>
t = 0.45 seconds
<u>Answer :</u>
(a) d = 0.25 m
(b) d = 0.5 m
<u>Explanation :</u>
It is given that,
Frequency of sound waves, f = 686 Hz
Speed of sound wave at 
is, v = 343 m/s
(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.
........(1)
Velocity of sound wave is given by :
Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.
(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.
( n = integers )
Let n = 1
So, 
Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.
