Answer:
The gauge pressure in Pascals inside a honey droplet is 416 Pa
Explanation:
Given;
diameter of the honey droplet, D = 0.1 cm
radius of the honey droplet, R = 0.05 cm = 0.0005 m
surface tension of honey, γ = 0.052 N/m
Apply Laplace's law for a spherical membrane with two surfaces
Gauge pressure = P₁ - P₀ = 2 (2γ / r)
Where;
P₀ is the atmospheric pressure
Gauge pressure = 4γ / r
Gauge pressure = 4 (0.052) / (0.0005)
Gauge pressure = 416 Pa
Therefore, the gauge pressure in Pascals inside a honey droplet is 416 Pa
Answer: 4
The mechanical advantage is the ratio of the force exerted by the object to the force applied to do work on it.
Here, Jeff tried to lift a rock weighing 600 pounds by wedging board under the rock. Jeff who weighs 150 pounds uses all his weight to exert force on lever and lift rock.
Mechanical advantage, 
Therefore, the mechanical advantage that lever provided to Jeff in lifting rock is 4.
Answer:
In free fall, mass is not relevant and there's no air resistance, so the acceleration the object is experimenting will be equal to the gravity exerted. If the object is falling on our planet, the value of gravity is approximately 9.81ms2 .
Answer:
The velocity of the other fragment immediately following the explosion is v .
Explanation:
Given :
Mass of original shell , m .
Velocity of shell , + v .
Now , the particle explodes into two half parts , i.e 
.
Since , no eternal force is applied in the particle .
Therefore , its momentum will be conserved .
So , Final momentum = Initial momentum
The velocity of the other fragment immediately following the explosion is v .
Answer: its linear momentum is 1.78 × 10²⁹ kg.m/s
Explanation:
Given that;
mass of Earth m = 5.972 x 10²⁴ kg
radius r = 1.496 x 10¹¹ m
period t = 3.15 x 10⁷ s
now we know that Earth rotates in a circular path so the distance travelled per rotation is;
d = 2πr we substitute
d = 2π × 1.496 x 10¹¹ m
= 9.4 × 10¹¹ m
Now formula for speed v is;
v = d/t
we substitute
v = 9.4 × 10¹¹ m / 3.15 x 10⁷ s
v = 2.98 × 10⁴ m/s
now we determine the linear momentum p
linear momentum p = mv
we substitute
p = (5.972 x 10²⁴ kg) × (2.98 × 10⁴ m/s)
p = 1.78 × 10²⁹ kg.m/s
Therefore its linear momentum is 1.78 × 10²⁹ kg.m/s
