Answer:
you must throw 3 snowballs
Explanation:
We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment
Initial. Before bumping that refrigerator
p₀ = n m v₀
Where n is the snowball number
Final. When the refrigerator moves
pf = (n m + M) v
The moment is preserved because the forces during the crash are internal
n m v₀ = (n m + M) v
n m (v₀ - v) = M v
n = M/m v/(vo-v)
Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton
F = m a
a = F / m
The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm
v₁² = v₀² + 2 a x
v₁² = 0+ 2 (100/1) 1
v₁ = 14.14 m / s
This is the initial speed for the crash
v₀ = v = 14.14 m / s
Let's calculate
n = M/m v/ (v₀-v)
n = 10/1 3 / (14.14 -3)
n = 2.7 balls
you must throw 3 snowballs
Answer:
a. 2.668 m/s
b. 0.00494
Explanation:
The computation is shown below:
a. As we know that


As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.
F = 3.70 cos 45 = 2.62 N


We know that
KE1 = Initial kinetic energy
KE2 = kinetic energy following 100 m
The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.
So, the equation is
KE2 = KE1 - W

Now solve for v2


= 2.668 m/s
b. Now the minimum value of Ug is
As we know that
Ff = force of friction
Us = coefficient of static friction
N = Normal force = weight of skater
So,

Now solve for Us


= 0.00494
Kinetic energy means fast move movement
The solution for this problem is:The charge would be now equal to:(electric constant) multiplied by the (field strength) multiplied by the (area) so plugging in our values, will give us:8.85 * 10^-12 As / (V * m) * 3 * 10^6 V/m * 0.055 m^2 = 1.46 e-6 amperes would be the answer
Answer:
Answer to the question is: 1837.65 millimeters of mercury are equal to 245 kPa.
Explanation:
1 kPa are equal to 7.50062 millimeters of mercury.