What will happen if an stop able item was to hit an unbreakable item

A cart moves with negligible friction or air resistance along a roller coaster track. The cart starts from rest at the top of a

lina2011 [118]

Answer:

hinit = 17.5 m

Explanation:

Assuming no friction present, the mechanical energy must be conserved, which means that at any point of the trajectory, the sum of the gravitational potential energy and the kinetic energy must keep the same.
At the top of the hill, since it starts from rest, all the energy must be potential, and we can express it as follows:

$E_{o} = U_{o} = m*g*h_{init} (1)$

When the car arrives to the top of the second hill, as we know that it is lower than the first one, the energy of the car, must be part gravitational potential energy, and part kinetic energy.
We can express this final energy as follows:

$E_{f} = U_{f} + K_{f} = m*g* h_{2} + \frac{1}{2} *m*v_{f} ^{2} (2)$

In order to find hinit, we need to make (1) equal to (2), and solve for it.
In (2) we have the value of h₂ (10 m), but we still need the value of the speed at the top of the second hill, vf.
Now, when the car is at the top of the hill, there are two forces acting on it, in opposite directions: the normal force (upward) and the weight (downward).
We know also that there is a force that keeps the car along the circular track, which is the centripetal force.
This force is just the net downward force acting on the car (it's vertical at the top), and is just the difference between the weight and the normal force.
If the cart just barely loses contact with the track at the top of the second hill, this means that at that point the normal force becomes zero.
So, the centripetal force must be equal to the weight.
The centripetal force can be expressed as follows:

$F_{c} = m*\frac{v_{f} ^{2}}{R} (3)$

We have just said that (3) must be equal to the weight:

$F_{c} = m*\frac{v_{f} ^{2}}{R} = m*g (4)$

Simplifying, and rearranging, we can solve for vf², as follows:

$v_{f}^{2} = R*g (5)$

Replacing (5) in (2), simplifying and rearranging in (1) and (2) we finally have:

$h_{init} = h_{2} + \frac{1}{2} R = 10m + 7.5 m = 17.5 m (6)$

7 0

3 years ago

Two 20kg spheres are placed with their

Maslowich

Answer:

F = 1.07 x 10⁻⁷ N

Explanation:

The gravitational force of attraction between two objects can be found by the use of Newton's Gravitational Law:

$F = \frac{Gm_{1}m_{2}}{r^2}\\\\$

where,

F = Gravitational Force of attraction = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = m₂ = mass of spheres = 20 kg

r = distance between the objects = 50 cm = 0.5 m

Therefore,

$F = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(20\ kg)(20\ kg)}{(0.5\ m)^2}\\\\$

5 0

3 years ago

A current is flowing in a wire in direction 3i + 4j where the direction of the magnetic field is 5j + 12k. The force on the wire

nasty-shy [4]

Answer:

F = 0.768 i ^ - 0.576 j ^ + 0.24 k ^

the correct answer is "b"

Explanation:

The magnetic force is

F = i l x B

The bold are vectors, in this case they give us the direction of the current and the magnetic field, for which we can solve as a determinant

$F = i \left[\begin{array}{ccc}x&y&z\\3&4&0\\0&5&12\end{array}\right]$

resolver

F = i ^ (4 12 - 0) + j ^ (0- 3 12) + k ^ (3 5 - 0)

F = i (48 i ^ - 36 j ^ + 15 k⁾

in this case i is the value of the current flowing through the cable

i = 16 mA = 0.016 A

F = 0.768 i ^ - 0.576 j ^ + 0.24 k ^

When reviewing the different answers, the correct answer is "b"

6 0

3 years ago

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for the force of static friction:

$f_s = \mu_s N$ --- (1)

where;

$f_s =$ static friction force

$\mu_s =$ coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

$N = ma_{min}$

From equation (1)

$f_s = \mu_s ma_{min}$

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

$F = f_s$