The distance an object falls, from rest, in gravity is
D = (1/2) (G) (T²)
'T' is the number seconds it falls.
In this problem,
0.92 meter = (1/2) (9.8) (T²)
Divide each side by 4.9 : 0.92 / 4.9 = T²
Take the square root
of each side: √(0.92/4.9) = T
0.433 sec = T
The horizontal speed doesn't make a bit of difference in
how long it takes to reach the floor. BUT ... if you want to
know how far from the table the pencil lands, you can find
that with the horizontal speed.
The pencil is in the air for 0.433 second.
In that time, it travels
(0.433s) x (1.4 m/s) = 0.606 meter
from the edge of the table.
It must be noted that during an elastic collision both the momentum and kinetic energy are conserved. For the kinetic energy, it can be solved through the equation,
KE = 0.5mv²
Equating the kinetic energies before and after collision,
0.5(60)(8 m/s)² = (0.5)(60)(x²) + (0.5)(0.45)(35 m/s)²
The value of x from the equation is approximately 7.40 m/s
Answer:
The time it will take for the object to hit the ground will be 4.
Explanation:
You have:
h(t)=−16t²+v0*t+h0
Being v0 the initial velocity (54 ft/s) and h0 the initial height (40 ft) and replacing you get:
h(t)=−16t²+54*t+40
To know how long it will take for the object to touch the ground, the height h(t) must be zero. So:
0=−16t²+54*t+40
Being a quadratic function or parabola: f (x) = a*x² + b*x + c, the roots or zeros of the quadratic function are those values of x for which the expression is 0. Graphically, the roots correspond to the points where the parabola intersects the x axis. To calculate the roots the expression is used:
In this case you have that:
Replacing in the expression of the calculation of roots you get:
Expresion (A)
and
Expresion (B)
Solving the Expresion (A):
Solving the Expresion (B):
These results indicate the time it will take for the object to hit the ground can be -5/8 and 4. Since the time cannot be negative, then <u><em>the time it will take for the object to hit the ground will be 4.</em></u>
I’m pretty sure that it’s D
Answer:
Explanation:
Force of attraction between two charges is 36N
Then from coulombs electrostatic law.
The force of attraction or repulsion between two charges (q1 and q2) is directly proportional to the product of the force and inversely proportional to the square of the distance apart.
F = kq1•q2 / r².
kq1•q2 / r² = 36
Now, we are told that the second charges has increased by a factor of 3
Then,
q2' = 3q2
So, since the first charge is contact, the radius is constant and k is constant,
Then,
F is directly proportional to the second charge, so if the second charge is triple then the force is tripled
Then,
F' = 3F = 3 × 36 = 108 units
Unit of force is newton
F' = 108N
Also,
F' = kq1•q2' / r²
F' = kq1•3q2 / r².
F' = 3 × kq1•q2 / r²
Since, kq1•q2 / r² = 36
Then,
F' = 3 × 36
F' = 108 units
F' = 108N