Answer:
44 N
Explanation:
The electrostatic forces between two charges is given by:
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is their separation
We notice that the force is directly proportional to the charges.
In this problem, initially we have a force of
F = 22 N
on a q2 = 4.0 C, exerted by a charge q1.
If the charge is doubled,
q2 = 8.0 C
This means that the force will also double, so it will be
Answer:
i know the questin but i got to try and find it
Explanation:
Answer:
m = 1.45 kg
Explanation:
For this exercise we look for size reduction in height
Reduction = y / y₀
Reduction = 2.15 / 6.75
Reduction = 0.3185
As the statue should not be deformed, all reduction has the same factor.
Let's use the concept of density
ρ = m / V
Initial statue
ρ = m₀ / V₀
It is reduced
V = x y z
V = 0.3185 x₀ 0.3185 y₀ 0.3185 z₀
V = 0.3185³ V₀
Density is
ρ = m / V
ρ = m / 0.3185³ V₀
As the density remains constant we can match them
m₀ / Vo = m / 0.3185³ V₀
m = 0.3185³ m₀
Let's calculate
m = 0.3185³ 45
m = 0.03231 45
m = 1.45 kg
Answer:
Explanation:
a )
momentum of baseball before collision
mass x velocity
= .145 x 30.5
= 4.4225 kg m /s
momentum of brick after collision
= 5.75 x 1.1
= 6.325 kg m/s
Applying conservation of momentum
4.4225 + 0 = .145 x v + 6.325 , v is velocity of baseball after collision.
v = - 13.12 m / s
b )
kinetic energy of baseball before collision = 1/2 mv²
= .5 x .145 x 30.5²
= 67.44 J
Total kinetic energy before collision = 67.44 J
c )
kinetic energy of baseball after collision = 1/2 x .145 x 13.12²
= 12.48 J .
kinetic energy of brick after collision
= .5 x 5.75 x 1.1²
= 3.48 J
Total kinetic energy after collision
= 15.96 J
s=600 m
t=12 s
s=0.5*a*t² (initial speed V0=0)
a=(2*s)/t²
a=(2*600)/12²
a≈8.33 m/s²
L= s(t2=12s)-s(t1=11s) -> (distance during the twelfth second)
L=0.5*a*(t2²-t1²)
L=0.5*((2*s)/t²)*(t2²-t1²)
L=0.5*((2*600)/12²)*(12²-11²)
L ≈ 95.83 m
