A softball has a (positive/negative) acceleration when it is thrown. A soft ball has a (positive/negative) acceleration when it
1 answer:
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a) The work done by the tree is
b) The amount of force applied is 2880 N
Explanation:
a)
According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:
where
W is the work done on the car
m is the mass of the car
u is its initial speed
v is its final speed
For the car in this problem, we have:
m = 2000 kg
u = 12.0 m/s
v = 0 (since the car comes to a stop, after the crash)
Therefore, the work done by the tree on the car is:
The work is negative because it is done in the direction opposite to the direction of motion of the car.
b)
The work done by the tree on the car can also be rewritten as
where
F is the force applied on the car
d is the displacement of the car during the collision
In this situation, we have:
is the work done
is the displacement of the car during the collision
Solving the equation for F, we find the force exerted by the tree on the car:
Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.
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Answer:
a) E =0, b) E = 1,129 10¹⁰ N / C , c) E = 3.33 10¹⁰ N / C
Explanation:
To solve this exercise we can use Gauss's law
Ф = ∫ E. dA = / ε₀
Where we must define a Gaussian surface that is this case is a sphere; the electric field lines are radial and parallel to the radii of the spheres, so the scalar product is reduced to the algebraic product.
E A = /ε₀
The area of a sphere is
A = 4π r²
E = q_{int} / 4πε₀ r²
k = 1 / 4πε₀
E = k q_{int} / r²
To find the charge inside the surface we can use the concept of density
ρ = q_{int} / V ’
q_{int} = ρ V ’
V ’= 4/3 π r’³
Where V ’is the volume of the sphere inside the Gaussian surface
Let's apply this expression to our problem
a) The electric field in center r = 0
Since there is no charge inside, the field must be zero
E = 0
b) for the radius of r = 6.0 cm
In this case the charge inside corresponds to the inner sphere
q_{int} = 5.0 4/3 π 0.06³
q_{int} = 4.52 10⁻³ C
E = 8.99 10⁹ 4.52 10⁻³ / 0.06²
E = 1,129 10¹⁰ N / C
c) The electric field for r = 12 cm = 0.12 m
In this case the two spheres have the charge inside the Gaussian surface, for which we must calculate the net charge.
The charge of the inner sphere is q₁ = - 4.52 10⁻³ C
The charge for the outermost sphere is
q₂ = ρ 4/3 π r₂³
q₂ = 8.0 4/3 π 0.12³
q₂ = 5.79 10⁻² C
The net charge is
q_{int} = q₁ + q₂
q_{int} = -4.52 10⁻³ + 5.79 10⁻²
q_{int} = 0.05338 C
The electric field is
E = 8.99 10⁹ 0.05338 / 0.12²
E = 3.33 10¹⁰ N / C
Answer:
Explanation:
The incident angle is so the reflected ray will also be
.
The normal divides the angle between the incident and reflected ray. So, the angle between them would be
So, the angle between the incident and reflected rays is .
