(a) If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.
(b) if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.
The force on an object is determined by applying Newton's second law of motion;
F = ma
(a)
when the force is halved, F₂ = 0.5F₁,
mass is doubled, m₂ = 2m₁
Thus, If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.
(b)
when the force is doubled, F₂ = 2F₁,
mass is halved, m₂ = 0.5m₁
Thus, if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.
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Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k 
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]
Answer:
51.96 m/s^-1
Explanation:
a) see the attachment
b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be
v_ox=v_o*cosФ
=60*cos (30)
= 51.96 m/s^-1
Answer:
emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise
Explanation:
Given data
initial circumference = 165 cm
rate = 12.0 cm/s
magnitude = 0.500 T
tome = 9 sec
to find out
emf induced and direction
solution
we know emf in loop is - d∅/dt ........1
here ∅ = ( BAcosθ)
so we say angle is zero degree and magnetic filed is uniform here so that
emf = - d ( BAcos0) /dt
emf = - B dA /dt ..............2
so area will be
dA/dt = d(πr²) / dt
dA/dt = 2πr dr/dt
we know 2πr = c,
r = c/2π = 165 / 2π
r = 26.27 cm
c is circumference so from equation 2
emf = - B 2πr dr/dt ................3
and
here we find rate of change of radius that is
dr/dt = 12/2π = 1.91 
cm/s
so when 9.0s have passed that radius of coil = 26.27 - 191 (9)
radius = 9.08 cm
so now from equation 3 we find emf
emf = - (0.500 ) 2π(9.08 ) 1.91
emf = - 0.005445
and magnitude of emf = 0.005445 V
so
emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise
Answer:
Bank angle = 35.34o
Explanation:
Since the road is frictionless,
Tan (bank angle) = V^2/r*g
Where V = speed of the racing car in m/s, r = radius of the arc in metres and g = acceleration due to gravity in m/s^2
Tan ( bank angle) = 40^2/(230*9.81)
Tan (bank angle) = 0.7091
Bank angle = tan inverse (0.7091)
Bank angle = 35.34o
