Answer:
the density of the wooden cube is 204.1 kg/m³
Explanation:
Given;
applied force, F = 54 N
length of each side of the solid wooden cube, L = 30 cm = 0.3 m
mass of the wooden cube is calculated as;
F = mg
m = F/g
m = 54/9.8
m = 5.51 kg
The volume of the wooden cube is calculated as;
V = L³
V = (0.3)³
V = 0.027 m³
The density of the wooden cube is calculated as;
ρ = m/V
ρ = (5.51 kg) / (0.027 m³)
ρ = 204.1 kg/m³
Therefore, the density of the wooden cube is 204.1 kg/m³
Answer:

Explanation:
Given data
Length L=2.9 cm=0.029m
Speed of sound v=335 m/s
to find
Fundamental frequency f
Solution
As we know that frequency is given as:
f=v/λ

Answer:
Incomplete question
The complete question is
A Ferris wheel is a vertical, circular amusement ride with radius 6.0 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 9.6 s. Consider a rider whose mass is 96 kg.
At the bottom of the ride, what is the rate of change of the rider's momentum?
Explanation:
Radius of wheel is 6m
Rider mass=96kg
He completes one revolution in 9.6s
Let get angular velocity (w)
1 Revolution =2πrad
θ=2πrad
w= θ/t
w=2π/9.6
w=0.654rad/s
Linear speed is give as
v=wr
v=0.654×6
v=3.93m/s
Centripetal acceleration a
a=rw²
a=6×0.654²
a=2.57m/s²
Acceleration due to gravity g=9.81m/s²
According to Newton's second law of motion net force acting on the rider at the bottom of the ride is given by: the two force acting at the bottom is the normal and the weight of the rider
ΣF = ma
N-W=ma
N-mg=ma
N=ma+mg
N=m(a+g)
N=96(2.57+9.81)
N=1188.48 N
Therefore the rate of change of momentum at the bottom of the ride is 1188.48 N.
Answer:
is correct
Explanation:
in my think, first this due to ray emitted from the light those ray may be affect our skin or party of body.
Answer:
B)
The magnitude of induced emf in the conducting loop is 0.99 mV.
Explanation:
Rate of increase in magnetic field per unit time = 0.090 T/s
Area of the conducting loop = 110 cm^2 = 0.0110 m^2
Electromagnetic induction is the production of an emf or voltage in a coil of wire due to a changing magnetic field through the coil.
Induced e.m.f is given as:
EMF = (-N*change in magnetic field/time)*Area
EMF = rate of change of magnetic field per unit time * Area
EMF = 0.090 * 0.0110
EMF = 0.00099 V
EMF = 0.99 mV