Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
Answer:
A (2066,6 N)
Explanation:
Use the Work formula
62.000J = F . 30
62.000/30 = 2066,6 N
The amout of time it took to move the rock doesn´t matter at all.
It is called a distraction variable, We don´t need it to solve the problem it is there just to confuse.
155Ω
Explanation:
R = R ref ( 1 + ∝ ( T - Tref)
where R = conduction resistance at temperature T
R ref = conductor resistance at reference temperature
∝ = temperature coefficient of resistance for conductor
T = conduction temperature in degrees Celsius
T ref = reference temperature that ∝ is specified at for the conductor material
T = 600 k - 273 k = 327 °C
Tref = 300 - 273 K = 27 °C
R = 50 Ω ( 1 + 0.007 ( 327 - 27) )
R = 155Ω
Part (a): Magnetic dipole moment
Magnetic dipole moment = IA, I = Current, A = Area of the loop
Then,
Magnetic dipole moment = 2.6*π*0.15^2 = 0.184 Am^2
Part (b): Torque acting on the loop
T = IAB SinФ, where B = Magnetic field, Ф = Angle
Then,
T = Magnetic dipole moment*B*SinФ = 0.184*12*Sin 41 = 1.447 Nm
Answer:
500cal
Explanation:
Given parameters:
Mass of water = 50g
Initial temperature = 22°C
Final temperature = 32°C
Specific heat of water = 1cal/g
Unknown:
Amount of heat absorbed by the water in calories = ?
Solution:
To solve this problem, we use the expression below:
H = m c Ф
H is the amount of heat absorbed
m is the mass
c is the specific heat capacity
Ф is the temperature change
H = 50 x 1 x (32 - 22) = 500cal