Answer:
The K.E of the bowling ball right before it hits the ground, K.E = 2450 J
Explanation:
Given data,
The mass of the bowling ball, m = 10 kg
The height of the building, h = 25 m
The total mechanical energy of the body is given by,
E = P.E + K.E
At height 'h' the P.E is maximum and the K.E is zero,
According to the law of conservation of energy, the K.E at the ground before hitting the ground is equal to the P.E at 'h'
Therefore, P.E at 'h'
P.E = mgh
= 10 x 9.8 x 25
= 2450 J
Hence, the K.E of the bowling ball right before it hits the ground, K.E = 2450 J
Answer:
10 A
Explanation:
τ = Maximum torque of the loop = 9 mN
N = Number of turns in the loop = 50
a = side of the loop = 15 cm = 0.15 m
A = Area of the loop = a² = 0.15² = 0.0225 m²
B = magnitude of magnetic field = 0.800 T
i = magnitude of current in the loop
Maximum torque of the loop is given as
τ = N B i A
Inserting the values
9 = (50) (0.8) (0.0225) i
i = 10 A
Given Information:
Wavelength of the red laser = λr = 632.8 nm
Distance between bright fringes due to red laser = yr = 5 mm
Distance between bright fringes due to laser pointer = yp = 5.14 mm
Required Information:
Wavelength of the laser pointer = λp = ?
Answer:
Wavelength of the laser pointer = λp = ?
Explanation:
The wavelength of the monochromatic light can be found using young's double slits formula,
y = Dλ/d
y/λ = D/d
Where
λ is the wavelength
y is the distance between bright fringes.
d is the double slit separation distance
D is the distance from the slits to the screen
For the red laser,
yr/λr = D/d
For the laser pointer,
yp/λp = D/d
Equating both equations yields,
yr/λr = yp/λp
Re-arrange for λp
λp = yp*λr/yr
λp = (5*632.8)/5.14
λp = 615.56 nm
Therefore, the wavelength of the small laser pointer is 615.56 nm.
Answer:
(a)10.5 rad/s2
(b) 20.9 rev
(c) 47.27 m
Explanation:
As the block of mass 53 kg is falling and pulling on the rope. The tension force on the rope must be equal to the gravity acting on the block according to Newton's 3rd law
T = mg = 53*9.81 = 519.93 N
Since this tension force would rotate the cylinder freely without any friction. The torque created by this tension force is
To = TR = 519.93 * 0.36 = 187.17 Nm
This solid cylinder would have a moment of inertia around it's rotating axis of:
(a)We can use Newton's 2nd law to calculate the angular acceleration exerted by such torque on the solid cylinder
(b) With such constant angular acceleration, the angle it would make after 5s is
Since each revolution equals to 
of angle, we can calculate the number of revolution it makes
(c) Assume the thickness of the rope is negligible (and its wounded radius is unchanging), we can calculate the rope length unwinded after rotating 131.3rad
Answer:
A model rocket is launched with an initial upward velocity of 215 ft/s.
Explanation:
