Answer:
u = 10.63 m/s
h = 1.10 m
Explanation:
For Take-off speed ..
by using the standard range equation we have
R = 9.1 m
θ = 26º,
Initial velocity = u
solving for u
u = 10.63 m/s
for Max height
using the standard h(max) equation ..
h = 1.10 m
Answer:
a)-1.014x 
J
b)3.296 x J
Explanation:
For Sphere A:
mass 'Ma'= 47kg
xa= 0
For sphere B:
mass 'Mb'= 110kg
xb=3.4m
a)the gravitational potential energy is given by
= -GMaMb/ d
= - 6.67 x 
x 47 x 110/ 3.4 => -1.014x J
b) at d= 0.8m (3.4-2.6) and =-1.014x J
The sum of potential and kinetic energies must be conserved as the energy is conserved.
+ = 
+ 
As sphere starts from rest and sphere A is fixed at its place, therefore is zero
= +
The final potential energy is
= - GMaMb/d
Solving for ' '
= + GMaMb/d => -1.014x + 6.67 x x 47 x 110/ 0.8
= 3.296 x J
The cube has 6 equal, square, foil faces. The mass of foil for each face is (380/6) milligrams.
The surface area of each piece is (380)/(6•11) cm^2.
The length of each side of the piece is √(380/6•11) cm
That's about 2.4 cm .
It's a cute little foil cube, just under 1-inch each way.
Answer:
1,3and4 just did and got correct
Explanation:
Answer:
1.19 m/s²
Explanation:
The frequency of the wave generated in the string in the first experiment is f = n/2l√T/μ were T = tension in string = mg were m = 1.30 kg weight = 1300 g , μ = mass per unit length of string = 1.01 g/m. l = length of string to pulley = l₀/2 were l₀ = lent of string. Since f is the second harmonic, n = 2, so
f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀ (1)
Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²
Equating (1) and (2) we ave
2(√mg/μ)/f = T²g/4π²
Making g subject of the formula
g = 2π√(2√(m/μ)/f)/T
The period T = 316 s/100 = 3.16 s
Substituting the other values into , we have
g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16
g = 2π√(2 × 35.877/200 Hz)/3.16
g = 2π√(71.753/200 Hz)/3.16
g = 2π√(0.358)/3.16
g = 2π × 0.599/3.16
g = 1.19 m/s²
