Answer:
The acceleration due to gravity of that planet is, gₐ = 1.25 m/s²
Explanation:
Given that,
Mass of the planet, m = 1/2M
Radius of the planet, r = 2R
Where M and R is the mass and radius of the Earth respectively.
The acceleration due to gravity of Earth, g = 10 m/s²
The acceleration due to gravity of Earth is given by the relation,
g = GM/R²
Similarly, the acceleration due to gravity of that planet is
gₐ = Gm/r²
where G is the Universal gravitational constant
On substituting the values in the above equation
gₐ = G (1/2 M)/4 R²
= GM/8R²
= 1/8 ( 10 m/s²)
= 1.25 m/s²
Hence, the acceleration due to gravity of that planet is, gₐ = 1.25 m/s²
Answer:
Explanation:
a ) Let the distance required in former case be d₁ .
initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 7.00 m /s²
v² = u² - 2 a s
0 = 30² - 2 x 7 x d₁
d₁ = 64.28 m
b) initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 5.00 m /s²
v² = u² - 2 a s
0 = 30² - 2 x 5 x d₂
d₂ = 90 m
c)
t = .5 s
s₁ = ut - .5 at²
= 30 x .5 - .5 x 7 x .5²
= 15 - .875
= 14.125 m
t = .5 s
s₂ = ut - .5 at²
= 30 x .5 - .5 x 5 x .5²
= 15 - .625
= 14.375 m
Answer:
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(89000/102000)×100
=87.25%
(92000/104000)×100
=88.46%
efficiency is (output/input)×100
if u get confused which way input and output should go, remember the smaller value is always output and it's above in the fraction, then only it's possible to get a efficiency lower than 100.
Answer:
71.4583 Hz
67.9064 N
Explanation:
L = Length of tube = 1.2 m
l = Length of wire = 0.35 m
m = Mass of wire = 9.5 g
v = Speed of sound in air = 343 m/s
The fundamental frequency of the tube (closed at one end) is given by
The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz
The linear density of the wire is
The fundamental frequency of the wire is given by
The tension in the wire is 67.9064 N
