Answer:
a
b
Explanation:
From the question we are told that
The mass of the rock is 
The length of the small object from the rock is 
The length of the small object from the branch 
An image representing this lever set-up is shown on the first uploaded image
Here the small object acts as a fulcrum
The force exerted by the weight of the rock is mathematically evaluated as
substituting values
So at equilibrium the sum of the moment about the fulcrum is mathematically represented as
Here 
is very small so 
and 
Hence
=> 
substituting values
The mechanical advantage is mathematically evaluated as
substituting values
The helium may be treated as an ideal gas, so that
(p*V)/T =constant
where
p = pressure
V = volume
T = temperature.
Note that
7.5006 x 10⁻³ mm Hg = 1 Pa
1 L = 10⁻³ m³
Given:
At ground level,
p₁ = 752 mm Hg
= (752 mm Hg)/(7.5006 x 10⁻³ mm Hg/Pa)
= 1.0026 x 10⁵ Pa
V₁ = 9.47 x 10⁴ L = (9.47 x 10⁴ L)*(10⁻³ m³/L)
= 94.7 m³
T₁ = 27.8 °C = 27.8 + 273 K
= 300.8 K
At 36 km height,
P₂ = 73 mm Hg = 73/7.5006 x 10⁻³ Pa
= 9.7326 x 10³ Pa
T₂ = 235 K
If the volume at 36 km height is V₂, then
V₂ = (T₂/p₂)*(p₁/T₁)*V₁
= (235/9.7326 x 10³)*(1.0026 x 10⁵/300.8)*94.7
= 762.15 m³
Answer: 762.2 m³
<span>The entire time the ball is in the air, its acceleration is 9.8 m/s2 down provided this occurs on the surface of the Earth. Note that the acceleration can be either 9.8 m/s2 or -9.8 m/s2.
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</span>
Answer:
0.4rad/s²
Explanation:
Angular acceleration is the time rate of change of angular velocity . In SI units, it is measured in radians per second squared (rad/s²)
w1 = 4rad/s, w2 =2rad/s, t = 5sec, r = 0.30m
a = ∆w/t
a = (w2 - w1)/t
a = (2 - 4)/5 = -2/5 =
a = - 0.4rad/s²
The -ve sign indicates a deceleration in the motion
Good luck
Answer:
1.96 m/s^2
Explanation:
angle of inclination, θ = 30.6°
coefficient of friction, μ = 0.36
Acceleration of the crate
a = g Sin θ - μ g Cos θ
a = 9.8 ( Sin 30.6 - 0.36 x Cos 30.6)
a = 9.8 ( 0.509 - 0.309)
a = 1.96 m/s^2
Thus, the acceleration f the crate is 1.96 m/s^2.
