Answer:
F = 5226.6 N
Explanation:
To solve a lever, the rotational equilibrium relation must be used.
We place the reference system on the fulcrum (pivot point) and assume that the positive direction is counterclockwise
F d₁ = W d₂
where F is the applied force, W is the weight to be lifted, d₁ and d₂ are the distances from the fulcrum.
In this case the length of the lever is L = 5m, t the distance desired by the fulcrum from the weight to be lifted is
d₂ = 200 cm = 2 m
therefore the distance to the applied force is
d₁ = L -d₂
d₁ = 5 -2
d₁= 3m
we clear from the equation
F = W d₂ / d₁
W = m g
F = m g d₂ / d₁
we calculate
F = 800 9.8 2/3
F = 5226.6 N
1h----------------> 70x3=210 bacteria
2h-----------------> 210*3=630 bactaeria
let be y the number of bacteria at the t=0h
it is y=70 3^0
for t= 1h
y=70*3^1=210
for t=2h
y=70*3^2=630
so we can write y=70*3^x, where x is the number of hour
Answer:
ΔP.E = 6.48 x 10⁸ J
Explanation:
First we need to calculate the acceleration due to gravity on the surface of moon:
g = GM/R²
where,
g = acceleration due to gravity on the surface of moon = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of moon = 7.36 x 10²² kg
R = Radius of Moon = 1740 km = 1.74 x 10⁶ m
Therefore,
g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²
g = 2.82 m/s²
now the change in gravitational potential energy of rocket is calculated by:
ΔP.E = mgΔh
where,
ΔP.E = Change in Gravitational Potential Energy = ?
m = mass of rocket = 1090 kg
Δh = altitude = 211 km = 2.11 x 10⁵ m
Therefore,
ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)
<u>ΔP.E = 6.48 x 10⁸ J</u>
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Answer:
Part a)

Part b)

Since the distance of other building is 15 m so YES it can make it to other building
Part c)

direction of velocity is given as
![[tex]\theta = 26.35 degree](https://tex.z-dn.net/?f=%5Btex%5D%5Ctheta%20%3D%2026.35%20degree)
Explanation:
Part a)
acceleration due to gravity on this planet is 3/4 times the gravity on earth
So the acceleration due to gravity on this new planet is given as


now the vertical displacement covered by the canister is given as

now by kinematics we have



Part b)
Horizontal speed of the canister is given as

now the distance moved by it



Since the distance of other building is 15 m so YES it can make it to other building
Part c)
Final velocity in X direction will remains the same

final velocity in Y direction



now magnitude of velocity is given as



direction of velocity is given as


![[tex]\theta = 26.35 degree](https://tex.z-dn.net/?f=%5Btex%5D%5Ctheta%20%3D%2026.35%20degree)