Answer:
Explanation:
F = ma. For us, this looks like
60 = 30a and
a = 2 m/s/s
If the force goes up to, say, 90, then
90 = 30a and
a = 3...if the force goes up, the acceleration also goes up.
If the mass goes up to say, 60, and the force stays the same, then
60 = 60a and
a = 1...if the mass goes up, the acceleration goes down.
Answer:
acceleration 8 km/h/s south
Explanation:
First of all, let's remind that a vector quantity is a quantity which has both a magnitude and a direction.
Based on this definition, we can already rule out the following two choices:
distance: 40 km
speed: 40 km/h
Since they only have magnitude, they are not vectors.
Then, the following option:
velocity: 5 km/h north
is wrong, because the car is moving south, not north.
So, the correct choice is
acceleration 8 km/h/s south
In fact, the acceleration can be calculated as

where
v = 40 km/h is the final velocity
u = 0 is the initial velocity
t = 5 s is the time
Substituting,

And since the sign is positive, the direction is the same as the velocity (south).
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Answer:
F = 351×10³lb
Explanation:
Given the density
ρg = 64.6lb/ft³
Diameter d = 12ft
The tank is horizontally cylindrical. The vertical distance from the top to the bottom of the tank is h = 12ft
The pressure in the tank is
P = ρgh = 64.6 × 12 = 775.2lb/ft²
The force exerted on one end of the tank is therefore F = PA = 775.2 × πd² = 775.2π×12²
F = 351×10³lb.
Answer:
a) v = √(v₀² + 2g h), b) Δt = 2 v₀ / g
Explanation:
For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.
The velocity of each ball is
ball 1. thrown upwards vo is positive
v² = v₀² - 2 g (y-y₀)
in this case the height y is zero and the height i = h
v = √(v₀² + 2g h)
ball 2 thrown down, in this case vo is negative
v = √(v₀² + 2g h)
The times to get to the ground
ball 1
v = v₀ - g t₁
t₁ =
ball 2
v = -v₀ - g t₂
t₂ = - \frac{v_{o} + v }{ g}
From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is
Δt = t₂ -t₁
Δt =
Δt = 2 v₀ / g