Answer:
2.35 m/s²
Explanation:
Given that
Mass of the smaller crate, m₁ = 21 kg
Mass of the larger crate, m₂ = 90 kg
Tensión of the rope, T = 261 N
We know that the sum of all forces for the two objects with a force of friction F and a tension T are:
(i) m₁a₁ = F
(ii) m₂a₂ = T - F, where m and a are the masses and accelerations respectively.
1) no sliding can also mean that:
a₁ = a₂ = a
This makes us merge the two equations written above together as:
m₂a = T - m₁a
If we then solve for a, we would have something like this
a = T / (m₁+m₂)
a = 261 / (21 + 90)
a = 261 / 111
a = 2.35 m/s²
Therefore, the needed acceleration of the small crate is 2.35 m/s²
Linear momentum (mass x speed) has to be conserved.
-- Momentum before the jump:
(boy's mass) x (boy's speed) = (25 kg) x (4.0 m/s) = 100 kg-m/s
(cart's mass) x (cart's speed) = (15 kg) x (zero) = zero
Total momentum before the jump: (100 kg-m/s) + (zero) = (100 kg-m/s)
-- Momentum after the jump:
(mass of boy+cart) x (speed of boy+cart) = (40 kg) x (speed)
-- Momentum after the jump = momentum before the jump
(40 kg) x (speed) = 100 kg-m/s
Divide each side by 40 kg:
Speed = (100 kg-m/s) / (40 kg)
<em>Speed = 2.5 m/s</em> (d)
Answer:
i hope this will help you :)
Explanation:
mass=19kg
density=800kg/m³
volume=?
as we know that
density=mass/volume
density×volume=mass
volume=mass/density
putting the values
volume=19kg/800kg/m³
so volume=0.02375≈0.02m³
Answer:
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