Answer:
The answer is 13 however make sure if they ask for a certain measurement like meter answer it by saying 13 meters.
Explanation:
This basically turns into basic algebra if you know the formula for work. The formula for work is W=F*d
Here are the variables that you know 650J=50N*d so you need d.
All you do is divide 650J by 50N and you get a total of 13 (meters since I don't know what they want you to put it in).
The freezing point is the same as the melting point.
If it freezes at -58°C, hence the melting point is also <span>-58°C.</span>
Answer:
98 m √
Explanation:
How about s = Vo * t + ½at² ?
s = h = Vo * 2s - 4.9m/s² * (2s)² = 2Vo - 19.6
and
h = Vo * 10s - 4.9m/s² * (10s)² = 10Vo - 490
Subtract 2nd from first:
0 = -8Vo + 470.4
Vo = 58.8 m/s
h = 58.8m/s * 2s - 4.9m/s² * (2s)² = 98 m
Ok, so adopting that the 2nd satellite is at rest and that we're not moving anywhere near the speed of light (so no special relativity considerations), we can just add the two speed together, and say the 1st satellite is moving at 0.9m/s at the 2nd satellite. We can then set up our conservation of momentum equation, m₁v₁+m₂v₂ = m₁v₃+m₂v₄, where I'm calling v 1 and 2 the initial velocities of satellite 1 and 2 and v 3 and 4 the final velocities of satellite 1 and 2 respectively. We know, based on our chosen frame, that v₂ = 0, so that falls out to leave m₁v₁ = m₁v₃+m₂v₄, but we don't know v₃ or v₄, so we need another equation. Let's set up conversation of energy (elastic collisions conserve energy), where we only have to worry about kinetic energy (K = 1/2mv²) for each satellite before and after the collision. So we get 1/2m₁v₁²+1/2m₂v₂² = 1/2m₁v₃²+1/2m₂v₄². Now we have 2 equations and two unknown variables so let's solve with substitution. Let's solve the momentum equation for v₃, v₃ = (m₁v₁ - m₂v₄)/m₁, sub that into the energy equation, cancel the 1/2's and let's drop the v₂ terms since it's zero and we get: m₁v₁² = m₁((m₁v₁ - m₂v₄)/m₁)²+m₂v₄², then after some algebra we get v₄ = sqrt(m₁v₁/((v₁ - m₂/m₁)²+m₂)), then we plug in numbers v₄ = sqrt((4.5*10³*0.9/((0.9-(7.5/4.5))²+7.5*10³) = 0.73 m/s for the 2nd satellite after the collision. Then go back to v₃ = (m₁v₁ - m₂v₄)/m₁ and plug in numbers now that we know v₄ and we get v₃ = (4.5*10³*0.9 - 7.5*10³*0.73)/(4.5*10³) = -0.3167 m/s for the 1st satellite.
Answer:
3 kg
Explanation:
Momentum before = momentum after
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
After jumping on the skateboard, the student and the skateboard have the same velocity, so v₁ = v₂.
m₁u₁ + m₂u₂ = (m₁ + m₂) v
(47.4 kg) (4.2 m/s) + m (0 m/s) = (47.4 kg + m) (3.95 m/s)
m = 3 kg
