Answer:
0.0268 m
Explanation:
Draw a free body diagram of the block. There are three forces: weight force mg pulling down, buoyancy of the oil B₁ pushing up, and buoyancy of the water B₂ pushing up.
Sum of forces in the y direction:
∑F = ma
B₁ + B₂ − mg = 0
ρ₁V₁g + ρ₂V₂g − mg = 0
ρ₁V₁ + ρ₂V₂ = m
ρ₁V₁ + ρ₂V₂ = ρV
ρ₁Ah₁ + ρ₂Ah₂ = ρAh
ρ₁h₁ + ρ₂h₂ = ρh
(930 kg/m³)h₁ + (1000 kg/m³)h₂ = (968 kg/m³) (4.93 cm)
Since the block is fully submerged, h₁ + h₂ = 4.93 cm.
(930 kg/m³) (4.93 cm − h₂) + (1000 kg/m³)h₂ = (968 kg/m³) (4.93 cm)
h₂ = 2.68 cm
h₂ = 0.0268 m
Answer:
The resistance of the tungsten coil at 80 degrees Celsius is 15.12 ohm
Explanation:
The given parameters are;
The resistance of the tungsten coil at 15 degrees Celsius = 12 ohm
The temperature coefficient of resistance of tungsten = 0.004/°C
The resistance of the tungsten coil at 80 degrees Celsius is found using the following relation;
R₂ = R₁·[1 + α·(t₂ - t₁)]
Where;
R₁ = The resistance at the initial temperature = 12 ohm
R₂ = The resistance of tungsten at the final temperature
t₁ = The initial temperature = 15 degrees Celsius
t₂ = The final temperature = 80 degrees Celsius
α = temperature coefficient of resistance of tungsten = 0.004/°C
Therefore, we have;
R₂ = 12×[1 + 0.004×(80 - 15)] = 15.12 ohm
The resistance of the tungsten coil at 80 degrees Celsius = 15.12 ohm.
Answer:
Explanation:
We can try writing the equation of the horizontal component of the length of the minute hand in terms of distance and the angle, that depends of time in this particular case.
The x-component of the length of the minute hand is:
(1)
- d is the length of the minute hand (d=D/2)
- D is the diameter of the clock
- t is the time (min)
Now, using the angular kinematic equations we can express the angle in term of angular velocity and time. As we know, the minute hand moves with a constant angular velocity, so we can use this equation:
(2)
Also we know, that the minute hand moves 90 degrees or π/2 rad in 15 min, so using the definition of angular velocity, we have:
Now, let's put this value on (2)
Finally the length x(t) of the shadow of the minute hand as a function of time t, will be:
I hope it helps you!
This is a perfect problem to use positive and negative velocities.
Let's call the positive direction the direction that points UP.
Then the negative direction is the one that points down.
The acceleration of gravity on Earth is 9.8 m/s² down. That means
that any object with no forces on it except gravity gains 9.8 m/s of
speed downward every second.
OK. Ready to go to work:
The ball's initial velocity is +(up)20.0 m/s .
Every second after the toss, the ball has -(down)9.8 m/s more velocity.
The ball's velocity after 3 seconds is
+20 + 3(-9.8) =
20 - 29.4 = - 9.4 m/s .
That means it's velocity at that time is 9.4 m/s pointing down.
The answer to your question is going to be living will.
