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DochEvi [55]
3 years ago
15

Where would you except to find the center of gravity of a plastic ruler

Physics
1 answer:
VARVARA [1.3K]3 years ago
3 0
You would expect to find the center of gravity in a ruler in the middle because if you were to cut a ruler in half, depending on the center of gravity, you would have two equal pieces, which mean there is equal weight, meaning the middle is the center of gravity
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What happens if you add additional,solid NaCl after the maximum has been reached?
azamat
<span>it would bond to the phosphate 

</span>
3 0
3 years ago
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Two cars are raised to the same elevation on service- station lifts. If one car is twice as massive as the other, how do their p
Brums [2.3K]

Answer:

The potential energy of the more massive one is twice that of the other.

Explanation:

Potential energy is given by

<em>PE</em> = <em>mgh</em>

where <em>m</em> = mass of body, <em>g</em> = acceleration of gravity and <em>h</em> = height or elevation.

For the less massive car, let the mass be m_1. Then its <em>PE</em> is

PE_1 = m_1gh

For the massive car, let the mass be m_2.  Its <em>PE</em> is

PE_2 = m_2gh

But m_2 =2m_1

\therefore PE_2 = 2m_1gh = 2(m_1gh) = 2PE_1

Hence, the potential energy of the more massive one is twice that of the other.

7 0
3 years ago
Which situation is an example of increasing potential energy? Question 4 options: A. a cat jumping from a tree B. pulling a wago
jeka94
Pulling an wagon uphill I believe.
4 0
3 years ago
Read 2 more answers
You just calibrated a constant volume gas thermometer. The pressure of the gas inside the thermometer is 294.0 kPa when the ther
Travka [436]

Answer: 361° C

Explanation:

Given

Initial pressure of the gas, P1 = 294 kPa

Final pressure of the gas, P2 = 500 kPa

Initial temperature of the gas, T1 = 100° C = 100 + 273 K = 373 K

Final temperature of the gas, T2 = ?

Let us assume that the gas is an ideal gas, then we use the equation below to solve

T2/T1 = P2/P1

T2 = T1 * (P2/P1)

T2 = (100 + 273) * (500 / 294)

T2 = 373 * (500 / 294)

T2 = 373 * 1.7

T2 = 634 K

T2 = 634 K - 273 K = 361° C

5 0
3 years ago
On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const
IceJOKER [234]

Answer:

a)

Explanation:

  • Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

        x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}   (1)

  • Since the car starts from rest, v₀ =0.
  • We know the value of t = 5 sec., but we need to find the value of a.
  • Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

       a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2  (2)

  • Replacing a and t in (1):

       x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}  = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m.  (3)

  • Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
  • Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

       a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2  (4)

  • Replacing v₀, at and t in (1), we have:

       x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m   (5)

  • Therefore, as the truck travels twice as far as the car, the right answer is a).
7 0
2 years ago
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