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blondinia [14]
3 years ago
12

A roller-coaster car rolls down a frictionless track, reaching speed v0 at the bottom. If you want the car to go twice as fast a

t the bottom, by what factor must you increase the height of the track
Physics
1 answer:
NeTakaya3 years ago
3 0

Answer:

h should become four times.

Explanation:

The speed at the bottom depends only on the vertical height difference from the top ... if there is no friction ... then the path you take from top to bottom doesn't matter in any way, only the difference in height between the two

mgh at top = 1/2 mv^2 at bottom, so that v=sqrt[2gh]

You see, the final velocity, v, depends on the square root of h, so double v, quadruple h, or four times higher

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Define "increased productivity" in terms of the number of tasks and the amount of time.
vovikov84 [41]

Answer:

The greater the amount of output for a given unit of input, the higher the overall productivity. Businesses generally aim to improve productivity over time to maintain competitiveness and increase the business's profitability. Individuals are familiar with the idea of productivity in their own lives.

6 0
3 years ago
Our favorite program runs in 10 seconds on computer A, which has a 4 GHz. clock. We are trying to help a computer designer build
Alona [7]

Answer:

The rate clock is about

F = 8 GHz

Explanation:

f₁ = 4 G Hz , t₁ = 10 s , t₂ = 6s , f₂ = 1.2 f₁

Can organize to find the rate clock the designer build to the target so

X / 4 Ghz = 10 s , 1.2 X /  Y = 6 s

X * Y = 10 s  ⇒ F = 10 s

1.2 * 4 G Hz = 6 s

F = 10 * ( 1.2 * 4 G Hz ) / 6

F = 10 * ( 1.2 * 4 x 10 ⁹ Hz )  /  6

F = 8 x 10 ⁹ Hz

F = 8 GHz

3 0
3 years ago
Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm an
bonufazy [111]

Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

          \frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}} ......... (1)

Hence, total charge will be as follows.

              q_{1} + q_{2} = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

and,    q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

                     = \frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}

                     = 82.714 nC

Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.

                       

5 0
3 years ago
The position of a particular particle as a function of is given by r=(8.5t i+5.6j-2tk )m (a) Determine the particle’s velocity a
Salsk061 [2.6K]

Answer:

use google to find answer

Explanation:

69 69 69 69 69 69

6 0
4 years ago
A5 kg frisbee is thrown from rest to a final speed of 12 m/s. What is the impulse of the frisbee?
lesya692 [45]

Answer:

60kgm/s

Explanation:

Given parameters:

Mass of frisbee  = 5kg

Final speed  = 12m/s

Unknown:

Impulse of the frisbee  = ?

Solution:

The impulse of the frisbee is the same as the change in momentum.

It is given as:

 Impulse  = mass (Final velocity  -  Initial velocity)

 Impulse  = 5(12  - 0)  = 60kgm/s

3 0
3 years ago
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