1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
olga2289 [7]
2 years ago
5

Speed of 26.7 m/s in 3.06 s. How far had the car traveled by the time the final speed was achieved?

Physics
1 answer:
Alik [6]2 years ago
6 0

Answer:

Below

Explanation:

You can use this equation to find the distance :

     distance = velocity x time

     distance = (26.7)(3.06)

                    = 81.702 m

Rounding to 3 sig figs

     = 81.7 m

Hope this helps

You might be interested in
True or False: The basketball should be dribbled below the waist.
zimovet [89]
True if you have proper stance and use your body the right way then the ball will be below your waist to allow for more control.
7 0
2 years ago
Calculate the kinetic energy in joules of a ball of mass 40g moving at a velocity of 4 metres per second​
AfilCa [17]
Given : A ball of mass 40 g moving at a velocity of 4 m/s.
To find : Calculate the kinetic energy in joules ?
Solution :
The kinetic energy formula is given by,

where, v is the velocity v=4 m/s
m is the mass m=40 g
Convert g into kg,



Substitute the values,



Therefore, the kinetic energy is 0.32 Joules.
6 0
2 years ago
Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin
tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of m_2 would be four times that of m_1.

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is T, then the angular velocity of the SHM would be

\displaystyle \omega = \frac{2\pi}{T}.

Assume that the mass starts with a zero displacement and a positive velocity. If A represent the amplitude of the SHM, then the displacement of the mass at time t would be:

\mathbf{x}(t) = A\sin(\omega\cdot t).

The velocity of the mass at time t would be:

\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t).

The acceleration of the mass at time t would be:

\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t).

Let m represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time t would be:

\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t),

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant k will be equal to:

\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}.

Since \displaystyle \omega = \frac{2\pi}{T}, it can be concluded that:

\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}.

For the first mass m_1, if the time period is T_1, then the spring constant would be:

\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2.

Similarly, for the second mass m_2, if the time period is T_2, then the spring constant would be:

\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Since the two springs are the same, the two spring constants should be equal to each other. That is:

\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Simplify to obtain:

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

6 0
3 years ago
In a very large closed tank, the absolute pressure of the air above the water is 6.46 x 105 Pa. The water leaves the bottom of t
a_sh-v [17]

Answer:

a) 35.94 ms⁻²

b) 65.85 m

Explanation:

Take down the data:

ρ = 1000kg/m3

a) First, we need to establish the total pressure of the water in the tank. Note the that the tanks is closed. It means that the total pressure, Ptot,  at the bottom of the tank is the sum of the pressure of the water plus the air trapped between the tank rook and water. In other words:

Ptot = Pgas + Pwater

However, the air is the one influencing the water to move, so elimininating Pwater the equation becomes:

Ptot  = Pgas

        = 6.46 × 10⁵ Pa

The change in pressure is given by the continuity equation:

ΔP = 1/2ρv²

where v is the velocity of the water as it exits the tank.

Calculating:

6.46 × 10⁵  =1/2 ×1000×v²

solving for v, we get v = 35.94 ms⁻²

b) The Bernoulli's equation will be applicable here.

The water is coming out with the same pressure, therefore, the equation will be:

ΔP = ρgh

6.46 × 10⁵  = 1000 x 9.81 x h

h = 65.85 meters

7 0
3 years ago
Someone stuck a piece of chewing gum on the outer edge of a blade on a ceiling fan. The radius (r) of where the gum is from the
maksim [4K]

Answer: why would u stick a gum there

Explanation:

3 0
2 years ago
Other questions:
  • An electron is moving at 7.4 x 10^5 m/s perpendicular to a magnetic field. It experiences a force of -2.0 x 10^-13 N. What is th
    12·1 answer
  • The weights in atwoods machine, starting at rest, attain a velocity of 2ft/sec in one sec. Find the ratio of the masses
    6·1 answer
  • A swimming pool heater has to be able to raise the temperature of the 40 000 gallons of water in the pool by 10.0 C°.
    6·1 answer
  • What has research determined about the orbit of an electron around a nucleus?
    9·2 answers
  • A 1000.kg car moving at north at 100. km/hr brakes to a stop in 50. m within a uniform acceleration. what are magnitude and dire
    9·1 answer
  • PLEASE HELP ASAP!!!!!!!!!!WILL GIVE BRAINLIEST...
    14·2 answers
  • Which of the following elements is in Group 2?
    8·1 answer
  • A rocket is fired vertically upwards starting frkm rest. It accelerates at 30m/s for 4secs. At the end of 4secs it runs out of f
    13·1 answer
  • Walk in a straight line. Now stop. Did you accelerate? Explain
    9·2 answers
  • A house is advertised as having 1580 square feet under roof. What is the area of this house in square meters?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!