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Elis [28]
3 years ago
14

Power equals work

Physics
2 answers:
iVinArrow [24]3 years ago
6 0
The answer is a. divided by time
yan [13]3 years ago
5 0

Easy the answer to your question is Obviously "A"

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A car traveling at 23 m/s starts to decelerate steadily. It comes to a complete stop in 5 seconds. What is its acceleration?
kykrilka [37]
We could determine the acceleration using this formula
\boxed{a= \dfrac{v_{1}-v_{0}}{t} }

Given from the question v₀ = 23 m/s, v₁ = 0 (the car stops), t = 5 s
plug in the numbers
a= \dfrac{v_{1}-v_{0}}{t}
a= \dfrac{0-23}{5}
a= \dfrac{-23}{5}
a = -4.6
The acceleration is -4.6 m/s²
8 0
3 years ago
Question 1 of 5
Elina [12.6K]

Answer:

solar energy warms most of the earths surface

3 0
3 years ago
Falling objects drop with an average acceleration of 9.8 m/s2. An arrow is shot with a velocity of 11.76 m/s straight down from
In-s [12.5K]

Answer:

3.8 secs

Explanation:

Parameters given:

Acceleration due to gravity, g = 9.8 m/s^2

Initial velocity, u = 11.76 m/s

Final velocity, v = 49 m/s

Using one of Newton's equations of linear motion, we have that:

v = u + gt

where t = time of flight of arrow

The sign is positive because the arrow is moving downward, in the same direction as gravitational force.

Therefore:

49 = 11.76 + 9.8*t\\\\\\\49 - 11.76 = 9.8t\\\\=> 9.8t = 37.24\\\\\\t = \frac{37.24}{9.8} \\\\\\t = 3.8 secs

The arrow was in flight for 3.8 secs

6 0
4 years ago
A computational model predicts the maximum potential energy a roller coaster car can have given its mass and its speed at the lo
andreyandreev [35.5K]

Answer:

The prediction for its maximum potential energy is 109,375 J

Explanation:

Given;

mass of the coaster car, m = 350 kg

speed of the coaster car at the lowest point, v = 25 m/s

The coaster car will have maximum kinetic energy at the lowest point and based on law of conservation of mechanical energy, the maximum kinetic energy of the coaster car at the lowest point will be equal to maximum potential energy at the highest point.

K.E_{max} = P.E_{max}

K.E_{max} = \frac{1}{2} mv^2\\\\K.E_{max} = \frac{1}{2} (350)(25)^2\\\\K.E_{max} =109,375 \ J

Thus, P.E_{max} = 109,375 \ J

Therefore, the prediction for its maximum potential energy is 109,375 J

3 0
3 years ago
A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a
Nimfa-mama [501]

Answer:

F_0 = 393 N

Explanation:

As we know that amplitude of forced oscillation is given as

A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}

here we know that natural frequency of the oscillation is given as

\omega_0 = \sqrt{\frac{k}{m}}

here mass of the object is given as

m = \frac{W}{g}

\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}

\omega_0 = 8.48 rad/s

angular frequency of applied force is given as

\omega = 2\pi f

\omega = 2\pi(10.5) = 65.97 rad/s

now we have

0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}

F_0 = 393 N

6 0
3 years ago
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