Answer:
is the time taken by the car to accelerate the desired range of the speed from zero at full power.
Explanation:
Given:
Range of speed during which constant power is supplied to the wheels by the car is
.
- Initial velocity of the car,

- final velocity of the car during the test,

- Time taken to accelerate form zero to 32 mph at full power,

- initial velocity of the car,

- final desired velocity of the car,

Now the acceleration of the car:



Now using the equation of motion:


is the time taken by the car to accelerate the desired range of the speed from zero at full power.
Answer:
A lens
Explanation:
A lens forms images when light passes Through it bending the rays of in the process.A phenomena called refraction and the speed of light changes in the process because it enters a medium since it's wavelength is reduced.
The velocity of the object s calculated as 22.1 m/s.
<h3>What is the speed of the object?</h3>
Given that we can write that;
v^2 = u^2 + 2gh
Now u = 0 m/s because the object was dropped from a height
v^2 = 2gh
v = √2 * 9.8 * 25
v = 22.1 m/s
Learn more about velocity:brainly.com/question/18084516
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Vas happenin!
The third one makes no since because the clouds carry the rain. It isn’t always cold when it’s going to rain
The fourth one is a good one
The second one again it’s not always cold when it’s raining
The first one could be it also
Hmmm I would go with the last one
Sorry if it’s wrong
Answer:
0.3 m
Explanation:
Initially, the package has both gravitational potential energy and kinetic energy. The spring has elastic energy. After the package is brought to rest, all the energy is stored in the spring.
Initial energy = final energy
mgh + ½ mv² + ½ kx₁² = ½ kx₂²
Given:
m = 50 kg
g = 9.8 m/s²
h = 8 sin 20º m
v = 2 m/s
k = 30000 N/m
x₁ = 0.05 m
(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²
x₂ ≈ 0.314 m
So the spring is compressed 0.314 m from it's natural length. However, we're asked to find the additional deformation from the original 50mm.
x₂ − x₁
0.314 m − 0.05 m
0.264 m
Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.