Answer:
![\boxed {\boxed {\sf 412.5 \ kilometers }}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%20412.5%20%5C%20kilometers%20%7D%7D)
Explanation:
Distance is the product of speed and time.
![d=s*t](https://tex.z-dn.net/?f=d%3Ds%2At)
The speed of the car is 75 kilometers per hour. It traveled for 5.5 hours.
![s= 75 \ km/hr \\t= 5.5 \ hr](https://tex.z-dn.net/?f=s%3D%2075%20%5C%20km%2Fhr%20%5C%5Ct%3D%205.5%20%5C%20hr)
Substitute the values into the formula.
![d= 75 \ km/hr * 5.5 \ hr](https://tex.z-dn.net/?f=d%3D%2075%20%5C%20km%2Fhr%20%2A%205.5%20%5C%20hr)
Multiply. Note that the hours will cancel each other out.
![d= 75 km * 5.5 \\d= 412.5 \ km](https://tex.z-dn.net/?f=d%3D%2075%20km%20%2A%205.5%20%5C%5Cd%3D%20412.5%20%5C%20km)
The car travelled <u>412.5 kilometers.</u>
Answer:
The surface of Mercury has landforms that indicate its crust may have contracted. They are long, sinuous cliffs called lobate scarps. These scarps appear to be the surface expression of thrust faults, where the crust is broken along an inclined plane and pushed upward.
Explanation:
I hope this helps a little bit.
(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².
(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.
(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.
<h3>
Work done in the spring</h3>
The work done in stretching the spring is calculated as follows;
W = ¹/₂kx²
W(1 to 2) = ¹/₂K₂Δx²
W(1 to 2) = ¹/₂(250)(0.65 - 0.35)²
W(1 to 2) = 11.25 J
W(0 to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃
W(0 to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)
W(0 to 3) = 64.28 J
Learn more about work done here: brainly.com/question/25573309
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Answer:
the speed after 3 seconds is 10 m/s
Explanation:
The computation of the speed is shown below:
As we know that
V = U + at
Here,
U = 34 m/s
a = - 8 m/s²
t = 3 Sec
V = velocity after 3 sec
V = 34 + (-8)3
= 34 - 24
V = 10 m/s
Hence, the speed after 3 seconds is 10 m/s