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Delicious77 [7]
3 years ago
15

a 1500.0 kg car is at rest on a level track. The driver gives the engine gas and the car begins to accelerate forward. The car m

oves from 0 m/s to 27 m/s in 8 s. What is the acceleration of the car. What is the force of the car moving forward?
Physics
1 answer:
padilas [110]3 years ago
4 0

Acceleration of the car is 3.375 m/s² and the force of the car moving forward is 5062.5 N

Explanation:

  • Acceleration is the rate of change of velocity.
  • It is given by the equation, a = change in velocity/time

Here, velocity changes from 0 to 27 m/s and time = 8

⇒ Acceleration = 27 - 0/8 = 27/8 = 3.375 m/s²

  • Force is calculated by the equation, F = Mass × Acceleration
  • This is based on Newton's second law of motion.

Here, mass of the car = 1500 kg and a = 3.375 m/s²

⇒ Force = 1500 × 3.375 = 5062.5 N

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3 years ago
(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup
KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx

Where

x_{o}, x_{f} - Initial and final position, respectively, measured in meters.

F(x) - Force as a function of position, measured in newtons.

Given that F = k\cdot x and the fact that F = 25\,N when x = 0.3\,m - 0.2\,m, the spring constant (k), measured in newtons per meter, is:

k = \frac{F}{x}

k = \frac{25\,N}{0.3\,m-0.2\,m}

k = 250\,\frac{N}{m}

Now, the work function is obtained:

W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx

W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]

W = 0.313\,J

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be r(\theta) = 2\cdot \sin 5\theta. The area of the region enclosed by one loop of the curve is given by the following integral:

A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta

A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta

By using trigonometrical identities, the integral is further simplified:

A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta

A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta

A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta

A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)

A = 4\pi

The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

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3 years ago
2 decaliters + 800 deciliters = __________ liters?
Stolb23 [73]
Answer is 100 liters
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4 years ago
Read 2 more answers
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