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scoray [572]
3 years ago
7

In this first example of constant accelerated motion, we will simply consider a car that is initially traveling along a straight

stretch of highway at 15 m/s. At t=0 the car begins to accelerate at 2.0 m/s2 in order to pass a truck. What is the velocity of the car after 5.0 s have elapsed?
Physics
1 answer:
algol133 years ago
4 0

Answer:

v_{f} =25m/s

Explanation:

Kinematics equation for constant acceleration:

v_{f}  =v_{o} + at=15+2*5=25m/s

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A taxi starts from Monument Circle and travels 5 kilometers to the east for 5 minutes. Then it travels 10 kilometers to the sout
lutik1710 [3]
If youre looking for speed the equation would be distance/time

3 0
3 years ago
Read 2 more answers
I need it due in 10 mins ​
ExtremeBDS [4]

Answer:

B. 14.4 N

Rotational speed (Angular Velocity) = 2

The Radius of the circle = 1.2 m

Velocity = Angular velocity × radius = 2×1.2 = 2.4 m/s

Centripetal force= mv²/r = 3 × 2.4×2.4/1.2 = 3 × 2.4 × 2

= 14.4 N

7 0
3 years ago
Help!!!
Taya2010 [7]
<span>c. contains different kind of matter</span>
3 0
3 years ago
An object with a mass of m = 3.85 kg is suspended at rest between the ceiling and the floor by two thin vertical ropes.
Aleksandr-060686 [28]

The tension in the upper rope is determined as 50.53 N.

<h3>Tension in the upper rope</h3>

The tension in the upper rope is calculated as follows;

T(u) = T(d)+ mg

where;

  • T(u) is tension in upper rope
  • T(d) is tension in lower rope

T(u) = 12.8 N + 3.85(9.8)

T(u) = 50.53 N

Thus, the tension in the upper rope is determined as 50.53 N.

Learn more about tension here: brainly.com/question/918617

#SPJ1

6 0
2 years ago
A frictionless piston-cylinder device contains 4.5 kg of nitrogen at 110 kPa and 200 K. Nitrogen is now compressed slowly accord
pochemuha

Answer:

427.392 kJ

Explanation:

m = Mass of gas = 4.5 kg

Initial temperature = 200 K

Final temperature = 360 K

R = Mass specific gas constant = 296.8 J/kgK

\gamma = Specific heat ratio = 1.5

Work done for a polytropic process is given by

W=\frac{mR\Delta T}{1-\gamma}\\\Rightarrow W=\frac{4.5\times 296.8(360-200)}{1-1.5}\\\Rightarrow W=-427392\ J\\\Rightarrow W=-427.392\ kJ

The work input during the process is -427.392 kJ

3 0
3 years ago
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