the slope of a line on a position vs time graph is the

a. velocity
b. time
c. distance
d. displacement

Answers

Answer 1

It's the velocity, but only the magnitude. It can't show the direction of the velocity. So it's better to call it speed.


Related Questions

Saul is testing an installation and discovers a short circuit what’s causing this
A.high current

B.high voltage

C.low resistance

D.low voltage

Answers

Answer:
C. Low resistance

Answer:

im pretty sure that it's (A.) High current

Explanation:

pf

not 100% sure, tho

why Fossil fuel has been used more in the existing world ?​

Answers

Fossil fuels are found most of the part of the world .It can be found easily by the people.It is free of cost or low cost.Thats why people can find it easily and in low cost.Also it is easy to useSo fossil fuel has been used more in existing world

derive an expression for resistivity of conductor of length l and area of cross section A​

Answers

Answer

Resistivity R = K * L / A    where resistivity is constant for material, L the length of the material and A the area of the material

K = R * A / L     where R is the resistance of the material

What do is mean by environment friendly behaviour?​

Answers

Answer:
Being environmentally friendly means having a lifestyle that is better for the environment

A car has a mass of 2000 kg. While it is traveling along a perfectly flat road, it goes around an unbanked turn that has a radius of 40.0 m. The coefficient of static friction between the car tires and the road is 0.500. The car travels successfully around the turn at a constant speed of 10.0 m/s. Calculate the magnitude of the car's acceleration as it goes around the turn. _______ m/s^2

Answers

Answer:

2.5 m/s²

Explanation:

The given parameters are;

The mass of the car, m = 2,000 kg

The radius of the car, r = 40.0 m

The coefficient of friction between the car tires and the road, μ = 0.500

The constant speed with which the car moves, v = 10.0 m/s

The normal reaction of the road on the car, N = The weight of the car;

∴ N = m × g

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

N ≈ 2,000 kg × 9.81 m/s² = 19,620 N

The frictional force, [tex]F_f[/tex] = μ × N

The centripetal force, [tex]F_c[/tex] = m·v²/r

The car moves without slipping when [tex]F_f[/tex] = [tex]F_c[/tex]

Therefore, [tex]F_f[/tex] = 0.500 × 19,620 N = 2,000 kg × [tex]v_{max}[/tex]²/40.0 m

∴ [tex]v_{max}[/tex] = √(0.500 × 19,620 N × 40.0 m/2,000 kg) ≈ 14.007 m/s

Therefore, the velocity with which the car moves, v < [tex]v_{max}[/tex]

The cars centripetal acceleration, [tex]a_c[/tex] = v²/r

∴ [tex]a_c[/tex] = (10.0 m/s)²/40.0 m = 2.5 m/s²

The cars centripetal acceleration as it goes round the turn, [tex]a_c[/tex] = 2.5 m/s².

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