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Whitepunk [10]
3 years ago
12

A car travels on a straight, level road. (a) Starting from rest, the car is going 38 ft/s (26 mi/h) at the end of 4.0 s. What is

the car's average acceleration in ft/s2? (b) In 3.0 more seconds, the car is going 76 ft/s (52 mi/h). What is the car's average acceleration for this time period?
Physics
1 answer:
lbvjy [14]3 years ago
8 0

Answer:

a)9.5\frac{ft}{s^2}\\ b) 12.66\frac{ft}{s^2}

Explanation:

A body has acceleration when there is a change in the velocity vector, either in magnitude or direction. In this case we only have a change in magnitude. The average acceleration represents the speed variation that takes place in a given time interval.

a)

a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{38\frac{ft}{s}-0}{4 s- 0}=9.5\frac{ft}{s^2}\\

b)

a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{76\frac{ft}{s}-38\frac{ft}{s}}{7 s- 4s}\\a_{avg}=\frac{38\frac{ft}{s}}{3s}=12.66\frac{ft}{s^2}

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A car enters a level, unbanked semi-circular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction betw
meriva

Answer:

As  28m/s = 28m/s

Explanation:

r = the radius of the curve

m =  the mass of the car

μ = the coefficient of kinetic friction

N = normal reaction

When rounding the curve, the centripetal acceleration is

a = \frac{v^{2}}{r}

therefore

\mu mg = m \frac{v^{2}}{r} \\\\ \mu =  \frac{v^{2}}{rg}

v = \sqrt{\mu rg}

\mu = \sqrt{0.8 \times 100\times9.8} \\\\= 28m/s

As  28m/s = 28m/s

8 0
3 years ago
According to newton's first law, what is required to make an object slow down?
Luden [163]
I believe it is friction

3 0
3 years ago
Read 2 more answers
Can someone explain which of Newton’s Law is demonstrated in part 1 and which is demonstrated in part 2? (Picture)
rewona [7]

Answer:

Every action has an equal and opposite reaction. If the student doesn't push, nothing moves, is one student pushes, both move which is an example of newtons third law.

Explanation:

3 0
3 years ago
At a stoplight, a truck traveling at 15 m/s passes a car as it starts from rest. The truck travels at constant velocity and the
Art [367]
The time that the car take to catch up to truck would be :

1/2 ar^2  = vt

1/2 (3) t^2 = 15t

1.5  t^2  =  15t

t ^2    =   10s

Hope this helps
4 0
3 years ago
In the diagram, q1= +8.0 C, q2= +3.5 C, and q3 = -2.5 C. q1 to q2 is 0.10 m, q2 to q3 is 0.15 m. What is the net force on q2? La
yulyashka [42]

Answer:

f(t) =  28,7 [N]

Explanation: IMPORTANT NOTE: IN PROBLEM STATEMENT CHARGES ARE IN C (COULOMBS) AND IN THE DIAGRAM IN μC. WE ASSUME CHARGES ARE IN μC.

The net force on +q₂  is the sum of the force of +q₁  on +q₂ ( is a repulsion force since charges of equal sign repel each other ) and the force of -q₃ on +q₂ ( is an attraction force, opposite sign charges attract each other)

The two forces have the same direction to the right of charge q₂, we have to add them

Then

f(t) = f₁₂ + f₃₂

f₁₂ = K * ( q₁*q₂ ) / (0,1)²

q₁  = + 8 μC     then   q₁ = 8*10⁻⁶ C

q₂ =  + 3,5 μC  then  q₂ = 3,5 *10⁻⁶ C

K = 9*10⁹  [ N*m² /C²]

f₁₂ = 9*10⁹ * 8*3,5*10⁻¹²/ 1*10⁻²   [ N*m² /C²]* C*C/m²

f₁₂ = 252*10⁻¹ [N]

f₁₂ = 25,2 [N]

f₃₂ =  9*10⁹*3,5*10⁻⁶*2,5*10⁻⁶ /(0,15)²

f₃₂ =  78,75*10⁻³/ 2,25*10⁻²

f₃₂ =  35 *10⁻¹

f₃₂ =  3,5 [N]

f(t) =  28,7 [N]

5 0
3 years ago
Read 2 more answers
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