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salantis [7]
3 years ago
14

A car slams on its brakes, coming to a complete stop in 4.0 s. The car was traveling south at 60.0 mph. Calculate the accelerati

on. mi/h/s
Physics
2 answers:
alexgriva [62]3 years ago
8 0

Answer:

The car is decelerating with the magnitude of 0.00416 mi/h².

Or

The car is decelerating with the magnitude of 15 mi/h/s.

Explanation:

Given that,

Time = 4.0 sec

Speed = 60.0 mph

We need to calculate the acceleration

Using equation of motion

v = u+at

Where, v = final speed

u = initial speed

a = acceleration

t = time

Put the value into the formula

a= \dfrac{v-u}{t}

a=\dfrac{0-60.0}{4.0\times3600}

a =-0.00416\ mi/h^2

We need to calculate the acceleration  in mi/h/s

a = \dfrac{60.0\ mph}{4.0\ s}

a= -15\ mi/h/s

Negative sign

Hence, The car is decelerating with the magnitude of 0.00416 mi/h².

Or

The car is decelerating with the magnitude of 15 mi/h/s.

lutik1710 [3]3 years ago
6 0
We need to apply the following expression

Vf = Vo + a.t

Where a is the acceleration
t is the time it takes to achieve Speed Vf
And Vo the initial speed 

In this problem Vo = 60 miles/hour
Vf =  0 (comes to a stop)
t = 4 seconds 

a = -60 [miles/hours] / 4 [s] = -15 miles/hour/second    (negative as it goes in                                                                                                  the opposite direction)
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3 years ago
PLEASE I NEED HELP I AM REALLY STUCK!!!
AlekseyPX

Answer: Force = Mass X Acceleration

F = 5 x 2

F = 10 N

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True or false. Students with a Learners License may not receive a motorcycle endorsement.
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3 years ago
A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference b
Lesechka [4]

Answer:

a)  Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F ,  V = 3.4375 V ,

c)  U = 54.7 nJ ,  d) ΔU = 54 nJ,

Explanation:

a) The capacity of a capacitor is defined

        C = Q / V

        Q = C V

         

can also be calculated using geometry consideration

        C = e or A / d

         

we reduce to the SI system

       A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²

       d = 1.53 cm = 1.53 10⁻² m

we substitute

         Q = eo A / d V

         Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275

         Q = 3.9757 10⁻¹⁰ C

         

let's reduce to pC

         Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

          Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

           Q = k Q₀

           Q = 80 397.57

           Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

           C = k C₀

           C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

           C = 1.157 10⁻¹⁰ F

           V = Vo / k

            V = 275/80

            V = 3.4375 V

c) The stored energy is

             U = ½ C V²

              U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²     275²

             U = 5.47 10⁻⁸ J

let's reduce to nJ

              109 nJ = 1 J

               U = 54.7 nJ

d) energy after submerging

             U = ½ (kCo) (Vo / k) 2

             U = ½ Co Vo2 / k

             U = U₀ / k

             U = 54.7 / 80 nJ

              U = 0.68375 nJ

the energy change is

         ΔU = U₀ -U

          ΔU = 54.7 - 0.687375

           

6 0
3 years ago
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A cons
Virty [35]

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

H_m=1.65m

H_E=1.16307m

Explanation:

From the question we are told that

Mass of ball M=2kg

Length of string L= 2m

Wind force F=13.2N

Generally the equation for \angle \theta is mathematically given as

tan\theta=\frac{F}{mg}

\theta=tan^-^1\frac{F}{mg}

\theta=tan^-^1\frac{13.2}{2*2}

\theta=73.14\textdegree

Max angle =2*\theta= 2*73.14=>146.28\textdegree

Generally the equation for max Height H_m is mathematically given as

H_m=L(1-cos146.28)

H_m=0.9(1+0.8318)

H_m=1.65m

Generally the equation for Equilibrium Height H_E is mathematically given as

H_E=L(1-cos73.14)

H_E=0.9(1+0.2923)

H_E=1.16307m

8 0
2 years ago
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