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Eva8 [605]
3 years ago
8

9) A policeman investigating an accident measures the skid marks left by a car. He determines that 9) the distance between the p

oint that the driver slammed on the brakes and the point where the car came to a stop was 28.0 m. From a reference manual he determines that the coefficient of kinetic friction between the tires and the road under the prevailing conditions was 0.300. How fast was the car going wh~er applied the brakes? (This car was not equipped with anti-lock brakes.)
Physics
1 answer:
sineoko [7]3 years ago
7 0

Answer:

Explanation:

Provided the car did not collide with anything during its slide,

Kinetic energy will convert to work of friction

½mv² = Fd

½mv² = μmgd

       v = √(2μgd)

       v = √(2(0.300)(9.81)(28.0))

       v = 12.8377...

       v = 12.8 m/s

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It is 5.5 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 km/h
stealth61 [152]

Explanation:

Displacement = 5 km

A.

Converting km/h to m/s,

10 km/h * 1000 m/1 km * 1 h/3600 s

= 25/9 m/s

Remember,

700 watt = 700 J/s

Velocity = displacement/time

Time = 5000/(25/9)

= 1800 s

Energy = power * time

= 700 * 1800

= 1,260,000

= 1260 kJ

B.

Converting km/h to m/s,

3 km/h * 1000 m/1 km * 1 h/3600 s

= 5/6 m/s

290 watt = 290 J/s

Velocity = displacement/time

Time = 5000/(5/6)

= 6000 s

Energy = power * time

= 290 * 6000

= 1,740,000

= 1740 kJ

C.

Walking burns more energy; 1,740,000 joules. It burns more because you walk for a greater period of time.

6 0
3 years ago
1. In Newton’s ring experiment, the diameter of the 5th ring is 0.30 cm and diameter of 15th the ring is 0.62 cm. Find the diame
IgorC [24]

Answer:

Diameter of Newton’s 5th ring = 0.30 cm

Diameter of Newton’s 15th ring = 0.62 cm

Diameter of Newton’s 25th ring = ?

From Newton’s rings experiment we infer that

D2n+m − D2n = 4λmR

For the 5th and 15th rings we have

D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)

For 15th and 25th rings

D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)

We equate the two derivatives

Equation (2) = Equation (1)

D225 − D215 = D215 − D25

D225 = 2D215 – D25

Substituting the values into the equation

D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2

D25 = 0.8239 cm

4 0
3 years ago
The difference between an experimental value and an accepted value is
likoan [24]

Answer:

HERE'S MY UNDERSTANDING OF THE DIFFERENCE

3 0
3 years ago
A particle, whose acceleration is constant, is moving in the negative x direction at a speed of 4.91 m/s, and 12.9 s later the p
Zinaida [17]

Answer:

The particle’s velocity is -16.9 m/s.

Explanation:

Given that,

Initial velocity of particle in negative x direction= 4.91 m/s

Time = 12.9 s

Final velocity of particle in positive x direction= 7.12 m/s

Before 12.4 sec,

Velocity of particle in negative x direction= 5.32 m/s

We need to calculate the acceleration

Using equation of motion

v = u+at

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

Where, v = final velocity

u = initial velocity

t = time

Put the value into the equation

a=\dfrac{7.12-(-4.91)}{12.9}

a=0.933\ m/s^2

We need to calculate the initial speed of the particle

Using equation of motion again

v=u+at

u=v-at

Put the value into the formula

u=-5.321-0.933\times12.4

u=-16.9\ m/s

Hence, The particle’s velocity is -16.9 m/s.

4 0
3 years ago
A car has a mass of 1600 kg. It is stuck in the snow and is being pulled out by a cable that applies a force of 7560 N due north
sergiy2304 [10]

Answer:

Acceleration of the car will be a=0.1375m/sec^2

Explanation:

We have given mass of the ball m = 1600 kg

Force in north direction F= 7560 N

Resistance force which opposes the movement of car F_R=7340N

So net force on the car F_{net}=F-F_R=7560-7340=220N

According to second law of motion we know that F=ma

So 220=1600\times a

a=0.1375m/sec^2

7 0
3 years ago
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