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The driver of a car traveling at a speed of 25.5 m/s slams on the brakes and comes to a stop in 3.4 s. If we assume that the car

marusya05 [52]

Answer:

(a). The average speed of the car is 12.75 m/s

(b). The distance is 43.35 m.

Explanation:

Given that,

Initial speed = -25.5 m/s

Final speed = 0

Time = 3.4 s

(a). We need to calculate the average speed

Using formula of average speed

$v_{avg}=\dfrac{v_{f}-v_{i}}{2}$

Put the value into the formula

$v_{avg}=\dfrac{0-(-25.5)}{2}$

$v_{avg}=12.75\ m/s$

(b). We need to calculate the acceleration

Using equation of motion

$v_{f}=v_{i}+at$

$a =\dfrac{v_{f}-v_{i}}{t}$

$a=\dfrac{-25.5-0}{3.4}$

$a=-7.5\ m/s^2$

We need to calculate the distance

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s=25.5\times3.4+\dfrac{1}{2}\times(-7.5)\times(3.4)^2$

$s=43.35\ m$

Hence, (a). The average speed of the car is 12.75 m/s

(b). The distance is 43.35 m.

5 0

4 years ago

Eugene is studying the levels of structural organization of an animal’s body. Which level would describe a dog’s eye

mrs_skeptik [129]

The levels of structural organization:
cell ---> tissue ---> organ ---> organ system ---> organism

A dog's eye would be an organ.

6 0

4 years ago

Read 2 more answers

Please help!!

Ronch [10]

Answer:

Explanation:

The distance will be the total distannce covered during the journey.

If you move 3 meters East and move 4 meters north, then the distance will be calculated as;

Distance = distnace through East+distance through north

Distance = 3m + 4m

<em>Distance = 7m</em>

Displacement is the distance covered in a specified direction. It is the shortest distance covered by me. This can be gotten using the Pythagoras theorem.

d² = 3²+4²

d² = 9+16

d² = 25

d = √25

d = 5m

<em>Hence the displacement of the object is 5metres</em>

5 0

4 years ago

a 5kg block on a rough horizontal surface is attached to a light spring (force constant=1.6kN/m). the block passes through its e

natima [27]

Answer:

2.12 J

Explanation:

Initial kinetic energy = final elastic energy + work by friction

KE = EE + W

KE = ½ kx² + W

5 J = ½ (1600 N/m) (0.06 m)² + W

W = 2.12 J

5 0

3 years ago

Contrast the force of gravity between these pairs of objects a 1 kg mass and a 2 kg mass that are 1 m apart and two 2 kg masses

Yuki888 [10]

Solution

Force between pair of objects of masses 1kg and 2kg that are 1m apart is given as

$F=G\times \frac{1\times 2}{1^2}$

here G is gravitational constant

G= $6.674\times10^{-11}Nm^2kg^{-2}$

therefore,

$F=6.674\times10^{-11}Nm^2kg^{-2}\times 2$

$F=13.348\times10^{-11}Nm^2kg^{-2}$

similarly Force between pair of objects of masses 2kg each that are 1m apart is given as

$F'=G\times \frac{2\times 2}{1^2}$

$F'=6.674\times10^{-11}Nm^2kg^{-2}\times 4$

$F'=2\times 13.348\times10^{-11}Nm^2kg^{-2}$

or F'=2F

it means Force between pair of objects of masses 2kg each that are 1m apart is equal to twice the Force between pair of objects of masses 1kg and 2kg that are 1m

7 0

3 years ago