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2 years ago

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A car is travelling to the right with a speed of 42\,\dfrac{\text m}{\text s}42 s m 42, space, start fraction, m, divided by,

s, end fraction when the driver slams on the brakes. The car skids for 4.0\,\text s4.0s4, point, 0, space, s with constant acceleration before it comes to a stop. How many meters did the car skid before coming to a stop? Answer using a coordinate system where rightward is positive. Round the answer to two significant digits.

2 answers:

Effectus [21]2 years ago

7 0

Answer:

d = 84 m

Explanation:

As we know that when an object moves with uniform acceleration or deceleration then we can use equation of kinematics to find the distance moved by the object

here we know that

initial speed $v_i = 42 m/s$

final speed $v_f = 0$

time taken by the car to stop

$t = 4s$

now the distance moved by the car before it stop is given as

$d = \frac{v_f + v_i}{2} \times t$

now we have

$d = \frac{42 + 0}{2} \times 4$

$d = 84 m$

wolverine [178]2 years ago

5 0

Answer:

84

Explanation:

khan academy

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Answer:

a

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b

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Explanation:

From the question we are told that

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Now given that we are considering just the straight line (i.e points along the line connecting the two sources ) then the order of the maxima m = 0

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=> $|2z - d| =\frac{\lambda}{2}$

substituting values

$|2z - 5| =\frac{6}{2}$

=> $z = \frac{5 \pm 3}{2}$

So

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and

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Explanation:

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For Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
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A 55-kg pilot flies a jet trainer in a half vertical loop of 1200-m radius so that the speed of the trainer decreases at a const

nataly862011 [7]

a) $-1.54 m/s^2$

b) 803.4 N

Explanation:

a)

At the point C (top position of the loop), the pilot feel weightless, so the normal reaction exerted by the seat is zero:

N = 0

Therefore, the equation of the forces at position C is:

$mg=m\frac{v^2}{r}$

where the term on the left is the weight of the pilot and the term on the right is the centripetal force, and where:

$g=9.8 m/s^2$ is the acceleration due to gravity

$v$ is the velocity of the jet at the top

$r=1200 m$ is the radius of the loop

Solving for v,

$v_C=\sqrt{gr}=\sqrt{(9.8)(1200)}=108.4 m/s$

So, this is the velocity of the jet at position C.

The velocity at position A (bottom) is

$v_A=550 km/h =152.8 m/s$

The distance covered by the jet is the length of a semi-circumference of radius r, so

$s=\pi r=\pi(1200)=3770 m$

Since the deceleration of the plane is constant, we can find it by using the following suvat equation:

$v_C^2-v_A^2=2as\\a=\frac{v_C^2-v_A^2}{2s}=\frac{108.4^2-152.8^2}{2(3770)}=-1.54 m/s^2$

b)

The force exerted on the pilot by the seat is equal to the normal force.

At point B (half of the loop), we have:

- The normal force exerted by the seat, N, acting towards the center of the loop

- There are no other forces acting towards the center of the loop, so N must be equal to the centripetal force:

$N=m\frac{v_B^2}{r}$ (1)

where $v_B$ is the velocity at position B.

To find the velocity at position B, we notice that the distance covered by the jet between position A and position B is a quarter of a circle:

$s=\frac{\pi r}{2}=\frac{\pi(1200)}{2}=1885 m$

Since we know the deceleration, we can use the suvat equation to find the velocity at point B:

$v_B^2-v_A^2=2as\\v_B=\sqrt{v_A^2+2as}=\sqrt{152.8^2+2(-1.54)(1885)}=132.4 m/s$

Therefore, we can now use eq.(1) to find the normal force exerted by the seat on the pilot at point B:

$N=(55)\frac{(132.4)^2}{1200}=803.4 N$

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