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Andru [333]
3 years ago
6

A car accelerates from rest at 5.75m/s2 for 4.4 sec when it runs out of pavement and runs into some mud. In the mud it accelerat

es to rest at -5.5m/s2. How far did it move from the beginning to the end? Give a variable legend for this problem. The model for this problem?
Physics
1 answer:
just olya [345]3 years ago
4 0

Answer:

113.85 m

Explanation:

When the car is on the pavement:

v₀ = 0 m/s

a = 5.75 m/s²

t = 4.4 s

Find: Δx and v

Δx = v₀ t + ½ at²

Δx = (0 m/s) (4.4 s) + ½ (5.75 m/s²) (4.4 s)²

Δx = 55.66 m

v = at + v₀

v = (5.75 m/s²) (4.4 s) + 0 m/s

v = 25.3 m/s

When the car is in the mud:

v₀ = 25.3 m/s

v = 0 m/s

a = -5.5 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (25.3 m/s)² + 2 (-5.5 m/s²) Δx

Δx = 58.19 m

The total displacement is therefore:

55.66 m + 58.19 m = 113.85 m

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A 50-cm-long spring is suspended from the ceiling. A 330 g mass is connected to the end and held at rest with the spring unstret
Nataly [62]

Answer:

a)32.34 N/m

b)10cm

c)1.6 Hz

Explanation:

Let 'k' represent spring constant

'm' mass of the object= 330g =>0.33kg

a) in order to find spring constant 'k', we apply Newton's second law to the equilibrium position 10cm below the release point.

ΣF=kx-mg=0

k=mg / x

k= (0.33 x 9.8)/ 0.1

k= 32.34 N/m

b) The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object.

Therefore, amplitude of the oscillation is 10cm

c)frequency of the oscillation can be determined by,

f= 1/2π \sqrt{\frac{k}{m} }

f= 1/2π \sqrt{\frac{32.34}{0.33} }

f= 1.57

f≈ 1.6 Hz

Therefore,  the frequency of the oscillation is 1.6 Hz

5 0
3 years ago
You are moving into an apartment and take the elevator to the 6th floor. Suppose your weight is 660 N and that of your belonging
Ivan

Answer:

Explanation:

Total weight

My weight+weight of belongings

660+1100=1760N.

a. Work done by the elevator to travel a total height of 15.2m

Using newton law of motion

ΣF = ma

There are only two forces acting upward, the weight and the reaction by the elevator

Also note it is moving at constant velocity then, a=0

N - W=0

Then, N=W

N=1760N

So, workdone is given as

Wordone, =force × distance

Work done=1760×15.2

W=26,752J

W=26.752KJ

b. Work done on me alone is still need to go through the same process but will remove the weight of the belonging

Therefore,

Weight now = 660N

And using the same equation of motion

ΣF = ma

Comstant velocity, a=0

N - W=0

N=W

N=660N

Then, workdone

W=F×d

W=660×15.2

W=10,032J

W=10.032KJ

6 0
3 years ago
Read 2 more answers
(please help ASAP!)
makkiz [27]

Answer:

It's C. Length

Explanation:

Just think of a ruler. Hope I helped! :)

8 0
2 years ago
Read 2 more answers
what happens to the period of revolution for the planets as they move farther away in position from the sun
Marina CMI [18]
It increases. Mercury takes 88 days to orbit the sun once. The Earth takes a year. Pluto takes 248 years.
3 0
3 years ago
A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same
Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

7 0
3 years ago
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