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iragen [17]
3 years ago
12

a 1210 kg roller coaster car is moving 6.33 m/s. as it approaches the station, brakes slow it down to 2.38 m/s over a distance o

f 4.20 m. what is the magnitude of the force applied by the brakes?(unit=N) PLEASE HELP
Physics
2 answers:
Ratling [72]3 years ago
8 0

Answer:

4,955.9 N

Explanation:

1/2 (m*v^2) = 1/2 (1210 x 6.33^2) = 24,241.7 J.

1/2 (1210 x 2.38^2) = 3,427 J.

(24,241.7 - 3,427) = 20,814.7J. of work done by the brakes.

(20,814.7/ 4.2m) = brake force of 4,955.9 N.

Stels [109]3 years ago
4 0

Answer:

4960 N

Explanation:

First, find the acceleration.

Given:

v₀ = 6.33 m/s

v = 2.38 m/s

Δx = 4.20 m

Find: a

v² = v₀² + 2aΔx

(2.38 m/s)² = (6.33 m/s)² + 2a (4.20 m)

a = -4.10 m/s²

Next, find the force.

F = ma

F = (1210 kg) (-4.10 m/s²)

F = -4960 N

The magnitude of the force is 4960 N.

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A rocket is launched from rest and moves in a straight line at 30.0 degrees above the horizontal with an acceleration of 35.0 m/
klemol [59]

Answer:

t = 123.59s

Explanation:

For the launch pad section:

Vf = Vo + a*t  where Vo=0.

Vf = 35*25 = 875m/s

The distance traveled during the launch:

d = Vo*t+\frac{a*t^2}{2} = 0+\frac{35*25^2}{2} = 10937.5m

Now the projectile motion, we know that its initial speed is the speed calculated previously and the initial height is the y-component of the previously calculated distance.

\Delta Y = Vo*sin(30)*t - \frac{g*t^2}{2}

-d*sin(30) = Vo*sin(30)*t - \frac{g*t^2}{2}  where d= 10937.5m; Vo=875m/s.

Solving for t:

t1 = -11.093s   t2 = 98.59s

So, the total time of flight will be:

t_{total} = t_{launch}+t_{projectile}=25+98.59 = 123.59s

6 0
3 years ago
Transverse waves are sent along a 4.50 m long string with a speed of 85.00 m/s. The string is under a tension of 20.00 N. What i
frutty [35]

Answer:

m = 0.0125 kg

Explanation:

Let us apply the formula for the speed of a wave on a string that is under tension:

v = \sqrt{\frac{F}{\mu} }

where F = tension force

μ = mass per unit length

Mass per unit length is given as:

μ  = m / l

where m = mass of the string

l = length of the string

This implies that:

v = \sqrt{\frac{F}{m/l} }\\ \\v = \sqrt{\frac{F * l}{m} }

Let us make mass, m, the subject of the formula:

v^2 = \frac{F * l}{m}\\\\m = \frac{F * l}{v^2}

From the question:

F = 20 N

l = 4.50 m

v = 85 m/s

Therefore:

m = \frac{20 * 4.5}{85^2}\\\\m = \frac{90}{7225}\\ \\m = 0.0125 kg

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F=ma
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