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Rasek [7]
3 years ago
6

When traveling through mountains, most highways periodically have emergency escape ramps to allow large, heavy vehicles to stop

safely, even when their brakes are not working properly. One such escape ramp is angled 13 ∘ above the horizontal. The brakes on one large truck (mass 1.4×104 kg ) fail when driving at a speed of 80 miles per hour (35.75 m/s), and the driver makes use of the escape ramp.
Part A: What is the minimum length of ramp necessary for this truck to come to a stop? Assume negligible friction and air resistance forces.

Part B: If instead dissipative forces (friction and air resistance) provide a constant force of 60,000N parallel to the ramp, what is the minimum length of ramp necessary for this truck to come to a stop?

my answer for part A is 290 m, but I am unsure how to get the answer for part B. please help
Physics
1 answer:
xeze [42]3 years ago
3 0

Answer:

hf vb sh ff shod hoke Spinx fishy

Explanation:

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Speakers A and B are vibrating in phase. They are directly facing each other, are 8.2 m apart, and are each playing a 78.0 Hz to
Stels [109]

Answer:6.298,4.1,1.9015

Explanation:

Wavelength=\frac{velocity of sound }{frequency}

=\frac{343}{78}=4.397 m

Distance of 3rd speaker from speaker A is x

From B 78-x

Difference between the distances must be a whole number of wavelengths

First

x-\left ( 8.2-x\right )=4.397    for 1 st wavelength

2x=8.2+4.397=12.597

x=6.298m

second

For zero wavelength

x-\left ( 8.2-x\right )=0

2x=8.2

x=4.1m

Third

\left ( 8.2-x\right )-x=4.397

x=1.9015 m

6 0
3 years ago
A 1.80-kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. The period for s
Evgen [1.6K]

Answer:

(a) I (Moment of inertia)=0.0987 kgm^{2}

(b) W(Angular Speed)=2.66 \frac{rad}{s}

Explanation:

Given data

m (Monkey mass)= 1.80 kg

d=2.50 m

T (Time Period)=0.940 s

Angle= 0.400 rad

(a) I (Moment of Inertia)=?

(b) W (Angular Speed)=?

For part (a) I (Moment of Inertia)=?

Time Period Formula is given as

T=2(3.14)\sqrt{\frac{I}{mgd} }

After Simplifying we get

I=\frac{mgdT^{2} }{4*(3.14)^{2}}

I=\frac{1.8*9.8*0.25*(0.94)^{2} }{4(3.14)^{2} }

I=0.0987 kgm^{2}

For Part (b) Angular Speed

From Kinetic Energy we get

KE=\frac{1}{2}IW^{2}

Pontential Energy

PE=mgd(1-Cosa)

KE=PE

\frac{1}{2}IW^{2}=mgd(1-Cosa)

W^{2}=\frac{2mgd(1-Cosa)}{I}

W=\sqrt{\frac{2*1.8*9.8*0.25*(1-Cos(0.4)rad)}{0.0987} }

W=2.66\frac{rad}{s}

5 0
3 years ago
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