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Elenna [48]
2 years ago
12

The figure below shows a motorcycle leaving the end of a ramp with a speed of 39.0 m/s and following the curved path shown. At t

he peak of the path, a maximum height h above the top of the ramp, the motorcycle's speed is 37.1 m/s. What is the maximum height h? Ignore friction and air resistance. (Enter your answer in m.)
_____m

Physics
1 answer:
lisov135 [29]2 years ago
7 0

The maximum height h of the curved path is 7.38 m.

<h3>Maximum height of the curved path</h3>

Apply the following kinematic equation;

v² = u² - 2gh

where;

  • v is the final velocity of the motorcycle
  • u is initial velocity of the motorcycle
  • h is the maximum height

(u² - v²)/2g = h

(39² - 37.1²)/(2 x 9.8) = h

7.38 m = h

Thus, the maximum height h of the curved path is 7.38 m.

Learn more about maximum height here: brainly.com/question/12446886

#SPJ1

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The number 3 in hydrogen in NH33 is the
frosja888 [35]

Answer:

no of atoms

Explanation:

for each amonia molecule one nitrogen atom bind with 3 hydrogen atoms

4 0
3 years ago
a female vocalist with a soprano voice can sing as high as 1000 hz. given that the speed of sound is 345 m/s what is the wavelen
Neporo4naja [7]

0.345 m.

<h3>Explanation</h3>

The wavelength is the distance that the wave travels in each cycle. The wave travels 345 meters in each second. Let the wavelength of this wave be \lambda. That's the distance the wave travels in one cycle.

The frequency of the sound wave is 1 000 Hz, meaning that there are 1 000 cycles in each second. The wave travels a distance of 1 000 wavelengths in one second. That would be a distance of 1,000\;\lambda.

From the speed of the wave, the wave travels 345 meters in one second. In other words,

1,000\;\lambda = 345.

\lambda = \dfrac{345}{1,000} = 0.345\;\text{m}.

To generalize:

\lambda = \dfrac{v}{f},

where

  • \lambda wavelength of the wave,
  • v the speed of the wave, and
  • f the frequency of the wave.
3 0
3 years ago
An excited 92 kg football player celebrates a touchdown by carelessly running straight into the goalpost at 9.4 m/s. He bounces
yKpoI14uk [10]

Answer:

i. 15.6 m/s

ii. I = 1.44 KNs

Explanation:

The impulse, I, on a body is the product of force applied on it and the time it acts.

i.e I = F x t

Impulse is sometimes expressed as the change in momentum of a body. It is measured in Ns.

i. mass, m, of the player = 92 kg

initial velocity of the player, u = 9.4 m/s

final velocity of the player, v = 6.2 m/s

Since he bounces back on hitting the pole, then the sign of initial and final velocities are of opposite sign.

So that,

change in velocity of the player = final velocity - initial velocity

                                          = 6.2 - (-9.4)

                                         = 6.2 + 9.4

                                         = 15.6 m/s

change in velocity of the player is 15.6 m/s

ii. Impulse, I = m(v - u)

                    = 92 x 15.6

                    = 1435.2

Impulse on the player is 1.44 KNs.

7 0
3 years ago
The froghopper Philaenus spumarius is supposedly the best jumper in the animal kingdom. To start a jump, this insect can acceler
Kay [80]

Answer:

Part a)

v_f = 4 m/s

Part b)

t = 0.001 s

Part c)

d = 0.815 m

Explanation:

Part a)

As we know that initially the grass hopper is at rest at the ground position

Now the acceleration is given as

a = 4000 m/s^2

distance of the legs that it stretched is given as

s = 2.0 mm

so we have

v_f^2 - v_i^2 = 2 a d

v_f^2 - 0 = 2(4000)(0.002)

v_f = 4 m/s

Part b)

time taken to reach this speed is given as

v_f - v_i = at

4 - 0 = 4000 t

t = 0.001 s

Part c)

as the grass hopper reach the maximum height its final speed would be zero

so we will have

v_f^2 - v_i^2 = 2 a d

0 - 4^2 = 2(-9.81) d

d = 0.815 m

5 0
3 years ago
The tension in the rope securing the upper pulley and the force that must be applied to keep the system in equilibrium ​
chubhunter [2.5K]

Answer:

Explanation:

In a frictionless system with no acceleration, the tension in the rope must be F along its entire length

FBD analysis of the lower pulley has two upward acting tension vectors F and one downward acting weight vector W

2F = W

F = W/2

FBD analysis of the upper pulley has one upward acting support vector T and three downward acting tension vectors F

T = 3F

T = 3(W/2)

T = 1.5W

8 0
3 years ago
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