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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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When is the universal theme of a story often revealed?
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It is often revealed <span>at the resolution of the story, when the reader can see how the story ends.</span>
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A 66.0 kg diver is 3.10 m above the water, falling at speed of 6.10 m/s. Calculate her kinetic energy as she hits the water. (Ne
castortr0y [4]

Kinetic energy as she hits the water is 3300 joule.

To find the answer, we need to know about the Newton's equation of motion.

<h3>What's the Newton's equation of motion to determine the final velocity?</h3>
  • The final velocity is determined as

V²=U²+2aS

  • V= final velocity, U= initial velocity, a= acceleration and S= distance
<h3>What's the final velocity of the driver falling from 3.10m with initial velocity of 6.10m/s?</h3>
  • Here, a= 9.8m/s², U= 6.10m/s and S= 3.10m
  • So, V²= 6.1²+2×9.8×3.10= 98
  • V= √98= 10m/s
<h3>What's the kinetic energy of the driver when touches the water?</h3>

Kinetic energy= 1/2×mass×velocity²

= 1/2 × 66 × 10²

= 3300J

Thus, we can conclude that the kinetic energy of the driver is 3300 Joule.

Learn more about the kinetic energy here:

brainly.com/question/25959744

#SPJ4

3 0
2 years ago
A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.
Sedaia [141]

Answer:

335°C

Explanation:

Heat gained or lost is:

q = m C ΔT

where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Heat gained by the water = heat lost by the copper

mw Cw ΔTw = mc Cc ΔTc

The water and copper reach the same final temperature, so:

mw Cw (T - Tw) = mc Cc (Tc - T)

Given:

mw = 390 g

Cw = 4.186 J/g/°C

Tw = 22.6°C

mc = 248 g

Cc = 0.386 J/g/°C

T = 39.9°C

Find: Tc

(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)

Tc = 335

7 0
3 years ago
Water is flowing at 4.0 m/s in a circular pipe. If the diameter of the pipe decreases to 1/2 of its former value, what is the ve
Montano1993 [528]

Answer:

v₂ = 16 m/s

Explanation:

We can use the continuity equation, which is as follows:

A_1v_1 = A_2v_2\\

where,

A₁ = Area of inlet = πd²/4

A₂ = Area of outlet = π(d/2)²/4 = πd²/16

v₁ = velocity at inlet = 4 m/s

v₂ = velocity at outlet = ?

Therefore,

(\frac{\pi d^2}{4})(4\ m/s)=(\frac{\pi d^2}{16})v_2\\\\

<u>v₂ = 16 m/s</u>

3 0
3 years ago
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