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Alex787 [66]
3 years ago
13

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

ment, consider a ball thrown off the top of a building with a velocity v at an angle θ with respect to the horizontal. Let the building be 44.0 m tall, the initial horizontal velocity be 8.60 m/s, and the initial vertical velocity be 10.5 m/s. Choose your coordinates such that the positive y-axis is upward, and the x-axis is to the right, and the origin is at the point where the ball is released.
Physics
1 answer:
Aleks04 [339]3 years ago
3 0

Explanation:

It is given that,

Height of the building, h = 44 m

Initial horizontal velocity, u_x=ucos\theta=8.6\ m/s

Initial vertical velocity, u_y=usin\theta=10.5\ m/s

It can be assumed to find the maximum height of the projectile and time taken to reach the maximum height.

The formula for the maximum height is given by :

H=\dfrac{(u\ sin\theta)^2}{2g}

g is the acceleration due to gravity

H=\dfrac{(10.5)^2}{2\times 9.8}=5.625\ m

The total maximum height, h' = h + H

h'=44+5.625=49.625\ m

The formula for the time of flight is given by :

t_{max}=\dfrac{u\ sin\theta}{g}

t_{max}=\dfrac{10.5}{9.8}=1.07\ s

Hence, this is the required solution.

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The velocity of sound on a particular day outside is 331 meters/second. What is the frequency of a tone if it has a wavelength o
N76 [4]

Frequency = (speed) / (wavelength)

Frequency = (331 m/s) / (0.6 m) = 551.7 Hz
3 0
3 years ago
The brakes of a 125 kg sled are applied while it is moving at 8.1 m/s, which exerts a force of 261 N to slow the sled down. How
sammy [17]

Answer:

15.7 m

Explanation:

m = mass of the sled = 125 kg

v₀ = initial speed of the sled = 8.1 m/s

v = final speed of sled = 0 m/s

F = force applied by the brakes in opposite direction of motion = 261

d = stopping distance for the sled

Using work-change in kinetic energy theorem

- F d = (0.5) m (v² - v₀²)

- (261) d = (0.5) (125) (0² - 8.1²)

d = 15.7 m

6 0
3 years ago
Which statement best explains the speed of light waves as they travel from gas to solid?
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Answer:

Light waves slow down as they travel from gas to solid.

Explanation:

The speed of light is often quoted with reference to a vacuum. Light travels fastest in a vacuum and in a gas.

When light travels through other media such as solids or liquids, the speed of light is decreased due to absorption and scattering of photons by molecules as well as remissions.

Hence the speed of light decreases from gas to solid.

8 0
3 years ago
The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the
kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0
3 years ago
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