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Inessa [10]
3 years ago
5

A projectile is launched with an initial speed of 21.8 m/s at an angle of 35º above the horizontal. Determine the time of flight

of the projectile.
Physics
1 answer:
katrin2010 [14]3 years ago
8 0

Answer:

2.55 seconds

Explanation:

Given that,

The initial speed of a projectile, u = 21.8 m/s

The angle of projection, \theta=35^{\circ}

We need to find the time of flight of the projectile. The formula for the time of flight is given by :

T=\dfrac{2u\sin\theta}{g}

Where

g is the acceleration due to gravity

T=\dfrac{2\times 21.8\sin(35)}{9.8}\\\\T=2.55\ s

So, the time of flight of the projectile is 2.55 seconds.

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6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri
mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0
3 years ago
Based on your observations of the six collisions, describe the physical difference between elastic and inelastic collisions.
Ahat [919]

Answer:

Collisions are basically two types: Elastic, and inelastic collision. Elastic collision is defined as the colliding objects return quickly without undergoing any heat generation. Inelastic collision is defined as the where heat is generated, and colliding objects are distorted.

In elastic collision, the total kinetic energy, momentum are conserved, and there is no wasting of energy occurs. Swinging balls is the good example of elastic collision. In inelastic collision, the energy is not conserved it changes from one form to another for example thermal energy or sound energy. Automobile collision is good example, of inelastic collision.

5 0
3 years ago
Newtons first law 1 to 5. <br>What is each of the net force for all of the 5 questions? ​
erica [24]

Answer:

1. 65 N.

2. 160 N.

3. 0 N.

4. 210 N.

5. 90 N.

Explanation:

1. Determination of the net force.

Force applied to the right (Fᵣ) = 80 N

Force applied to the left (Fₗ) = 145 N

Net force (Fₙ) =?

Fₙ = Fₗ – Fᵣ

Fₙ = 145 – 80

Fₙ = 65 N

Thus, the net force is 65 N

2. Determination of the net force.

Force 1 applied to the left (F₁) = 35 N

Force 2 applied to the left (F₂) = 125 N

Net force (Fₙ) =?

Fₙ = F₁ + F₂

Fₙ = 35 + 125

Fₙ = 160 N

Thus, the net force is 160 N.

3. Determination of the net force.

Force applied to the right (Fᵣ) = 75 N

Force applied to the left (Fₗ) = 75 N

Net force (Fₙ) =?

Fₙ = Fₗ – Fᵣ

Fₙ = 75 – 75

Fₙ = 0

Thus, the net force is 0 N

4. Determination of the net force.

Force 1 applied to the right (F₁) = 150 N

Force 2 applied to the right (F₂) = 60 N

Net force (Fₙ) =?

Fₙ = F₁ + F₂

Fₙ = 150 + 60

Fₙ = 210 N

Thus, the net force is 210 N.

5. Determination of the net force.

Force applied to the right (Fᵣ) = 115 N

Force applied to the left (Fₗ) = 25 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 115 – 25

Fₙ = 90 N

Thus, the net force is 90 N

8 0
3 years ago
How does Piggy feel about being called "Piggy"?
denpristay [2]

Answer:

good because piggy loves peppa pig :3

Explanation:

4 0
3 years ago
3. A model rocket takes 0.05 seconds to speed up from rest to its maximum velocity of 80 m/s.
nikklg [1K]

Answer:

1600 \frac{m}{s^2}

Explanation:

Acceleration is defined as the change in velocity divided by the time it took to produce such change. The formula then reads:

a = \frac{change-in-velocity}{time} = \frac{Vf-Vi}{t}

Where Vf is the final velocity of the object, (in our case 80 m/s)

Vi is the initial velocity of the object (in our case 0 m/s because the object was at rest)

and t is the time it took to change from the Vi to the Vf (in our case 0.05 seconds.

Therefore we have:

a = \frac{80 m/s - 0 m/s}{0.05 sec} = 1600 \frac{m}{s^2}

Notice that the units of acceleration in the SI system are \frac{m}{s^2} (meters divided square seconds)

7 0
3 years ago
Read 2 more answers
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