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yKpoI14uk [10]
4 years ago
15

Two golfers each hit a ball at the same speed, but one at 60 degrees with the horizontal and the other 30 degrees. a. Which ball

goes farther? b. Which hits the ground first?
Physics
1 answer:
likoan [24]4 years ago
5 0

Answer:

Therefore, both balls will hit the ground at the same distance from their respective starting points and at the same time.

Explanation:

Both balls will hit the ground at the same distance and at the same time assuming they both start at the same initial elevation from the ground.

We know that range in the projectile motion given as

R=\dfrac{v_o^2sin2\theta}{g}

sin 60° = sin 120° =0.866

Therefore, the maximum range will be identical assuming both golfers hit the ball at the same elevation from the ground.

As for the time...

Where v is 0 m/s (final velocity at highest point), v0 is the initial velocity (for both golfers) and is the acceleration due to gravity (-9.8 m/s^²). As you can see, the time it takes to get to the highest point is independent of the angle; it is only dependent on the initial velocity. Since both golfers hit the ball with the same speed, the time for the ball to reach the highest point will be the same. Also, since it's a parabola, you can multiply the time by 2 for each golfer to get the time it takes to hit the ground.

Therefore, both balls will hit the ground at the same distance from their respective starting points and at the same time.

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Siven an element's atomic number and mass number, how can you tell the number of protons and neutrons in its nucleus?
sukhopar [10]
It would be the first option.

Explanation-
The number of protons is equal to the atomic number the number of neutrons is the mass minus the atomic number.
3 0
3 years ago
Read 2 more answers
The prop blades of an airplane spin with a linear velocity of 875 m/s and have a centripetal acceleration on the farthest edge o
storchak [24]

The radius of the prop blade of an airplane is determined as 4.25 m.

<h3>Radius of the prop blade</h3>

The radius of the prop blade of an airplane is calculated as follows;

a = v²/r

where;

  • v is the linear speed
  • r is the radius of the prop blade
  • a is the centripetal acceleration

r = v²/a

r = (875²)/(180,000)

r = 4.25 m

Thus, the radius of the prop blade of an airplane is determined as 4.25 m.

Learn more about centripetal acceleration here: brainly.com/question/79801

#SPJ1

4 0
2 years ago
You are designing an elevator for a hospital. The force exerted on a passenger by the floor of the elevator is not to exceed 1.7
HACTEHA [7]

Answer:

5.62 m/s

Explanation:

Newton's law of motion can be used to determine the maximum speed of the elevator. In the question, we are given:

Force exerted by the elevator (R) = 1.7 times the weight of the passenger (m*g)

Thus: R = 1.7*m*g

Distance (s) = 2.3 m

Newton's second law of motion: R - m*g = m*a

1.7*m*g - m*g = m*a

a = 0.7*m*g/m = 0.7*g = 0.7*9.8 = 6.86 m/s²

To determine the maximum speed:

v^{2} _{f} = v^{2} _{i} + 2as= 0 + 2(6.86)(2.3) = 31.556

v_{f} = \sqrt{31.556} = 5.62 m/s

Therefore, the elevator maximum speed is equivalent to 5.62 m/s.

8 0
3 years ago
A car has a momentum of 20,000 kg. m/s. What would the car's momentum be if its velocity doubles?
Savatey [412]

Answer:

40,000

Explanation:

Momentum is defined as mass*velocity, so a doubling of velocity means a doubling of momentum

8 0
3 years ago
An aircraft is in level flight at 225 km/hr through air at standard conditions. The lift coefficient at this speed is 0.45 and t
mojhsa [17]

Answer:

- the effective lift area for the aircraft is 8.30 m²

- the required engine thrust is 1275 N

- required power is 79.7 kW

Explanation:

Given the data in the question;

Speed V = 225 km/hr = 62.5 m/s

The lift coefficient CL = 0.45

drag coefficient CD = 0.065

mass = 900 kg

g = 9.81 m/s²

a)  the effective lift area for the aircraft

we know that for a steady level flight, weight = lift and thrust = drag

Using the equation for the lift force

F_L = C_L\frac{1}{2}ρV²A = W

we substitute

0.45 × \frac{1}{2} × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )

1081.05 × A = 8829

A = 8829 / 1081.05

A = 8.30 m²

Therefore, the effective lift area for the aircraft is 8.30 m²

b) the required engine thrust and power to maintain level flight.

we use the expression for drag force

F_D = T = C_D\frac{1}{2}ρV²A

we substitute

= 0.065 × \frac{1}{2} × 1.21 × ( 62.5 )² × 8.30

T = 1275 N

Since drag and thrust force are the same,

Therefore, the required engine thrust is 1275 N

Power required;

P = TV

p = 1275 × 62.5

p = 79687.5 W

p = ( 79687.5 / 1000 )kW

p = 79.7 kW

Therefore, required power is 79.7 kW

8 0
3 years ago
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