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3 years ago

Explain how a projectile might be modified to decrease the air resistance impacting its trajectory.

Physics

1 answer:

Vitek1552 [10]3 years ago

5 0

Answer:

Explanation:

Explain how a projectile might be modified to decrease the air resistance impacting its trajectory.

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What is used to calculate or determine intensity during exercise

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3 years ago

The driver of a pickup truck accelerates from rest to a speed of 37 mi/hr over a horizontal distance of 215 ft with constant acc

ZanzabumX [31]

Answer:

Maximum shearing force developed in each of the two pegs during acceleration is 1830 lbf

Explanation:

First we will find the acceleration of pickup truck.

As, the acceleration is uniform, therefore we can use Newton's third equation of motion:

2as = $V_{f}^{2}-V_{i}^{2}$

First convert speed into ft/sec

1 mile/hr = 1.47 ft/sec

therefore,

37 mile/hr = 37 x 1.47 ft/sec

37 mile/hr = 54.39 ft/sec

with initial speed 0 ft/sec (starting from rest), using in equation of motion:

a = [(54.39 ft/sec)² - (0 ft/sec)²]/2(215 ft)

a = 6.88 ft/sec²

Now, the total shear force will be given by Newton's second law of motion:

F = ma

F = (460 lbm +72 lbm)(6.88 ft/sec²)

F = 3660 lbf

Now for the max shear force in each of the two pegs we divide total fore by 2:

Force in each peg = F/2 = (3660 lbf)/2

<u>Force in each peg = 1830 lbf</u>

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3 years ago

A ball whose mass is 0.4 kg hits the floor with a speed of 8 m/s and rebounds upward with a speed of 6 m/s. Collapse question pa

Colt1911 [192]

Answer:

The the average magnitude of the force exerted on the ball by the floor $11.2 \times 10^{3}$ N

Explanation:

Given:

Mass of ball $m = 0.4$ kg

Initial speed $v_{i} = -8 \frac{m}{s}$

Rebound speed $v_{f} = 6 \frac{m}{s}$

Contact time interval $\Delta t = 0.5 \times 10^{-3}$ sec

For finding the average magnitude of the force on the ball by the floor is given by,

$F_{avg} = \frac{\Delta P}{\Delta t}$

Here $\Delta P = m (v_{f}- v_{i} )$

$F_{avg} = \frac{m (v_{f} -v_{i} )}{\Delta t}$

$F_{avg} = \frac{0.4 \times (6 -( -8 ) )}{0.5 \times 10^{-3} }$

$F_{avg} = 11.2 \times 10^{3}$ N

Therefore, the the average magnitude of the force exerted on the ball by the floor $11.2 \times 10^{3}$ N

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2 years ago

9. Which of the following is true regarding current in an external circuit?

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3 years ago

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