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6 days ago

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How, if at all, would the equations written in Parts C and E change if the projectile was thrown from the cliff at an angle abov

e the horizontal? Explain your answer.

1 answer:

Keith_Richards [3.1K]6 days ago

7 0

Answer:

x = v₀ cos θ t, y = y₀ + v₀ sin θ t - ½ g t2

Explanation:

This pertains to a projectile motion scenario. Here, we will express the equations for both the x and y dimensions.

Now, we will apply trigonometry to determine the initial velocity components.

sin θ = $v_{oy}$ / v₀

cos θ = v₀ₓ / v₀

v_{y} = v_{oy} sin θ

v₀ₓ = v₀ cos θ

Next, let's formulate the equations of motion.

X axis

x = v₀ₓ t

x = v₀ cos θ t

vₓ = v₀ cos θ

Y axis

y = y₀ + $v_{oy}$ t - ½ g t2

y = y₀ + v₀ sin θ t - ½ g t2

v_{y} = v₀ - g t

v_{y} = v₀ sin θ - gt

$v_{y}^{2}$ = v_{oy}^2 sin² θ - 2 g y

It is evident that the major distinction lies in the fact that in an inclined launch compared to a horizontal one, the velocity comprises different components

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Albert presses a book against a wall with his hand. As Albert gets tired, he exerts less force, but the book remains in the same

Maru [3280]

Answer:

the maximum static friction force of the wall acting on the book (Increasing)

the normal force of the wall acting on the book (Decreasing)

the weight of the book (Constant)

Explanation:

According to Newton's third law of motion:

"Every action has an equal and opposite reaction"

In the scenario provided, Albert is pressing the book against the wall and subsequently decreases the force applied against the wall.

Let's evaluate all forces influencing the book in this situation.

1. Weight of the book acting downwards (y-axis)

2. Friction from the book against the wall acting upwards (y-axis)

3. Albert’s force exerted on the book against the wall (x-axis)

4. Normal force of the wall reacting to Albert’s applied force (x-axis)

As Albert eases off his force, the new scenario reads:

1. The weight remains constant as represented by W = mg

Since neither mass nor gravitational acceleration has changed, the weight exerted on the book remains the same.

2. As Albert reduces his force, the wall’s normal reaction force decreases correspondingly, following Newton's third law of motion.

3. Friction operates in response to the force applied to it. With a box resting on the floor, no friction acts upon it until it is dragged, at which point friction begins to manifest and rise until it reaches its maximum. Therefore, when Albert diminishes his force, the weight's pull will influence the book and the maximum static friction will rise to resist the book’s downward movement.

It should be noted that the maximum static friction is working to prevent movement of the book. With Albert's force reduced, but the weight of the book unchanged, maximum static friction increases to prevent downward movement.

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1 month ago

In the past, salmon would swim more than 1130 km (700 mi) to spawn at the headwaters of the Salmon River in central Idaho. The t

Keith_Richards [3157]

Answer:

$E_t_o_t_a_l=7.603MJ$

Explanation:

The overall energy expenditure of the salmon, which corresponds to its swimming upstream effort, $W$ , is linked to its specific mechanical power. $Mechanical \ power$ calculated per unit mass can be derived from the following equation:

$\frac{p}{m}=\frac{1}{m}|\frac{dW}{dt}|=2W/kg\\\frac{1}{2}|\frac{W}{22\times 24 \times 60\times 60}|=2\\\\W=7.603MJ$

As a result, the total energy utilized during the 22-day journey is 7.603 MJ

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1 month ago

Sully is riding a snowmobile on a flat, snow-covered surface with a constant velocity of 10 meters/second. The total mass of the

inna [2995]

Result: 168N

The calculation shows 16 - 10 equals 6
and 6 divided by 10 equals 0.6
. Therefore, F equals 280 multiplied by 0.6 equals 168.

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14 days ago

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In 2014, the Rosetta space probe reached the comet Churyumov Gerasimenko. Although the comet's core is actually far from spheric

ValentinkaMS [3372]

To tackle this issue, we will utilize concepts related to gravity based on Newtonian definitions. To find this value, we'll apply linear motion kinematic equations to determine the required time. Our parameters include:

Comet mass $M = 1.0*10^{13} kg$

Radius $r = 1.6km = 1600 m$

The rock is released from a height 'h' of 1 m above the surface.

The relationship for gravity's acceleration concerning a body with mass 'm' and radius 'r' is described by:

$g = \frac{GM}{R^2}$

Where G represents the gravitational constant and M denotes the mass of the planet.

$g = \frac{(6.67408*10^{-11})(1*10^{13})}{1600^2}$

$g = 2.607*10^{-4} m/s^2$

Now, let’s compute the time value.

$h = \frac{1}{2} gt^2$

$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2(1)}{2.607*10^{-4}}}$

$t = 87.58s$

Ultimately, the time for the rock to hit the surface is t = 87.58s.

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1 month ago

A 55-kg pilot flies a jet trainer in a half vertical loop of 1200-m radius so that the speed of the trainer decreases at a const

Keith_Richards [3157]

a) $-1.54 m/s^2$

b) 803.4 N

Explanation:

a) At point C (top of the loop), the pilot experiences weightlessness, leading to the normal force from the seat being zero:

N = 0

Consequently, the force balance equation at position C becomes:

$mg=m\frac{v^2}{r}$ where the left term signifies the pilot's weight and the right term represents the centripetal force, with:

= acceleration due to gravity

= jet's velocity at the top $g=9.8 m/s^2$

= loop radius

$v$ By solving for v,

$r=1200 m$

Thus, this is the jet's speed at C.

The speed at position A (bottom) can be derived from $v_C=\sqrt{gr}=\sqrt{(9.8)(1200)}=108.4 m/s$

The distance traveled by the jet corresponds to half the circumference of the circle with radius r, therefore

$v_A=550 km/h =152.8 m/s$

Given the plane's deceleration is consistent, we can obtain it using the following equation:

$s=\pi r=\pi(1200)=3770 m$

b) The pilot experiences a force equal to the normal force from the seat. At point B (half-way through the loop), we find:

$v_C^2-v_A^2=2as\\a=\frac{v_C^2-v_A^2}{2s}=\frac{108.4^2-152.8^2}{2(3770)}=-1.54 m/s^2$ - The normal force from the seat, N, directed towards the center of the loop

- As there are no further forces acting toward the central axis, N must equal the centripetal force:

(1)

where

represents the speed at position B.

To deduce the velocity at B, we note that the distance covered by the jet between positions A and B is a quarter of a circle:

With knowledge of the deceleration, we can implement the equation of motion to find the velocity at the midway point B: $N=m\frac{v_B^2}{r}$

$v_B$

Thus, we then apply eq.(1) to determine the normal force acting on the pilot at B:

$s=\frac{\pi r}{2}=\frac{\pi(1200)}{2}=1885 m$

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21 day ago