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1 month ago

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The position of an object at any time t is given by s(t)=3t4−40t3+126t2−9. i. Determine the velocity of the object at any time t

. [5 marks] ii.Does the object ever stop changing? [5 marks] iii.When is the object moving to the right and when is the object moving to the left?

1 answer:

Sav [3.1K]1 month ago

0 0

Answer:

1.) V = 12t^3 - 120t^2 + 252t

2.) No

3.) The object has positive V and negative V.

Explanation: The position of the object at any time t is expressed as s(t)=3t^4−40t^3+126t^2−9.

1.) To find the velocity V, we differentiate this function concerning t.

V = ds/dt

ds/dt yields V = 12t^3 - 120t^2 + 252t.

2.) This equation showcases an exponential form indicating that the object's state is constantly changing.

3.) The object progresses to the right at a positive velocity V and shifts left at a negative velocity V.

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1 month ago

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A 55 kg gymnast wedges himself between two closely spaced vertical walls by pressing his hands and feet against the walls. Part

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2 months ago

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

Softa [3030]

Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle $\theta$ to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

where $y_o$ represents the initial height from ground level, $v_{oy}$ is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

$v_y=v_{oy}-gt$

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of $v_o$ .

The peak height is attained when $v_y=0$ , which allows us to compute the time to reach that height.

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum height reached is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is equal to 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Continuing with the calculations for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Substituting known values yields

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

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2 months ago

In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we

Yuliya22 [3333]

Answer:

The typical weight of a human heart is approximately 0.93 lbs.

Explanation:

Based on this,

the heart's weight constitutes about 0.5% of total body mass.

Total human weight = 185 lbs

Let the entire body weight be represented as w and the heart's weight as $w_{h}$ .

We aim to determine the heart's weight for a human

Using the provided information

$w_{h}=0.5\times w$

Where, h = heart weight

w = human weight

$w_{h}=\dfrac{0.5}{100}\times 185$

$w_{h}=0.93\ lbs$

The final weight of a human heart is 0.93 lbs.

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2 months ago

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Softa [3030]

Bernoulli's equation at a point on the streamline is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = air density, 0.075 lb/ft³ (under standard conditions)
g = 32 ft/s²

Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s

Point 2 (stagnation):
The velocity at the stagnation point is zero.

The density stays constant.
Let p₂ denote the pressure at the stagnation location.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
= 314.37 lb/ft²
= 314.37/144 = 2.18 lb/in²

Thus, the answer is 2.2 psi

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1 month ago