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3 months ago

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A running mountain lion can make a leap 10.0 mlong, reaching a maximum height of 3.0 m. What is the speed of the mountain lion j

ust as it leaves the ground? At what angle does it leave the ground?

1 answer:

ValentinkaMS [3.4K]3 months ago

5 0

To tackle this question we will apply the kinematic equations that describe the motion of a projectile, where both maximum height and distance traveled are defined. This scenario demonstrates a lion achieving a height (H) of 3m and covering a horizontal distance (R) of 10m. The equations governing this kind of motion are expressed as follows:

$H = \frac{v_0^2sin^2\theta}{2g}$

$R = \frac{v_0^2 sin 2\theta}{g}$

By dividing the two equations, we determine:

$\frac{H}{R}=\frac{\frac{v_0^2sin^2\theta}{2g}}{\frac{v_0^2 sin 2\theta}{g}}$

$\frac{H}{R}= \frac{sin^2\theta}{2}*\frac{1}{sin2\theta}$

$\frac{H}{R}= \frac{sin^2\theta}{2}*\frac{1}{2sin\theta cos\theta}$

$\frac{H}{R}= \frac{1}{4} \frac{sin\theta}{cos\theta}$

$\frac{H}{R}= \frac{1}{4} tan\theta$

Plugging in the values for H and R yields:

$\frac{3}{10} = \frac{1}{4} tan\theta$

$\theta = tan^{-1} \frac{12}{10}$

$\theta = 50.2\°$

After substituting \theta into the relevant equation, we find:

$H = \frac{v_0^2sin^2\theta}{2g}$

$v_0^2 = \frac{H 2g}{sin^2\theta}$

$v_0^2 = \frac{3*2*9.8}{sin^2(50.2)}$

$v_0^2 = 99.62$

$v_0 = \sqrt{99.62}$

$v_0 = 9.98m/s$

In conclusion, the mountain lion's launch speed upon takeoff is approximately 9.98m/s at an angle of 50.2°.

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A 2.0 kg bird lands on a 1.0 x 10^1 kg bit of tree bark sitting on a frictionless ice-covered pond. The bird’s initial horizonta

ValentinkaMS [3465]

In this scenario, the principles of momentum conservation can be applied since there are no external forces acting on the system. Consequently, the conservation of momentum principle is applicable here. After the bird lands on it, both the bird and the bark will have a unified final speed. Thus, this final speed will be 1 m/s.

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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

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

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

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

$y=7.39\ m$

5 0

3 months ago

A 1.97-pF capacitor with a plate area of 5.86 cm2 and separation between the plates of 2.63 mm is connected to a 9.0-V battery a

inna [3103]

If the plate separation is modified after the battery is disconnected, the updated distance between plates is 9.21 mm

If changes are made while the battery remains connected, the new separation becomes 0.11 mm

The capacitance for an air-filled parallel plate capacitor can be expressed as:

$C=\frac{\epsilon_0A}{d}$

In this equation, $\epsilon_0$ refers to the permittivity of free space, A stands for the plate area, and D represents the separation distance.

Thus,

$C \alpha \frac{1}{d}$ .......(1)

Therefore, should the distance between the plates shift from d₁ to d₂, the capacitance ratio in both scenarios can be represented as:

$\frac{C_1}{C_2} =\frac{d_2}{d_1}$ ......(2)

Scenario (i)

When the capacitor is fully charged and then disconnected from the battery before adjusting the plate distance, the charge will remain steady while the capacitance varies.

The initial energy E₁ stored in the capacitor can be expressed as:

$E_1=\frac{Q^2}{2C_1}$ ......(3)

Once the separation changes to d₂, capacitance becomes C₂, but the charge Q remains unchanged.

Thus,

$E_2=\frac{Q^2}{2C_2}$ ......(4)

By dividing equation (4) by (3),

$\frac{E_2}{E_1} =\frac{C_1}{C_2}$

According to equation (2),

$\frac{E_2}{E_1} =\frac{C_1}{C_2}=\frac{d_2}{d_1}$

This results in a 3.5 fold increase in energy.

$\frac{E_2}{E_1} =\frac{d_2}{d_1}=3.5\\ d_2=3.5*2.63 mm\\ =9.205 mm=9.21 mm$

Scenario (2)

If the capacitor is kept connected to the power source, the voltage V across the plates will remain unchanged.

The initial energy is described as

$E_1=\frac{1}{2} C_1V^2$ ......(5)

The final energy when the plate separation transitions to d₂ can be written as:

$E_2=\frac{1}{2} C_2V^2$ .....(6)

Referencing equations (5) and (6)

$\frac{E_2}{E_1} =\frac{C_2}{C_1}$

From equation (2),

$\frac{E_2}{E_1} =\frac{C_2}{C_1}=\frac{d_1}{d_2}$

Thus, in this particular scenario,

$\frac{E_2}{E_1} =\frac{d_1}{d_2}\\d_2=\frac{d_1}{3.5} \\ =\frac{2.63 mm}{3.5} \\ =0.109 mm=0.11 mm$

Therefore,

Adjusting plate separation after battery disconnection yields 9.21 mm

If modified while connected, the new separation measures 0.11 mm

6 0

2 months ago

To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th

Softa [3030]

Answer:

$H = 10.05 m$

Explanation:

The stone reaches the top of the flagpole at both t = 0.5 s and t = 4.1 s

therefore, the total duration of the upwards motion above the peak of the pole is provided as

$\Delta t = 4.1 - 0.5 = 3.6 s$

now we have

$\Delta t = \frac{2v}{g}$

$3.6 = \frac{2v}{9.8}$

$v = 17.64 m/s$

this indicates the speed at the flagpole's top

at this point we have

$v_f - v_i = at$

$17.64 - v_i = (-9.8)(0.5)$

$v_i = 22.5 m/s$

the height of the flagpole is stated as

$H = \frac{v_f + v_i}{2}t$

$H = \frac{22.5 + 17.64}{2} (0.5)$

$H = 10.05 m$

5 0

3 months ago

If a 50.0-kg mass weighs 554 n on the planet saturn, calculate saturn’s radius

ValentinkaMS [3465]

Answer:

17.35 × 10^(-6) m

Explanation:

Mass; m = 50 kg

Weight; W = 554 N

From the formula:

W = mg

This simplifies to; 554 = 50g

g = 554/50

g = 11.08 m/s²

Also, using the formula;

mg = GMm/r²

hence; g = GM/r²

Rearranging gives;

r = √(GM/g)

With G as a known constant of 6.67 × 10^(-11) Nm²/kg²

r = √(6.67 × 10^(-11) × 50/11.08)

r = 17.35 × 10^(-6) m

8 0

3 months ago