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GenaCL600 [577]

3 years ago

11

A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate

r at the bottom of the well 5.00 seconds after being dropped from the tower. Calculate the depth of the well. Given: g = -9.81 meters/second2.

1 answer:

Elan Coil [88]3 years ago

7 0

Make the base of the building zero. Then the initial distance is 100m, final distance unknown x. Use gravity, time and initial velocity to solve for final distance.
x - 100 = (0)(5) +(1/2)(-9.81)(5^2)
x - 100 = 0 - 122.625
x = -122.625 + 100
x = -22.625 m below ground

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A point P1 is located by the vector A = (3.74)î + (1.64)ĵ and a point P2 is located by the vector B = (1.60)î + (3.66)ĵ. The vec

seropon [69]

Answer:

$\vec{C} = ( \ - 2.14 \ , \ 2.02 \ )$

Explanation:

The vector that point from point P1 to point P2 its found simply by taking the vector at which point P2 its located and subtracting the vector at which point P1 its located:

$\vec{C} = \vec{B} - \vec{A}$

So:

$\vec{C} = ( \ 1.60 \ , \ 3.66 \ ) - ( \ 3.74 \ , \ 1.64 \ )$

$\vec{C} = ( \ 1.60 \ - \ 3.74 \ , \ 3.66 \ - \ 1.64 \ )$

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3 years ago

A racecar is equipped with a computer that records the reading on its speedometer every second during a race. If you graph this

Tanya [424]

Because it records speed of the car at a certain time, the independent variable should be time and dependent would be speed or velocity. Since it's taken every second, it would be considered instantaneous velocity, which is D.

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2 years ago

The AC voltage source is connected to an inductor and a resistor in series. If the frequency of the source is increased the curr

DaniilM [7]

Answer:

If the frequency of the source is increased the current in the circuit will decrease.

Explanation:

The current through the circuit is given as;

$I = \frac{V}{Z}$

Where;

V is the voltage in the AC circuit

Z is the impedance

$Z = \sqrt{R^2 + X_L^2}$

Where;

R is the resistance

$X_L$ is the inductive reactance

$X_L$ = ωL = 2πfL

where;

L is the inductance

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Finally, the current in the circuit is given as;

$I = \frac{V}{\sqrt{R^2 + (2\pi fL)^2} }$

From the equation above, an increase in frequency (f) will cause a decrease in current (I).

Therefore, If the frequency of the source is increased the current in the circuit will decrease.

5 0

3 years ago

A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

Evgen [1.6K]

Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

<h3>b) Time </h3>

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :

$V_{2}=V_{1}+at$ (6)

$V_{2}^{2}=V_{1}^{2}+2ad$ (7)

Where:

$a$ is the acceleration

$d=0.01 m$ is the length the ball was compressed

$t$ is the time

Finding $a$ from (7):

$a=\frac{V_{2}^{2}-V_{1}^{2}}{2d}$ (8)

$a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)}$ (9)

$a=90.25 m/s^{2}$ (10)

Substituting (10) in (6):

$57 m/s=-38 m/s+(90.25 m/s^{2})t$ (11)

Finding $t$ :

$t=1.052 s$ (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force $F$ is proportional to the variation of momentum $\Delta p$ in time $\Delta t$ :

$F=\frac{\Delta p}{\Delta t}$ (13)

$F=\frac{65.075 kgm/s}{1.052 s}$ (14)

Finally:

$F=61.82 N$

6 0

3 years ago

Read 2 more answers

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Answer:

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mass = 1.6 g = 1.6 x $10^{-3}$ kg

weight of the bird under acceleration due to gravity = mg

where g = acceleration due to gravity = 9.81 m/s^2

weight of the bird = 1.6 x $10^{-3}$ x 9.806 = 0.0156 ≅ 0.016 N

for this bird to maintain flight, the least lift upward, it must generate must be equal to its weight downwards, i.e

lift = weight

therefore,

lift = <em>0.016 N</em>

<em></em>

For the bustard of Europe and Asia,

mass = 21 kg

weight of the bird under acceleration due to gravity = mg

weight of the bird = 21 x 9.806 = 205.9 N

lift = weight = <em>205.9 N</em>

<em></em>

<em>lift generated is proportional to the wing surface area according to the lift equation</em>

L = Cs x p x $\frac{v}{2}$ x S

where L = lift

C = lift coefficient

p = density of air

v = relative velocity of bird and air

S = surface are of the wing.

<em>The great bustard will have a proportionally larger wing area to hold its weight in flight</em>

<em></em>

7 0

3 years ago