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V125BC [204]
3 years ago
15

Your roommate drops your wallet down to you from the third-floor window of your apartment, which is 11.5 m from the ground. What

is the velocity of the wallet when you catch it? (assume you catch the wallet right before it hits the ground)
Physics
1 answer:
Aneli [31]3 years ago
6 0

Answer:

15 m/s

Explanation:

Using the law of conservation of energy, potential energy equals kinetic energy hence

mgh=0.5mv^{2}

Therefore

v=\sqrt{2gh}

where g is the acceleration due to gravity, m is the mass of the object, h is the height and v is the speed of the wallet

Taking g as 9.81 then

v=\sqrt{2\times 9.8\times 11.5}=15.02098532  m/s\approx 15 m/s

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a pillow , a textbook and a paper airplane are dropped from the top of a tall building at the same time. consider what you have
MAVERICK [17]

A textbook would hit the ground first


Factors:

-Textbook weighs most

-Pillow is flat and fluffy not very aerodynamic) also is very light

-Paper airplane will glide to the ground do to its wings and will hit the ground last

3 0
3 years ago
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Dont use me for points please i need help, (circuits for physics)
zhannawk [14.2K]

Answer:

B, A, A, B

Explanation:

Just trust me on this one.

3 0
2 years ago
57:07
Reptile [31]

Answer:

v (speed) = S / t = 4 * 400 m / (6 * 60 sec) = 4.4 m/s

The average velocity  is zero because there is no net vector displacement.

5 0
3 years ago
A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th
svp [43]

a) x(t)=2.0 sin (10 t) [m]

The equation which gives the position of a simple harmonic oscillator is:

x(t)= A sin (\omega t)

where

A is the amplitude

\omega=\sqrt{\frac{k}{m}} is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s

The amplitude, A, can be found from the maximum velocity of the spring:

v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m

So, the equation of motion is

x(t)= 2.0 sin (10 t) [m]

b)  t=0.10 s, t=0.52 s

The potential energy is given by:

U(x)=\frac{1}{2}kx^2

While the kinetic energy is given by:

K=\frac{1}{2}mv^2

The velocity as a function of time t is:

v(t)=v_{max} cos(\omega t)

The problem asks as the time t at which U=3K, so we have:

\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}

However, \frac{m}{k}=\frac{1}{\omega^2}, so we have

(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\

with two solutions:

\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s

\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s

c) 3 seconds.

When x=0, the equation of motion is:

0=A sin (\omega t)

so, t=0.

When x=1.00 m, the equation of motion is:

1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s

The period of a pendulum is:

T=2 \pi \sqrt{\frac{L}{g}}

where L is the length of the pendulum. By using T=0.628 s, we find

L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m






5 0
3 years ago
Why is the light bulb not a closed system?
Nana76 [90]

Answer:An incandescent light bulb gives only energy of the system in the form of heat. ... The roof of a house is stable but it receives energy from the surroundings and transfers energy through it towards lower temperature side which includes no mass transfer. So, it is best considered as closed system.

Explanation:

4 0
2 years ago
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