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zhannawk [14.2K]
3 years ago
8

You accidentally slide a glass of milk off a table that is 0.9 m tall. How long does it take the milk to hit the ground.

Physics
1 answer:
labwork [276]3 years ago
7 0

Fine, lets do a retry of this.

Δd = -0.9m

v₁ = 0

v₂ = ?

a = -9.8 m/s²

Δt = ?

We can use the following kinematic equation and solve for Δt.

                             Δd = v₁Δt + 0.5(a)(Δt)²

                             Δd = 0.5(a)(Δt)²

                           2Δd = a(Δt)²

                     √2Δd/a = Δt

√2(-0.9m)/(-9.8 m/s²) = Δt

0.<u>4</u>28571428574048 = Δt

Therefore, it takes 0.4 seconds for the glass to hit the ground, or 0.43s as you said (even though I don't believe it follows significant digit rules)

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Two speakers in a stereo emit identical pure tones. As you move around in front of the speakers, you hear the sound alternating
BigorU [14]

Answer:

Interference

Explanation:

When two traveling waves traveling waves along the same path are superimposed(combine). The superimposition of these two waves results in the production of a resultant wave which is defined by the net effect of the two waves. Wave interference occurs most types of waves including radio wave, light, acoustic waves and other wave types. Alternating sound between loud and Zero is heard as the two speakers emit identical pure tones because the resultant amplitude after the interference of the two sound waves is the vector sum of each of their amplitudes. A loud sound is heard, when the crest of both waves meets each other and a zero is heard if the crest of one meets the trough of the other as they cancel out.

3 0
2 years ago
What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your an
AveGali [126]

Complete Question

A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the wheel starts turning, you are at the very top.

What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your answer as an angle between 0∘ and 360∘. Express your answer in degrees.

Answer:

\phi=123.75

Explanation:

From the question we are told that:

Height h=27m

Period T=32sec

Time t=75sec

Generally the equation for angular velocity is mathematically given by

\omega=\frac{2 \pi}{T}

\omega=\frac{2 \pi}{32}

\omega=0.196rad/s

Therefore

\theta=\omega t

\theta=0.196rad/s*75sec

\theta=843.75 \textdegree

Therefore

\phi=\theta-2(360)

\phi=123.75

6 0
3 years ago
A 100 Kg man is diving off a 50 meter cliff. What is his kinetic energy when he is 20 meters from the water?
iren2701 [21]

Answer:

K.E=29.403125J

Explanation:

From the question we are told that

Mass M=100

Height 50-20=30m

Generally the equation for velocity before impact is is is mathematically given by

v=\sqrt{2gh}

v=\sqrt{2*9.8*30}

v=24.25

Generally the equation for Kinetic Energy is is mathematically given by

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

K.E=\frac{1}{2}*100*(24.25)^2\\

K.E=29403.125J

K.E=29.403125J

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3 years ago
How are motors and generators different?
saw5 [17]
A generator converts mechanical energy into electrical energy, while a motor does the opposite - it converts electrical energy into mechanical energy
4 0
3 years ago
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
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