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Alex73 [517]
3 years ago
7

A tennis ball is tossed upward with a speed of 3.0 m/s. We can ignore air resistance. What is the velocity of the ball .40 secon

ds after the toss?
Physics
2 answers:
alexira [117]3 years ago
6 0

Answer:

<em>Hello, The velocity of the ball is 0.92 m/s in the downward direction (-0.92 m/s).</em>

Explanation:

The equation for the velocity of an object thrown upward is the following:

v = v0 + g · t

Where:

v = velocity of the ball.

v0 = initial velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

t = time.

To find the velocity of the ball at t = 0.40 s, we have to replace "t" by 0.40 s in the equation:

v = v0 + g · t

v = 3.0 m/s - 9.8 m/s² · 0.40 s

v = -0.92 m/s

The velocity of the ball is 0.92 m/s in the downward direction (-0.92 m/s).

<em>Hope That Helps!</em>

aksik [14]3 years ago
4 0

Answer:

It’s actually -0.92

Explanation:

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Answer:

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Explanation:

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Now,\\For\ Average\ Velocity\ we\ need\ to\ consider\ displacement\ as:\\Average\ Velocity\ =\frac{Total\ Displacement}{Total\ Time} \\Now,\\As\ we\ already\ know\ that\ displacement\ is\ the\ shortest\ distance\\ from\ the\ initial\ to\ the\ final\ point.\\We\ observe\ that, \\The\ car\ forms\ a\ right\ triangle\ during\ its\ complete\ journey.\\Hence,\\As\ we\ already\ know\ that,\\Distance\ travelled\ Eastwards= 120\ km\\Distance\ travelled\ Northwards= 160\ km\\Hence,\\We\ may\ apply\ Pythagoras\ Property\ to\ find\ the\ net\ displacement.\\Hence,\\a^2+b^2=c^2\\120^2+160^2=c^2\\14400+25600=c^2\\40000=c^2\\c=\sqrt{40000}\\c=200\\Hence,\\Total\ displacement=200\ km\\Total\ Time\ taken=2+8=10\ hours\\Hence,\\Average\ Velocity\ Of\ the\ Car=\frac{200}{10}=20\ km/h

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A 2.0-kg block is on a perfectly smooth (frictionless) ramp that makes an angle of 30^\circ30 ​∘ ​​ with the horizontal. What is
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Answer:

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In response to mgcosθ ramp will apply a normal force to the block which will be of equal magnitude to that of mgcosθ.

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An imaginary line perpendicular to a reflecting surface is called ____refrac_____.
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3 years ago
For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}

T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

8 0
3 years ago
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