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2 years ago

Which statement is true about a falling object? (Assume no

1 answer:

vladimir2022 [97]2 years ago

5 0

I would say the answer is 3 because by falling technically the ball would be kind of moving in the air. Plus potential energy is when for example a soccer ball isnt moving

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Which symbol represents a type of electromagnetic radiation released during radioactive decay?

Tanya [424]

The answer for <span>electromagnetic radiation released during radioactive decay i</span>s C. He

8 0

3 years ago

During an experiment, Ellie records a measurement of 0.0034 m. How would

Goshia [24]

Answer:

(A) She needs to move the decimal point by 3 places

8 0

3 years ago

What is the slope of the line if the rise of a line on a distance versus-time graph is 900 meters and the run is 3 minutes?

Alexandra [31]

600/3 = 200
the slope is 200m/min

OR

600/ (3/60) =
600 x 60/3 =
600 x 20 = 12,000 meters per hour

6 0

2 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)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

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

⇒A₀ $e^{\frac{-bt}{2m}}$ =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

Jet fighter planes are launched from aircraft carriers with the aid of their own engines and a catapult. If in the process of be

ANTONII [103]

Answer:

W=315 x 10⁵ J

Explanation:

Given that

F= 2.5 x 10⁵ N

d= 90 m

K.E.=5.4 x 10⁷ J

We know that work done by all force is equal to the change in kinetic energy

Lets take work done by catapult is W

W + F.d= K.E.

W= 5.4 x 10⁷ - 2.5 x 10⁵ x 90 J

W= (540 - 25 x 9) 10⁵ J

W=315 x 10⁵ J

5 0

3 years ago