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4 years ago

PLEASE HELP ASAP!!

1 answer:

3 0

Velocity is displacement over time, so do 60/8 to get 7.5 yards per second, and remember to add the direction the ball is traveling in.

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The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the dist

Sidana [21]

Complete Question

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.08m. She sets the pendulum swinging, and her collaborators carefully count 101 complete cycles of oscillation during 2.00×102 s. What is the result? acceleration due to gravity:acceleration due to gravity: m/s2

Answer:

The acceleration due to gravity is $g = 167.2 \ m/s^2$

Explanation:

From the question we are told that

The length of the simple pendulum is $L = 1.081.08 \ m$

The number of cycles is $N = 101$

The time take is $t = 2.00 *10^{2 \ }s$

Generally the period of this oscillation is mathematically evaluated as

$T = \frac{N}{t }$

substituting values

$T = \frac{101}{2.0*10^2 }$

$T = 0.505 \ s$

The period of this oscillation is mathematically represented as

$T = 2 \pi \sqrt{\frac{l}{g} }$

making g the subject of the formula we have

$g = \frac{L}{[\frac{T}{2 \pi } ]^2 }$

$g = \frac{4 \pi ^2 L }{T^2 }$

Substituting values

$g = \frac{4 * 3.142 ^2 * 1.08 }{505.505^2 }$

$g = \frac{4 * 3.142 ^2 * 1.08 }{0.505^2 }$

$g = 167.2 \ m/s^2$

7 0

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)=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

4 years ago

What is the main function of an earthworm organ

Annette [7]

Answer:

brain

Explanation:

brain and blood

7 0

3 years ago

The cheetah, the fastest of land animals, can run a distance of 274 m in 8.65 s at its top speed. What is the cheetahs tips spee

ycow [4]

Answer:

Top speed is 31.68 m/s

Explanation:

Speed of a body is the distance traveled by the body in unit time.

Speed is generally expressed in meter per second or kilometer per hour.

Here, the cheetah travels a distance of 274 m in 8.65 s.

Therefore, speed of cheetah is the distance traveled by it in one second.

For 8.65 s, distance traveled is 274 m

For 1 s, distance traveled will be $\frac{274}{8.65}$ meters per second. This value is called speed.

So, speed is given as the ratio of the distance traveled to that of the time taken.

∴ Top speed of cheetah is given as:

$v_{max}=\frac{274}{8.65}=31.68 \textrm{ m/s}$

Here, $v_{max}$ is the top speed of cheetah.

7 0

4 years ago

Pulsars are neutron stars that emit X rays and other radiation in such a way that we on Earth receive pulses of radiation from t

stira [4]

Answer:

(a) $a_{c} = 5.41\times 10^{9} m/s^{2}$

(b) $a_{t} = 2.99\times 10^{- 5} m/s^{2}$

Given:

Time period of Pulsar, $T_{P} = 33.085 ms == 33.085\times 10^{- 3} s$

Equatorial radius, R = 15 Km = 15000 m

Spinning time, $t_{s} = 9.50\times 10^{10}$

Solution:

(a) To calculate the value of the centripetal acceleration, $a_{c}$ on the surface of the equator, the force acting is given by the centripetal force: