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3 years ago

An object accelerates from rest, with a constant acceleration of 5.2 m/s2, what will its velocity be after 3 s?

Physics

1 answer:

topjm [15]3 years ago

4 0

Answer:

V = 15.6 m/s

Explanation:

using the equation of motion

v = u + at

v = 0 + 5.2(3)

v = 15.6 m/s

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Answer:

-353.16

Explanation:

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3 years ago

On earth, the solar constant is approximately I = 1367 W/m2 . This corresponds to the power per unit area that a detector at the

d1i1m1o1n [39]

Answer:

Explanation:

for earth -sun system

E = σ T⁴ , E is radiation emitted by sun at temperature T

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4 years ago

a particle is moving with shm of period 8.0s and amplitude 5.0cm. find (a) the speed of particle when it is 3.0m from the centre

Fudgin [204]

Answer:

a) $speed=\pi cm/s$

b) $v_{max}=\frac{5\pi}{4} cm/s$

c) $a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

Explanation:

The very first thing we must do in order to solve this problem is to find an equation for the simple harmonic motion of the given particle. Simple harmonic motion can be modeled with the following formula:

$y=Asin(\omega t)$

where:

A=amplitude

$\omega$ = angular frequency

t=time

we know the amplitude is:

A=5.0cm

and the angular frequency can be found by using the following formula:

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

so our angular frequency is:

$\omega=\frac{2\pi}{8s}$

$\omega=\frac{\pi}{4}$

so now we can build our equation:

$y=5sin(\frac{\pi}{4} t)$

we need to find the speed of the particle when it is 3m from the centre of its motion, so we need to find the time t when this will happen. We can use the equation we just found to get this value:

$y=5sin(\frac{\pi}{4} t)$

$3=5sin(\frac{\pi}{4} t)$

so we solve for t:

$sin(\frac{\pi}{4} t)=\frac{3}{5}$

$\frac{\pi}{4} t=sin^{-1}(\frac{3}{5})$

$t=\frac{4}{\pi}sin^{-1}(\frac{3}{5})$

you can directly use this expression as the time or its decimal representation:

t=0.81933

since we need to find the speed of the particle at that time, we will need to get the derivative of the equation that represents the particle's position, so we get:

$y=5sin(\frac{\pi}{4} t)$

$y'=5cos(\frac{\pi}{4} t)*\frac{\pi}{4}$

which simplifies to:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

and we can now substitute the t-value we found previously, so we get:

$y'=\frac{5\pi}{4}cos(\frac{\pi}{4} (0.81933))$

$y'=\pi$

so its velocity at that point is $\pi$ cm/s

b) In order to find the maximum velocity we just need to take a look at the velocity equation we just found:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

its amplitude will always give us the maximum velocity of the particle, so in this case the amplitude is:

$A=\frac{5\pi}{4}$

so:

$v_{max}=\frac{5\pi}{4} cm/s$

c) we can use a similar procedure to find the maximum acceleration of the particle, we just need to find the derivative of the velocity equation and determine its amplitude. So we get:

$y'= \frac{5\pi}{4}cos(\frac{\pi}{4} t)$

We can use the chain rule again to find this derivative so we get:

$y" =-\frac{5\pi}{4}sin(\frac{\pi}{4} t)*(\frac{pi}{4})$

so when simplified we get:

$y"=-\frac{5\pi^{2}}{16}sin(\frac{\pi}{4} t)$

its amplitude is:

$A=\frac{5\pi^{2}}{16}$

so its maximum acceleration is:

$a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

7 0

3 years ago

A man tries to push a 200 kg Car that moves at a acceleration 0.50 m/s2. The man is able to displace the car 10 m. How much work

yawa3891 [41]

The work done by the man pushing the car over the given distance is 1000J.

Given the data in the question;

Mass of car; $m = 200kg$

Acceleration of the car; $a = 0.5m/s^2$

Distance covered by the car; $d = 10m$

Work done; $W = \ ?$

Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

$Work\ done = f * d$

Where f is force applied and d is distance travelled.

To determine the work done by the man, we first solve for the force applied F.

From Newton's Second Law; $Force \ F = m * a$

We substitute our given values into the expression

$F = m * a \\\\F = 200kg * 0.5m/s^2\\\\F = 100kg.m/s^2$

Next we substitute our values into the expression of work done above.

$Work \ done = f * d\\\\Work \ done = 100kg.m/s^2 * 10m\\\\Work \ done = 1000kgm^2/s^2\\\\Work \ done = 1000J$

Therefore, the work done by the man pushing the car over the given distance is 1000J.

Learn more about work done: brainly.com/question/26115962

7 0

2 years ago

What is the weight, in pounds, of a 205-kg object on jupiter?

Crazy boy [7]

<span>a 205 kg mass object will weight 1143.43 lbs on Jupiter. Looking up the surface gravity of Jupiter, you can find that it's 2.53 times that of earth. So the 205 kg object will weigh 205 * 2.53 = 518.65 kg on Jupiter. Now we need to convert from kg to pounds. This is done by multiplying by 2.20462 518.65 kg * 2.20462 lb/kg = 1143.43 lb</span>

4 0

3 years ago