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2 years ago

Calculate the work done by a force of 30 N in lifting a load of 2 kg to a height of 10 m (g = 10 m/s2)

Physics

1 answer:

Mars2501 [29]2 years ago

3 0

Answer:

300 J

Explanation:

Given :

Applied force = 30 N
Mass = 2 kg
Height = 10 m
g = 10 m/s²

Work done by the force :

Work done = Force x Displacement
Work done = mg x h
Work done = 30 N x 10 m
Work done = 300 J

Note :

What you have calculated is the work done by gravitational force on the object (that too, incorrectly)
But in the end, it asks for work done by the force of 30N
Hence, the given answer ~

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Which is one piece of information that astronomers use to calculate the age of the universe?

Valentin [98]

Answer:

the age of the oldest stars
patterns of background radiation
how fast distant galaxies are moving away from us

Explanation:

According to the NASA, astronomers can estimate how old the universe is by comparing the age of the oldest stars and the rate of expansion of the universe. The rate of expansion of the universe is based on the Big Bang theory and it states that our universe is continuously expanding since it first developed. Astronomers measure Cosmic Microwave Background radiation resulting from the Big Bang.

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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:

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

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3 years ago

A block with mass M = 3 kg is moving on a flat surface with constant speed v1 = 12 m/s. A bullet with mass m = 0,1 kg is shot at

adelina 88 [10]

The speed does the block move after it is hit by the bullet that remains stuck inside the block will be 23.7 m/sec and it takes 12.07 seconds to stop.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

Apply the law of conservation of momentum principle;

m₁v₁+m₂v₂cosΘ =(m₁+m₂)V

3 kg × 12 m/s + 0,1 kg × 400 m/s cos 20° = (3+0.1)V

36 + 40 cos 20° = 3.1 V

V=23.7 m/sec

The time it takes to stop when the friction coefficient between the block and the surface is 0.2 is found as;

V = u +at

V = 0+ μgt

23..7=0.2× 9.81 ×t

t=12.07 sec

Hence, it takes 12.07 seconds to stop.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

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2 years ago

An unwary football player collides with a padded goalpost while running at a velocity of 5.70 m/s and comes to a full stop after

ollegr [7]

The relationship between the initial velocity, final velociy, distance, and deceleration can be expressed in the following equation.

2(a)(0.270 m) = 0² - (5.70 m/s)²

The value of a (which is the deceleration) is 0.06 m/s². Thus, the answer is that the deceleration value is approximately 0.06 m/s².

6 0

4 years ago