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NNADVOKAT [17]
3 years ago
11

A box of mass 3.6 kg is lifted 5.4 m above the

Physics
1 answer:
horsena [70]3 years ago
7 0

So, the energy change that occurs is 190.512 J.

<h3>Introduction</h3>

Hello ! I am Deva from Brainly Indonesia will help you regarding energy and its transformation. In this case, it's the use of energy from the lifter to be equivalent to the change in the object's potential energy. Why potential energy? Because the box undergoes a change in height and the potential energy specializes at a certain height. Work (W) due to change in potential energy (\sf{\Delta PE}) can be realized in the equation :

\sf{W = \Delta PE}

\sf{W = m \cdot g \cdot h_2 - m \cdot g \cdot h_2}

\boxed{\sf{\bold{W = m \cdot g \cdot (h_2 - h_1)}}}

With the following condition :

  • W = work of subject (J)
  • \sf{\Delta PE} = change of potential energy (J)
  • m = mass (kg)
  • g = acceleration of the gravity (m/s²)
  • \sf{h_2} = final height (m)
  • \sf{h_1} = initial height (m)

<h3>Problem Solving</h3>

We know that :

  • m = mass = 3.6 kg
  • g = acceleration of the gravity = 9.8 m/s²
  • \sf{h_2} = final height = 5.4 m
  • \sf{h_1} = initial height = 0 m

What was asked :

  • W = work of subject = ... J

Step by step :

\sf{W = \Delta PE}

\sf{W = m \cdot g \cdot (h_2 - h_1)}

\sf{W = 3.6 \cdot 9.8 \cdot (5.4 - 0)}

\sf{W = 3.6 \cdot 9.8 \cdot 5,4}

\boxed{\sf{W = 190.512 \: J}}

So, the energy change that occurs is 190.512 J.

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Answer: 6.24 km

Explanation:

Given

The magnitude of the first vector(say) \left |  a\right |=5\ km

the magnitude of the second vector(say) \left |  b\right |=7\ km

the angle between them is 120^{\circ}

The resultant vector magnitude is given by

\left |  \vec{R}\right |=\sqrt{a^2+b^2+2ab\cos \theta}

\left |  \vec{R}\right |=\sqrt{5^2+7^2+2\times 5\times 7\cdot \cos 120^{\circ}}\\\left |  \vec{R}\right |=\sqrt{74-35}=\sqrt{39}\\\left |  \vec{R}\right |=6.24\ km

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3 years ago
A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr
KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s

Time the parachutist falls without friction is 3.19 seconds

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2

Now the initial velocity of the last half height will be the final velocity of the first half height.

v=u+at\\\Rightarrow v=at

Since the height are equal

\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0

t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2

Magnitude of acceleration is -43200 ft/s²

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Answer:

A) Its density will decrease

Explanation:

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At the same time, the mass of the object does not change, because the mass just represents the amount of matter contained in the object, so it does not increase/decrease at different temperatures.

The density of an object is defined as the ratio between the mass (m) and the volume (V):

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We said that the mass remains unchanged while the volume increases: since the density is inversely proportional to the volume, this means that the density decreases.

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<u>Given the following data:</u>

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To determine the X and Y components of the force:

<h3>The horizontal component (X) of a force:</h3>

Mathematically, the horizontal component of a force is given by this formula:

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Fx = 90.63 Newton.

<h3>The vertical component (Y) of tensional force:</h3>

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