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3 years ago

How much work is it to lift a 6.35 kg Simon 2.14 meters to the top of his cat tree

Physics

1 answer:

Sloan [31]3 years ago

8 0

Answer:

135.89J

Explanation:

weight = mass × gravitational field strength

= 6.35kg × 10N/kg

= 63.5N

work done = force × distance

= 63.5N × 2.14m

= 135.89J

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a tungsten and coil has a resistance of 12 ohm at 15 degree celsius is the temperature coefficient of resistance of tungsten is

Dennis_Churaev [7]

Answer:

The resistance of the tungsten coil at 80 degrees Celsius is 15.12 ohm

Explanation:

The given parameters are;

The resistance of the tungsten coil at 15 degrees Celsius = 12 ohm

The temperature coefficient of resistance of tungsten = 0.004/°C

The resistance of the tungsten coil at 80 degrees Celsius is found using the following relation;

R₂ = R₁·[1 + α·(t₂ - t₁)]

Where;

R₁ = The resistance at the initial temperature = 12 ohm

R₂ = The resistance of tungsten at the final temperature

t₁ = The initial temperature = 15 degrees Celsius

t₂ = The final temperature = 80 degrees Celsius

α = temperature coefficient of resistance of tungsten = 0.004/°C

Therefore, we have;

R₂ = 12×[1 + 0.004×(80 - 15)] = 15.12 ohm

The resistance of the tungsten coil at 80 degrees Celsius = 15.12 ohm.

6 0

3 years ago

I will gib brainlyest or whatever.

astraxan [27]

Answer:

Range of the projectile: approximately $1.06 \times 10^{3}\; {\rm m}$ .

Maximum height of the projectile: approximately $80\; {\rm m}$ (approximately $45.0\; {\rm m}$ above the top of the cliff.)

The projectile was in the air for approximately $7.07\; {\rm s}$ .

The speed of the projectile would be approximately $155\; {\rm m \cdot s^{-1}}$ right before landing.

(Assumptions: drag is negligible, and that $g = 9.81\; {\rm m\cdot s^{-1}}$ .)

Explanation:

If drag is negligible, the vertical acceleration of this projectile will be constantly $a_{y} = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}$ . The SUVAT equations will apply.

Let $\theta$ denote the initial angle of elevation of this projectile.

Initial velocity of the projectile:

vertical component: $u_{y} = u\, \sin(\theta) = 153\, \sin(11.2^{\circ}) \approx 29.71786\; {\rm m\cdot s^{-1}}$
horizontal component: $u_{x} = u\, \cos(\theta) = 153\, \cos(11.2^{\circ}) \approx 150.086\; {\rm m\cdot s^{-1}}$ .

Final vertical displacement of the projectile: $x_{y} = (-35)\; {\rm m}$ (the projectile landed $35\: {\rm m}$ below the top of the cliff.)

Apply the SUVAT equation $v^{2} - u^{2} = 2\, a\, x$ to find the final vertical velocity $v_{y}$ of this projectile:

${v_{y}}^{2} - {u_{y}}^{2} = 2\, a_{y}\, x_{y}$ .

$\begin{aligned} v_{y} &= -\sqrt{{u_{y}}^{2} + 2\, a_{y} \, x_{y}} \\ &= -\sqrt{(29.71786)^{2} + 2\, (-9.81)\, (-35)} \\ &\approx (-39.621)\; {\rm m\cdot s^{-1}}\end{aligned}$ .

(Negative since the projectile will be travelling downward towards the ground.)

Since drag is negligible, the horizontal velocity of this projectile will be a constant value. Thus, the final horizontal velocity of this projectile will be equal to the initial horizontal velocity: $v_{x} = u_{x}$ .

The overall final velocity of this projectile will be:

$\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \sqrt{(150.086)^{2} + (-39.621)^{2}} \\ &\approx 155\; {\rm m\cdot s^{-1}} \end{aligned}$ .

Change in the vertical component of the velocity of this projectile:

$\begin{aligned} \Delta v_{y} &= v_{y} - u_{y} \\ &\approx (-39.621) - 29.71786 \\ &\approx 69.3386 \end{aligned}$ .

Divide the change in velocity by acceleration (rate of change in velocity) to find the time required to achieve such change:

$\begin{aligned}t &= \frac{\Delta v_{y}}{a_{y}} \\ &\approx \frac{69.3386}{(-9.81)} \\ &\approx 7.0682\; {\rm s}\end{aligned}$ .

Hence, the projectile would be in the air for approximately $7.07\; {\rm s}$ .

Also the horizontal velocity of this projectile is $u_{x} \approx 150.086\; {\rm m\cdot s^{-1}}$ throughout the flight, the range of this projectile will be:

$\begin{aligned}x_{x} &= u_{x}\, t \\ &\approx (150.086)\, (7.0682) \\ &\approx 1.06 \times 10^{3}\; {\rm m} \end{aligned}$ .

When this projectile is at maximum height, its vertical velocity will be $0$ . Apply the SUVAT equation $v^{2} - u^{2} = 2\, a\, x$ to find the maximum height of the projectile (relative to the top of the $35\; {\rm m}$ cliff.)

$\begin{aligned}x &= \frac{{v_{y}}^{2} - {u_{y}}^{2}}{2\, a} \\ &\approx \frac{0^{2} - 29.71786^{2}}{2\, (-9.81)} \\ &\approx 45.0\; {\rm m}\end{aligned}$ .

Thus, the maximum height of the projectile relative to the ground will be approximately $45.0\; {\rm m} + 35\; {\rm m} = 80\; {\rm m}$ .

5 0

1 year ago

USATESTPREP

natulia [17]

Answer:

C, D, and E

Explanation:

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2 years ago

Explain each <br> Mechanical weathering:<br> Chemical weathering:

marissa [1.9K]

Mechanical weathering is the process of breaking big rocks into little ones. This process usually happens near the surface of the planet. Temperature also affects the land. The cool nights and hot days always cause things to expand and contract. That movement can cause rocks to crack and break apart.

Chemical weathering occurs when chemical reactions weaken and decompose rocks, often acting alongside the physical breakdown of rock, aka mechanical weathering. This process involves a chemical change, which actually alters the rock's or mineral's chemical composition.

3 0

3 years ago

Use the graph below to answer the following question: What is happening to the object's velocity?

hjlf

The object's velocity is decreasing.

Explanation:

From graph is the attached image, we can clearly point that the velocity of this motion is decreasing with time.

Velocity is a vector quantity.

The y-axis represent displacement.
The x-axis depicts time
Using the graph, we know that the slope of the line on the graph gives us the velocity as it denotes the change of displacement with time.
When we find the slope, it will give us a negative value which shows that the body is slowing down and not increasing speed.

learn more:

Velocity brainly.com/question/4460262

#learnwithBrainly

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4 years ago