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2 years ago

An object on Earth weighs 150 N. What is its mass?

Physics

1 answer:

Alex777 [14]2 years ago

3 0

Answer:

$\tt \: mass \: = \frac{weight}{acceleration \: due \: to \: gravity}$

$\longrightarrow \tt \: \frac{150}{10}$

$\longrightarrow \boxed{ \tt{15 \: kg}}$

Our final answer is 15 kg .

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I believe the answer is a1

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3 years ago

The Burj Khalifa is the tallest building in the world at 828 m. How much work would a man with a weight of 700 N do if he climbe

Hoochie [10]

Answer:

579600J

Explanation:

Given parameters:

Height of the building = 828m

Weight of the man = 700N

Unknown:

Work done by the man = ?

Solution:

The work done by the man is the same as the potential energy expended.

Work done:

Work done = Weight x height = 700 x 828

Work done = 579600J

6 0

2 years ago

Three-fourths of the area of a rectangular lawn 30 feet wide by 40 feet long is to be enclosed by a rectangular fence. If the en

satela [25.4K]

Answer:

The fence is 5feet less.

Explanation:

We need to determine

The less amount of fence required, if the enclosure has full width and reduced length, compared to full length and reduced width.

Approach & WorkingArea of lawn = 30 × 403/4th of the area of lawn = ¾(30 × 40) = 30 * 30

When full width will be fenced, and reduced length will be fenced.

Width = 30 feet30 * L = 30 * 30Hence, length = 30 feetLength of fence needed = 2(30 + 30) = 120 feet

When full length will be fenced, and reduced width will be fenced

Length = 40 feet40 * W = 30 * 30W = 22.5 feetLength of fence needed = 2(40 + 22.5) = 125 feet

Difference in length of fence needed = 125 – 120 = 5 feet.

4 0

3 years ago

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,