A bowling ball with a mass of 4.5 kg travels at a velocity of 37 m/s for 2.5 s until it stopped at the end of the lane by the ball return. What additional info is required to determine the weight of the bowling ball?

Answers

Answer 1

Answer: None. We already have more than we need ... provided only that the lane is on Earth.

Weight = (mass) (gravity).

On Earth, 'gravity' = 9.8 m/s^2.

Weight of the ball =

(4.5 kg) (9.8 m/s^2).= 44.1 newtons.

(roughly 9.9 pounds)

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Which of these is the unit of measure for work?a. watts
b. joules
c. pounds
d. newtons

Answers

The SI unit of measure for work, as well as
for all other kinds of energy, is the "joule".

Which of these is the unit of measure for work?

B. Joules

Suppose you decide to cut the wire into two pieces (not necessarily the same length) to shape into two circles. Write a function A ( x ) which models the total area of the two circles in terms of x , the length of one of the pieces of wire. (It may be helpful to use for a circle A = πr 2 and C = 2 πr ). Then find the length x that will minimize the total area.

Answers

Answer:

$2\pi r(r+rx)$

Explanation:

Let the area of one circular side be given by the formula : $A_(1) = \pi r^(2)$

However, the wire is a solid cylinder, then it means that the total area is 2 × $\pi r^(2)$ = $2\pi r^(2)$

However, there is the surface area to consider. This is the curved area of the wire. This is given as:

$A_(2) = lb$

The length is x.

The breadth is calculated as follows - the length of the circle = $\pi D = 2\pi r$

Then the area = lb

= $2\pi rx$

Therefore, the total area is given as $A_(1) + A_(2)$

= $2\pi r^(2) + 2\pi rx\n 2\pi r(r+rx)$

The force of attraction between things that have mass is called ________________.

Answers

I would assume gravitation.

You throw a rock out of a 7th story window. you time that it takes 3.7 seconds to hit the ground, and measure that it hit the ground 115 m from the base of the building? How fast must you have thrown the rock?
How high up must the 7th floor be?
If you had thrown it at the same speed but at the 28th floor, how far from the base of the building would it land?

Answers

1) 31.1 m/s

The rock has been thrown straight out of the window: its motion on the horizontal direction is simply a uniform motion, with constant speed $v$ , because no forces act in the horizontal direction. The speed in a uniform motion is given by

$v=(S)/(t)$

where S is the distance traveled and t the time taken.

In this case, the distance by the rock before hitting the ground is $S=115 m$ and the time taken is $t=3.7 s$ , so the initial speed is given by

$v=(115 m)/(3.7 s)=31.1 m/s$

2) 67.1 m

In this part of the problem we are only interested in the vertical motion of the rock. The vertical motion is a uniformly accelerated motion, with constant acceleration $a=9.8 m/s^2$ (acceleration of gravity) towards the ground. In a uniformly accelerated motion, the distance traveled by the object is given by

$S=(1)/(2)at^2$

where t is the time. Substituting a=9.8 m/s^2 and t=3.7 s, we can find S, the vertical distance covered by the rock, which corresponds to the height of the 7th floor:

$S=(1)/(2)(9.8 m/s^2)(3.7 s)^2=67.1 m$

3) 230.1 m

The height of the 7th floor is 67.1 m. So we can assume that the height of each floor is

$h=(67.1 m)/(7)=9.6 m$

And so, the height of the 28th floor is

$h=28\cdot 9.6 m=268.8 m$

We can find the total time of the fall in this case by using the same formula of the previous part:

$S=(1)/(2)at^2$

In this case, S=268.8 m, so we can re-arrange the formula to find t

$t=\sqrt{(2S)/(g)}=\sqrt{(2(268.8 m))/(9.8 m/s^2)}=7.4 s$

And now we can consider the motion of the rock on the horizontal direction: we know that the rock travels at a constant speed of v=31.1 m/s, so the distance traveled is

$S=vt=(31.1 m/s)(7.4 s)=230.1 m$

And this is how far from the building the rock lands.

speed = distance/time taken
115/3.7 =31.08m/s^2

Our solar system formed from a huge cloud of dust and gas called a _____.a. disk
b. protostar
c. solar nebula
d. planetesimal

Answers

The correct answer is C. Solar nebula

Explanation:

According to scientists, the Solar nebula was rotating disk or cloud of dust and gas (mainly hydrogen and helio) that is believed was the origin of our solar system. Indeed, it is estimated the formation of our solar system began 4.6 billion years ago as the solar nebula began contracting forming first the sun in the center and then other elements such as planets. This theory was first proposed in 1734 by Emanuel Swedenborg; however, from the first proposal, multiple models have emerged related to this theory and nowadays, the solar nebula theory is the one that scientist believe explain the formation of our solar system and other solar systems. Thus, our solar system formed from a huge cloud of dust and gas called a solar nebula.

Our solar system formed from a huge cloud of dust and gas called a
c. solar nebula

Based on the nebular hypothesis, our solar system formed from hydrogen gas and interstellar dust. The gas and the dust contracted and formed the early stage of the sun.

The density of mercury is 13.6 g/mL. What is the volume of a 155-gram sample of mercury?

Answers

The volume of a 155-gram sample of mercury is 11.397 mL if the density of mercury is 13.6 g/mL.

What is density?

It is defined as the mass-to-volume ratio. The density indicates the object's density and is represented by the symbol. The density is measured in kilograms per cubic meter.

It is given that:

The density of mercury is 13.6 g/mL.

The mass of the mercury is 155 grams.

As we know,

Volume can be defined as a three-dimensional space enclosed by an object or thing.

The density = mass/volume

13.6 = 155/volume

volume = 155/13.6

volume = 11.397 mL

Thus, the volume of a 155-gram sample of mercury is 11.397 mL if the density of mercury is 13.6 g/mL.

Learn more about the density here:

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