During a home run, the batter only needs to run around all 4 bases if he wants to, since the ball cleared the outfield fence.

In the park, there are 25 pigeons, 15 squirrels, 5 rabbits, and 5 stray cats. Why

Debora [2.8K]

Answer: Line graph should be used to show how one variable changes over time not to show multiple categories or variables are at one specific point in time.

Explanation:

In maths, statistics, and related fields, graphs are used to visually display variables and their values. In the case of line graphs, these are mainly used to display evolution or change of a variable over time. For example, a line graph can show how the number of divorces changed from 1920 to 2010.

In this context, the number of different animals in the park cannot be represented through a line graph because this situation does not imply a variable changing over time. Moreover, this situation includes multiple variables or categories of animals and the data shows only one specific point in time, which can be better represented through a bar graph.

8 0

2 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

2 years ago

What are some resources used to generate electricity in energy stations?

Setler79 [48]

-- Coal
-- Oil
-- Natural gas
-- Falling water
-- Sunlight
-- Nuclear fission of Uranium

3 0

2 years ago

A 0.245 kg ball is thrown straight up from 2.07 m above the ground. Its initial vertical speed is 8.00 m/s. A short time later,

iris [78.8K]

Answer:

The work done by gravity is $4.975 \: Joules$

Explanation:

The data given in the question is :

Mass is $0.245 kg$

Height from ground is $2.07 m$

As we know , the work done is state function , it depends on initial and final position not on the path followed.

So, work done by gravity = change in potential energy

Work done = Initial potential energy - final potential energy

Insert values from question

Work done = $mass \times gravity \times (change \: in \: height)$

Work done = $0.245 kg \times 9.81 m/s^{2} \times 2.07 m$

So, work done = $4.975 Joules$

Hence the work done by gravity is $4.975 \: Joules$

3 0

2 years ago

All atomic nuclei except those of ordinary hydrogen contain neutrons<br> true or false

MatroZZZ [7]

The answer to this question is False

8 0

2 years ago