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3 years ago

Which one of these birds is an experimental population

Physics

2 answers:

natta225 [31]3 years ago

8 0

Whooping crane is the answer

kkurt [141]3 years ago

4 0

Answer Whooping Crane

An experimental population is a geologically depicted gathering of reintroduced plants or creatures that is segregated from other existing populaces of the species. It is done to protect the endangered species from getting extinct. Whooping crane is also one of those birds which were endangered and experimental population technique was used.

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Which statement correctly explains mechanical advantage?

Shtirlitz [24]

I'm not entirely sure but I would think it's A.

4 0

3 years ago

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Which way(s) can increase the c

Sav [38]

Answer:

A, B and F are the right answers

Explanation:

The formula is $a_{c} = v^{2} / r$

A) Increasing both the radius and the speed by a factor of 9

$a_{c} = (9v)^{2} / (9r)$ = $9v^{2} / r$

which means the acceleration will increase by a factor of 9

B) Keeping the speed fixed and decreasing the radius by a factor of 9

$a_{c} = (v)^{2} / (r/9)$ = $9v^{2} / r$

which means the acceleration will increase by a factor of 9

C) Keeping the radius fixed and increasing the speed by a factor of 9

$a_{c} = (9v)^{2} / (r)$ = $81v^{2} / r$

It will increase the acceleration 81 times

D) Keeping the speed fixed and increasing the radius by a factor of 9

$a_{c} = (v)^{2} / (9r)$ = $v^{2} / 9r$

It will decrease the acceleration by 9 times

E) decreasing both the radius and the speed by a factor of 9

$a_{c} = (v/9)^{2} / (r/9)$ = $v^{2} / 9r$

It will decrease the acceleration by 9 times

F) Keeping the radius fixed and increasing the speed by a factor of 3

$a_{c} = (3v)^{2} / (r)$ = $9v^{2} / r$

It will increase the acceleration by 9 times

4 0

3 years ago

The storage method used for radioactive wastes generated from fission nuclear reactors must be designed to last for how long?

Phantasy [73]

Answer:

The storage method is originally designed to last years. The exact amount of time comes down to what type of waste it is.

4 0

3 years ago

Read 2 more answers

A blue ball is thrown upward with an initial speed of 24.1 m/s, from a height of 0.5 meters above the ground. 2.9 seconds after

Aleks04 [339]

Answer:

1. Speed=0

2. 2.46 s

3.30.1 m

4. 22.0 m

5.1.004 s

Explanation:

We are given that

Initial speed of blue ball, u=24.1 m/s

Height of blue ball from ground y_0=0.5 m

Initial speed of red ball , u'=7.2 m/s

Height of red from ground=y'0=32 m

Gravity, g= $9.81ms^{-2}$

1.When the ball reaches its maximum height then the speed of the blue ball is zero.

2.v=0

$v=u+at$

Using the formula and substitute the values

$0=24.1-9.81t$

Where g is negative because motion of ball is against gravity

$24.1=9.81t$

$t=\frac{24.1}{9.81}=2.46s$

3. $y=y_0+ut+\frac{1}{2}at^2$

Using the formula

$y=0.5+24.1(2.46)-\frac{1}{2}(9.81)(2.46)^2$

$y=30.1 m$

4.Time of flight for red ball=3.77-2.9=0.87s

$y'=32-7.2(0.87)-\frac{1}{2}(9.81)(0.87)^2$

$y'=22.0m$

Hence, the height of red ball 3.77 s after the blue ball is 22.0 m.

5.According to question

$0.5+24.1(t+2.9)-\frac{1}{2}(9.81)(2.9+t)^2=32-7.2t-\frac{1}{2}(9.81)t^2$

$0.5+24.1t+69.89-4.905(t^2+5.8t+8.41)=32-7.2t-4.905t^2$

$0.5+24.1t+69.89-4.905t^2-28.449t-41.25105=32-7.2t-4.905t^2$

$0.5+69.89-41.25105-32=-24.1t+28.449t-7.2t$

$-2.86105=-2.851t$

$t=\frac{2.86105}{2.851}=1.004 s$

Hence, 1.004 s after the blue ball is thrown are the two balls in the air at the same height.

8 0

3 years ago

The average diameter of an atom is ________ (a) 10“8 m (b) 1 m (c) 1A° (d) 10

Thepotemich [5.8K]

Answer

0.1 - 0.5 nano meters

Explanation:

Here ya go buddy! :D

5 0

3 years ago