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3 years ago

Who else has a problem with ads not working for answers??

Physics

2 answers:

Ber [7]3 years ago

8 0

Answer:

me‍♀️

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tigry1 [53]3 years ago

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Answer:

me it's annoying like just let us get the answer

You might be interested in

The Inner Galactic Plane Survey conducted with the Allen Telescope Array in northern California is an example of a

xz_007 [3.2K]

Given our understanding of Galactic space searches and techniques, we can confirm that the Inner Galactic Plane Survey conducted with the Allen Telescope Array is an example of a targeted search.

A targeted search is when a technique is used not to sweep for general information, but to identify points at which specific processes occur. In the case of the Inner Galactic Plane Survey, the goal is to find points of star formations.

Therefore, we can say that the Inner Galactic Plane Survey conducted with the Allen Telescope Array in northern California is an example of a targeted search.

To learn more visit:

brainly.com/question/6272572?referrer=searchResults

7 0

3 years ago

2. Grace drives her car 40 km in 75 minutes. What is her average speed in kilometers per hour?

Agata [3.3K]

Answer:

Explanation:

3 0

3 years ago

You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi

Mama L [17]

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

6 0

3 years ago

If the mass of each ball were the same, but the velocity of ball A were twice as much as ball B, what do you think would happen

Margaret [11]

Given momentum is conserved and is mv

if
mass is the same and va=2vb
then momentum is m(2vb + vb)

if collision is elastic (bounces off)
then m(2vb + vb) = m(vfa + vfb)
3vb = vfa + vfb
meaning final velocities would be different

if collision is inelastic (sticks together)
then m(2vb + vb) = (m+m)vf
meaning final velocity would be same

is this what you wanted?

5 0

3 years ago

A ladder placed up against a wall is sliding down. The distance between the top of the ladder and the foot of the wall is decrea

Kitty [74]

Answer:

distance changing at rate of 3.94 inches/sec

Explanation:

Given data

wall decreasing at a rate = 9 inches per second

ladder L = 152 inches

distance h = 61 inches

to find out

how fast is the distance changing

solution

we know that

h² + b² = L² ..................1

h² + b² = 152²

Apply here derivative w.r.t. time

2h dh/dt + 2b db/dt = 0

h dh/dt + b db/dt = 0

db/dt = - h/b × dh/dt .............2

and

we know

h = 61

so h² + b² = L²

61² + b² = 152²

b² = 19383

so b = 139.223

and we know dh/dt = -9 inch/sec

so from equation 2

db/dt = -61/139.223 (-9)

db/dt = 3.94 inches/sec

distance changing at rate of 3.94 inches/sec

3 0

3 years ago