Four radioactive 10-gram (g) samples with their remaining isotope mass are listed in the table that spans 50 years.

Two hockey pucks, labeled A and B, are initially at rest on asmooth ice surface and are separated by a distance of18.0 m. Simult

zlopas [31]

Answer:

8.505 m

Explanation:

Let V1 and V2 be velocities of puck A and B respectively

Since A and B move in the same direction, so the relative velocity will be V1+V2=3.5+3.9=7.4m/s

Or

Vr=7.4 m/s

Distance=S= 18 m

Time =t=?

S=Vr×t

==> t=S/Vr

==> t= 18/7.4=2.43 sec

At this time both will strike together

Distance by puck A

V1=3.5 m/s

Time=t= 2.43 sec

Distance covered=d=?

d=V1×t=3.5×2.43=8.505 m

So, puck A will cover 8.505 meters before collision

6 0

3 years ago

When an element tends to lose its valence electrons in chemical reactions, it behaves more like a _______ . When an element tend

vampirchik [111]

the answers are

metal

nonmetal

lower

higher

in the middle

7 0

4 years ago

Read 2 more answers

A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

6 0

3 years ago

Read 2 more answers

It took Eric 11 hours to drive to a football game. On the way home, he was able to increase his average speed by 12mph and make

yarga [219]

Answer:

54mph

Explanation:

Speed is defined as the change in distance of a body with respect to time.

Let x be the speed of Eric on his way to the football game, if he used 11hours to the football game, his distance will be the product of his speed and time taken i.e

Distance = speed × time

Distance= x × 11

Distance = 11x... (1)

During his return journey, his speed increases by 12mph with a return drive in only 9hours i.e speed = x+12 and time is 9hours

Since distance = speed × time (return Journey)

Distance = (x+12)×9

Distance = 9x+84...(2)

Note that the distance to and fro the journey is the same as such we will equate equations 1 and 2 to get his initial speed "x" to have;

11x = 9x+84

11x-9x = 84

2x = 84

x = 42mph

This means that his initial speed on his way to the football game is 42mph. Since his speed increases by 12mph during his return drive, his return speed will be 42mph + 12mph which gives 54mph.

6 0

3 years ago

A 1 kg ball and a 10 kg ball are dropped from a height of 10 m at the same time. In the absence of air resistance,

koban [17]

Answer: A

Explanation: Neglecting air resistance every object will return to the ground at the same interval regardless of their weights, we can say that the objects are undergoing free falls, they are falling under the sole of gravity.

No matter the weight of the objects or the elevation, without significant air resistance, they will reach the ground at the same time.

7 0

4 years ago