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NikAS [45]
3 years ago
6

Two water balloons of mass 0.75 kg collide and bounce off of each other without breaking. Before the collision, one water balloo

n moved at a velocity of 3.5 m/s east, while the other moved at a velocity of 2.75 m/s west. After the collision, one balloon moves at a velocity of 1.5 m/s west. What is the velocity of the other water balloon?
Physics
1 answer:
bagirrra123 [75]3 years ago
5 0
By the law of momentum conservation:-
=>m¹u¹ + m²u² = m1v1 + m²v² {let East is +ve}
=>u¹ + u² = v¹ + v² {as m1=m2}
=>3.5 - 2.75 = v1-1.5
<span> =>v¹ = 2.25 m/s (East) </span>
You might be interested in
Calculate A, E, μ, cv and S for 1 mole of Kr at 298 K and 1 atm (assuming ideal behavior)
vaieri [72.5K]

Answer:

internal energy E 3716.35 j

cv = 12.47 J/K

S = 12.47 J/K

A =  0.29 J

\mu =3716.35 J/mole

Explanation:

given data:

Kr  atomic number = 36

degree of freedom = 3

1) internal energy E = \frac{f}{2} n R T

                                = \frac{3}{2} *1*8.314*298 = 3716.35 j

2) cv = \frac{E}{T}

         = \frac{3716.35}{298} = 12.47 J/K

3) S = cv =\frac{E}{T} = 12.47 J/K

4) A, Halmholtz free energy = E -TS = 37146.35 - 12.47*298 = 0.29 J

5)chemcial potential \mu = \frac{energy}{mole} = 3716.35 J/mole

6 0
3 years ago
I've been on this for a while please help me
True [87]

Answer: Student 2

Explanation: Iron nail and a paperclip are conductors because they are made of metal. A rock, rubber band, and wooden stick are insulators because they cannot conduct electricity.

7 0
2 years ago
A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force
Tamiku [17]

Answer:

(a) 91 kg (2 s.f.)    (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

                                                   s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}

     Subsequently,

                                                  F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

                                                    v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}

     According to Newton's First Law which states objects will remain at rest

     or in uniform motion (moving at constant velocity) unless acted upon by

     an external force. Hence, the block of ice by the end of the first 5

     seconds, experiences no acceleration (a = 0) but travels with a constant

     velocity of 4.4 m \ s^{-1}.

                                                    s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}

      Therefore, the ice block traveled 22 m in the next 5 seconds after the

      worker stops pushing it.

4 0
2 years ago
What times what equals 400
Radda [10]

If you're willing to consider fractions or decimals,
then there are an infinite number of answers. 
Like (2.5 x 160), and (15 x 26-2/3).

If you want to stick to only whole numbers,
then these 8 combinations do:

1, 400
2, 200
4, 100
5, 80
8, 50
10, 40
16, 25
20, 20

7 0
3 years ago
An object falls from rest on a high tower and takes 5.0 s to hit the ground. Calculate the object's position from the top of the
Lena [83]

Answer:

After 1 sec = 4.9 m

After 2 sec = 19.6 m

After 3 sec = 44.1 m

After 4 sec =  78.4 m

After 5 sec = 122.5 m

Explanation:

After 1 sec:

<em>u=0m/s   t=1 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(1) + (1/2)(9.8)(1²) = 4.9m

After 2 sec:

<em>u=0m/s   t=2 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(2) + (1/2)(9.8)(2²) = 19.6m

After 3 sec:

<em>u=0m/s   t=3 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(3) + (1/2)(9.8)(3²) = 44.1m

After 4 sec:

<em>u=0m/s   t=4 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(4) + (1/2)(9.8)(4²) = 78.4m

After 5 sec:

<em>u=0m/s   t=5 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(5) + (1/2)(9.8)(5²) = 122.5m

7 0
2 years ago
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