2. The nucleus holds two particles:

A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a

katen-ka-za [31]

Answer:b

Explanation:

Given

mass of first cart $m_1=6 kg$

mass of second cart $m_2=3 kg$

velocity of first cart $v_1=3 m/s$

conserving momentum

$m_1v_1+m_2v_2=(m_1+m_2)v$

$6\times 3+3\times 0=(9)\cdot v$

$v=2 m/s$

Initial kinetic Energy $K.E._1=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$

$K.E._1=\frac{1}{2}\cdot 6\cdot 3^2+0$

$K.E._1=27 J$

Final Kinetic Energy

$K.E._2=\frac{1}{2}(m_1+m_2)v^2$

$K.E._2=\frac{1}{2}(6+3)\cdot 2^2=18 J$

Ratio of initial Kinetic Energy to the Final Kinetic Energy

$=\frac{27}{18}=1.5$

6 0

4 years ago

The stone, which weighs 400 g, is thrown upwards at a speed of 20 m / s. Climbed to a height of 12 m. Determine: what is equal t

maxonik [38]

Given that,

Mass of the stone, m = 400 g = 0.4 kg

Initial speed, u = 20 m/s

It is climbed to a height of 12 m.

To find,

The work done by the resistance force.

Solution,

Let v is the final speed. It can be calculated by using the conservation of energy.

$v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 12} \\\\v=15.33\ m/s$

Work done is equal to the change in kinetic energy. It can be given as follows :

$W=\dfrac{1}{2}m(v^2-u^2)\\\\=\dfrac{1}{2}\times 0.4\times (15.33^2-20^2)\\\\=-32.99\ J$

So, the required work done is 32.99 J.

3 0

3 years ago

If the price of gasoline at a particular station in Europe is 5 euros per liter. An American student in Europe is allowed to use

Mashcka [7]

Answer:

Number of gallons =2 gallon

Explanation:

given data:

rate of gasoline ineurope = 5 euro per liter

total money to buy gasoline = 40 euro

total gasoline an american can buy in europe $= \frac{40}{5}$

= 8 litres of gasoline

As given in the question 1 ltr is 1 quarts therefore

Total no. of quarts is 8 quarts

As from question 4 quarts is equal to one gallon, hence

Number of gallons $= \frac{8}{4} = 2 gallon$

5 0

4 years ago

When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for

sveta [45]

Answer:

F = -4567.40 N

Explanation:

Given that,

The power developed by the engine, P = 196 hp

1 hp = 746 W

196 hp = 146157 W

Speed of the car, v = 32 m/s

Let F is the total friction force acting on the car. The product of force and velocity is called the power developed by the engine. It is given by :

$P=-F\times v$

$F=\dfrac{-P}{v}$

$F=\dfrac{-146157\ W}{32\ m/s}$

F = -4567.40 N

So, the total frictional force acting on the car is 4567.40 N. Hence, this is the required solution.

7 0

4 years ago

Simplify 6.25 − 8.<br><br>please help

tatyana61 [14]

6.25 - 8 = -1.75

Hope this helps

-AaronWiseIsBae

7 0

3 years ago