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Doss [256]
3 years ago
8

Two 10 kg pucks head straight towards each other with velocities of 10 m/s and -20 m/s. They collide and stick together. Calcula

te the resulting velocity.
Physics
1 answer:
RUDIKE [14]3 years ago
8 0

The final velocity of the two pucks is -5 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum.

In fact, in absence of external force, the total momentum of the two pucks before and after the collision must be conserved - so we can write:

m_1 u_1 + m_2 u_2 = (m_1 +m_2)v

where

m_1 = m_2 = m = 10 kg is the mass of each puck

u_1 = 10 m/s is the initial velocity of the 1st puck

u_2 = -20 m/s is the initial velocity of the 2nd puck

v is the final velocity of the two pucks sticking together

Re-arranging the equation and solving for v, we find:

mu_1 + mu_2 = (m+m)v\\u_1 + u_2 = 2v\\v=\frac{u_1+u_2}{2}=\frac{10-20}{2}=-5 m/s

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

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The dimensions of a cylinder are changing, but the height is always equal to the diameter of the base of the cylinder. If the he
VashaNatasha [74]

Answer:

dV/dt  = 9 cubic inches per second

Explanation:

Let the height of the cylinder is h

Diameter of cylinder = height of the cylinder = h

Radius of cylinder, r = h/2

dh/dt = 3 inches /s

Volume of cylinder is given by

V = \pi r^{2}h

put r = h/2 so,

V = \pi \frac{h^{3}}{4}

Differentiate both sides with respect to t.

\frac{dV}{dt}=\frac{3h^{2}}{4}\times \frac{dh}{dt}

Substitute the values, h = 2 inches, dh/dt = 3 inches / s

\frac{dV}{dt}=\frac{3\times 2\times 2}{4}\times 3

dV/dt  = 9 cubic inches per second

Thus, the volume of cylinder increases by the rate of 9 cubic inches per second.

6 0
3 years ago
Two Force one of 12 Newton and another 24 Newton acts at 90 degree with each other find the resultant of two force and its direc
Leona [35]

Answer:

Fr = 26.83 [N]

Explanation:

To solve this problem we must use the Pythagorean theorem, since the forces are vector quantities, that is, they have magnitude and density. Therefore the Pythagorean theorem is suitable for the solution of this problem.

F_{r}=\sqrt{(12)^{2}+(24)^{2}  } \\F_{r}=26.83[N]

3 0
3 years ago
1. What are the things needed by plants to make their own food?
iren2701 [21]

Answer:

1-D(carbon dioxide, water and sunlight)

2-D(parasitism)

3-C(competition)

Explanation:

hope it helps

4 0
3 years ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level.
BARSIC [14]

Answer:

There is an interval of 24.28s in which the rocket is above the ground.

Explanation:

y_{i}=0m

v_i=80\frac{m}{s}

a=4\frac{m}{s^2}

y_{e}=1000m

g=9.8\frac{m}{s^2}

From Kinematics, the position y as a function of time when the engine still works will be:

y(t)=v_it+\frac{1}{2}at^2

At what time the altitud will be y_{e}=1000m?

v_it+\frac{1}{2}at^2=y_{e} ⇒ \frac{1}{2}at^2+v_it-y_{e}=0

Using the quadratic formula: t_1=10s.

How much time does it take for the rocket to touch the ground? No the function of position is:

y(t)=y_{e}+v_et-\frac{1}{2}gt^2

Where our new initial position is y_{max}, the velocity when the engine breaks is v_e=v_i+at=120\frac{m}{s} and the only acceleration comes from gravity (which points down).

Now, when the rocket tounches the ground:

y_{e}+v_et-\frac{1}{2}gt^2=0

Again, using the quadratic ecuation:

t_2=24.49s

Now, the total time from the moment it takes off and the moment it tounches the ground will be:

t_T=t_1+t_2=34.49s.

6 0
3 years ago
Which of the following statements about the motion of an object is correct?
xxMikexx [17]

Answer:

B.

Velocity describes how fast something is going, whereas Speed describes how fast something is going and in which direction.

Explanation:

. If the above statement is true, then describe an example of such a motion. ... b. What was the displacement for the entire trip? 0 miles (You finish where you started) ... Speed is a quantity that describes how fast or how slow an object is moving.

4 0
3 years ago
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