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emmainna [20.7K]
3 years ago
9

Helllp me will give brainlist

Physics
1 answer:
Mandarinka [93]3 years ago
7 0

The top row of boxes is " F O R C E " .

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A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a
sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

5 0
3 years ago
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
3 years ago
Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1360 N. Assume that the play
Alona [7]

Answer:

7.59Ns

Explanation:

Given parameters:

Force  = 1360N

Time of contact  = 5.85 x 10⁻³s

Unknown:

Impulse  = ?

Solution:

The impulse of the ball is given as:

        Impulse  = Force x time

       Impulse  = 1360 x 5.85 x 10⁻³ = 7.59Ns

4 0
3 years ago
An empty can of soda is yeeted (thrown) into a crowd. Then, a full can of soda is yeeted into a crowd. Throwing the full can too
4vir4ik [10]

Answer:

Newton's third law of motion.

Explanation:

We are told the force needed to throw the full soda can was more than that needed to throw the empty can.

Now, the weight of the full soda can will be more than that of the empty can. Therefore, the full can will demand more force than that of the empty can due to Newton's third law of motion which states that to every action, there is an equal and opposite reaction.

3 0
3 years ago
_____ is the kinetic energy of an object, proportional to the object's moment of inertia and the square of its angular velocity.
Juli2301 [7.4K]
Rotational kinetic energy <span>is the kinetic energy of an object, proportional to the object's moment of inertia and the square of its angular velocity.</span>
7 0
3 years ago
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