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lozanna [386]
2 years ago
10

What type of stimulus do we tend to be particularly aware of ?

Physics
1 answer:
Volgvan2 years ago
7 0
Emotional stimuli is the answer to your question.
You might be interested in
A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.
Bas_tet [7]

Answer:

Explanation:

Total momentum of the system before the collision

.5 x 3 - 1.5 x 1.5 = -0.75 kg m/s towards the left

If v be the velocity of the stuck pucks

momentum after the collision = 2 v

Applying conservation of momentum

2 v = -  .75

v =  - .375 m /s

Let after the collision v be the velocity of .5 kg puck

total momentum after the collision

.5 v + 1.5 x .231 = .5v +.3465

Applying conservation of momentum law

.5 v +.3465 = - .75

v = - 2.193 m/s

2 ) To verify whether the collision is elastic or not , we verify whether the kinetic energy is conserved or not.

Kinetic energy before the collision

= 2.25 + 1.6875

=3.9375 J

kinetic energy after the collision

= .04 + 1.2 =1.24 J

So kinetic energy is not conserved . Hence collision is not elastic.

3 ) Change in the momentum of .5 kg

1.5 - (-1.0965 )

= 2.5965

Average force applied = change in momentum / time

= 2.5965 / 25 x 10⁻³

= 103.86 N

5 0
3 years ago
13. Momentum is an object’s mass times velocity. Which has more momentum? A paper airplane with a mass of 10g flying at a veloci
Ivenika [448]
Since the velocity of the real plane is 0, p=mv=0.  So the paper airplane actually has more momentum since it's value is not 0.
5 0
3 years ago
Tarzan (who has mass 80.0 kg) is running across the jungle floor with speed 7.00 m/s as
Lesechka [4]

<u>We are given:</u>

Mass of Tarzan before swinging = 80 kg

Mass of Tarzan when swinging = 80 + 15 = 95 kg

Velocity of Tarzan = 7 m/s

The height of the rock Tarzan's monkey is sitting on = 3 m

__________________________________________________________

<u>Momentum of Tarzan before swinging:</u>

We know that:

Momentum = Mass*Velocity

Momentum = 80 * 7

Momentum = 560 kg m/s

__________________________________________________________

<u>Speed of Tarzan after grabbing the bananas:</u>

The momentum of Tarzan will remain the same but his mass will increase

So, Since Momentum = New Mass* velocity

560 = 95 * v                           [where v is the velocity of Tarzan]

v = 5.9 m/s

<u>__________________________________________________________</u>

<h3><u>Finding the Initial and Final KE and PE:</u></h3>

<em>Here, KE = Kinetic Energy and PE = Potential Energy</em>

<u>Initial and Final KE:</u>

We know that KE = 1/2*(mv²)

<u>Initial KE:</u>

Initial KE = 1/2*(mv²)          [where v is the velocity after picking the bananas]

Initial KE = 1/2*(95*5.9²)

Initial KE = 1653.5 Joules

<u>Final KE:</u>

Final KE = 1/2*(mv²)            

<em>[where v is the velocity at the maximum point of the swing]</em>

Since Tarzan will be at rest at the maximum point of the swing, v = 0 m/s

Final KE = 1/2*(95*0²)

Final KE = 0 Joules

<u>Initial and Final PE:</u>

We know that:

PE = mgh                

<em>[where g is the acceleration due to gravity and h is the height]</em>

<u>Initial PE:</u>

Since the height of Tarzan from the ground was 0 m at the beginning of the swing, h = 0

Initial PE = 95*10*0

Initial PE = 0 Joules

<u>Final PE:</u>

Let the maximum height of Tarzan be h m

Final PE = 95*10*h

Final PE = 950(h)

__________________________________________________________

<h3><u>Finding the maximum height Tarzan will reach:</u></h3>

<em>Here, KE = Kinetic Energy and PE = Potential Energy</em>

From the law of conservation of momentum, we know that:

Initial KE + Initial PE = Final KE + Final PE

Replacing the variables:

1653.5 + 0 = 0 + 950h

1653.5 = 950h

h = 1653.5/950         [dividing both sides by 950]

h = 1.74 m

Therefore, the maximum height reached by Tarzan is 1.74 m

but since his monkey is sitting 3 m high, he will NOT be able to reach his monkey

6 0
3 years ago
A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7
maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
From which we can find the value of the coefficient of kinetic friction:
\mu =  \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49
3 0
3 years ago
The solar nebula was a
tangare [24]

Solar Nebula

Our solar system began forming within a concentration of interstellar dust and hydrogen gas called a molecular cloud. The cloud contracted under its own gravity and our proto-Sun formed in the hot dense center. The remainder of the cloud formed a swirling disk called of the solar nebula.

4 0
3 years ago
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