All categories

4 years ago

Jim wants to learn more about the feeding habits of ants. What steps should he take in order to best study the patterns of ants?

1 answer:

3 0

They have any farm kits. You can observe the ants and their patterns for as long as you wish or at any time you wish keep a journal

You might be interested in

A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci

vovikov84 [41]

According to the law of conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.

m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$
$(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'$
$1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')$
$-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')$

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.

$-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')$
$7.5kg.m/s=(235kg)(v_{2}')$
$(7.5kg.m/s)/(235kg)=(v_{2}')$
$0.03m/s=(v_{2}')$

The velocity of the 2nd car after the collision is 0.03m/s.

5 0

4 years ago

What is the net force on a truck if the force of friction is 31 N and the force of the engine is 79 N?

Evgen [1.6K]

Answer:

Fr = 48 [N] forward.

Explanation:

Suppose the movement is on the X axis, in this way we have the force of the engine that produces the movement to the right, while the force produced by the brake causes the vehicle to decrease its speed in this way the sign must be negative.

∑F = Fr

$F_{engine}-F_{brake} =F_{r}\\F_{r}=79-31\\F_{r}=48[N]$

The movement remains forward, since the force produced by the movement is greater than the braking force.

3 0

3 years ago

Question 2 of 34

tatiyna

You can find the magnitude of the resultant vector : (B). By adding the magnitudes of the two vectors

<h3>Meaning of Vectors</h3>

A vector can be defined as any quantity which possesses magnitude and also has direction.

A Vector quantity is very useful because This type of quantity gives more details to the student or teacher analyzing it.

Lear more about Vectors: brainly.com/question/25705666

#SPJ1

8 0

2 years ago

The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total

pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass $m$ travelling at a velocity $\vec{v}$ . The momentum $\vec{p}$ of this object would be:

$\vec{p} = m \cdot \vec{v}$ .

For the law of conservation of momentum, consider two objects: object $\rm a$ and object $\rm b$ . Assume that these two objects collided with each other.

Let $m_{\rm a}$ and $m_{\rm b}$ denote the mass of the two objects.
Let $\vec{v}_{\rm a}(\text{initial})$ and $\vec{v}_{\rm b}(\text{initial})$ denote the velocity of the two object right before the interaction.
Let $\vec{v}_{\rm a}(\text{final})$ and $\vec{v}_{\rm b}(\text{final})$ denote the velocity of the two objects right after the interaction.

The momentum of the two objects right before the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ , respectively.
The momentum of the two objects right after the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ , respectively.

The sum of the momentum of the two objects would be:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ right before the collision, and
$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ .

4 0

4 years ago

In a certain electrolysis experiment involving Al3+ ions, 60.2 g of Al is recovered when a current of 0.352 A is used. How many

yawa3891 [41]

Answer:

30.5 x 10³ min.

Explanation:

Atomic weight of aluminium = 27

For the reaction

Al⁺³ + 3e = Al

3 mole of electron is required by 1 mole of aluminium

3 x 96500 C of charge is required for 27 gram of aluminium

60.2 g aluminium will require

$\frac{3\times96500}{27} \times60.2$

645.47 x 10³ C

So charge required = 645.47 x 10³ C

Now Charge = Current x time

Current ( given ) = 0.352A

Time = charge / current = 645.47 x 10³ / .352

=1833 x 10³ s

30.5 x 10³ minutes

6 0

4 years ago