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Tamiku [17]
3 years ago
11

A 4 kg object moving to the left collides with and sticks to a 3 kg object moving to the right. Which of the following is true o

f the motion of the combined objects immediately after the collision?
a. They must be moving to the left.
b. They must be moving to the right.
c. They must be at rest.
d. The motion cannot be determined without knowing the speeds of the objects before the collision.
Physics
1 answer:
madam [21]3 years ago
4 0

Answer:

D. The motion cannot be determined without knowing the speeds of the objects before the collision.

Explanation:

This question is tricky! We know the object moving to the left has a greater mass than the one moving to the right. We'd <em>assume</em> they would move to the left because the leftwards object has a greater mass, right?

Not. So. Fast.

We can solve for the objects' final velocity using the formula for momentum, m₁v₁ + m₂v₂ = (m₁ + m₂)v .

Now here's where the trap is sprung: <em>we don't think about the equation</em>. This shows that the final velocity of the objects and the direction depends on both the mass of the objects <em>and</em> their initial velocity.

Basically, what if the 3 kg object is moving at 1 m/s and the 4 kg object is moving at –0.5 m/s? The objects would move to the <em>right</em> after the collision!

Do we know the velocity of these objects? No, right?

That means we <em>can't</em> determine the direction of their motion <u>unless we know their initial, pre-collision velocity</u>. This question is tricky because we look at the 4 kg vs. 3 kg and automatically assume the 4 kg object would dictate the direction of motion. That's not true. It depends on velocity as well.

I hope this helps you! Have a great day!

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Water drips from a shower head (the sprayer at the top of the shower) and falls onto the floor 2.3 m below. The droplets are fal
Maurinko [17]

Answer:

0.767m

Explanation:

We are given that the time interval between each droplet is equal.

We are also given that the fourth drop is just dripping from the shower when the first hits the floor.

If they fall at the same time interval and we know that the distance between the shower head and floor are the same, they must therefore fall at the same velocity.

The distance between each drop has to be the same given that they fall at equal time intervals.

Let this distance be x.

We can then partition the entire height of the system into three parts (as shown in the diagram).

Hence, we can say that:

x + x + x = 2.3m

3x = 2.3m

=> x = 2.3/3 = 0.767m

Therefore, at the time the first drop hits the floor, the third drop is only 0.767 m below the shower head.

8 0
3 years ago
QUESTION 10
Elena L [17]

The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

\theta =37.01^{\circ}

We can apply the first Newton's law in x and y-direction.

If we do a free body diagram of the system we will have:

x-direction

All the forces acting in this direction are:

T_{1}sin(\theta)-T_{2}sin(\theta)=0    (1)

Where:

  • T(1) is the tension due to the rope 1
  • T(2) is the tension due to the rope 2

Here we just conclude that T(1) = T(2)

y-direction

The forces in this direction are:

T_{1}cos(\theta)+T_{2}cos(\theta)-W=0   (2)

Here W is the weight of the steel beam.

We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.

Knowing that T(1) = T(2) and W = mg, we have:

T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)=m_{steel}g

T(1) must be equal to 5479 N, so we have:

cos(\theta)=\frac{m_{steel}g}{2T_{1}}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=0.80

Therefore, the maximum angle allowed is θ = 37.01°.

You can learn more about tension here:

brainly.com/question/12797227

I hope it helps you!

8 0
3 years ago
A spring with a force constant of 5.3 n/m has a relaxed length of 2.60 m. when a mass is attached to the end of the spring and a
olganol [36]
This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant. 
<span> U = (1/2)kx^2 
</span><span> U = (1/2)(5.3)(3.62-2.60)^2 
</span> U = <span> <span>2.75706 </span></span>J
6 0
3 years ago
What is the length of a simple pendulum with a period of 11.5 s?
givi [52]
Using the equation for period length for a pendulum, you get 32.829 meters.
4 0
3 years ago
A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. The wire ful
balu736 [363]

Answer:

FB = 0.187 N

Explanation:

To find the magnetic force FB in the wire you use the following formula:

|\vec{F_B}|=ILBsin\theta\\\\L=0.25m\\\\|\vec{B}|=\sqrt{(0.3y)^2+(0.4y)^2}=0.5y \ T

the angle between B and L is given by:

\vec{L}\cdot\vec{B}=LBcos\theta\\\\\theta=cos^{-1}(\frac{\vec{L}\cdot\vec{B}}{LB})=cos^{-1}(\frac{0*0.3y+0.25*0.4y}{0.25*0.5y})=36.86\°

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:

|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N

hence, the magnitude of the magnetic force is 0.187N

4 0
3 years ago
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