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3 years ago

When two objects collide and stick together, what will happen to their speed, assuming momentum is conserved?

Physics

2 answers:

sladkih [1.3K]3 years ago

7 0

Answer:

When two objects collide and stick together, what will happen to their speed, assuming momentum is conserved? They will move at the same velocity as whichever object was fastest initially. They will move at the same velocity of whichever object was slowest initially.

Explanation:

AveGali [126]3 years ago

5 0

The correct answer is that when two objects collide and stick together, BOTH objects will change speed.
If they are moving in the same direction, the faster moving object will slow down and the slower moving object will speed up.
If they are moving in opposite directions the faster will slow down and the slower one will change direction.

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According to Ohm's law, a circuit with a high resistance _____. will have a low electric current will have a high electric curre

lyudmila [28]

Remark
When you are asked a question like this, the first thing to do is search out a formula and put some limits on it.

Formula
I = E/R which comes from E = IR. To get to the derived formula, divide both sides by R
E/R = I*R/R
E/R = I

Discussion
This is an inverse relationship. That means that as one goes up the other one will go down.

So in this case you keep E constant and you manipulate R and look at your results for I

Case 1
Let us say that E = 10 volts
Let us also say the R = 10 ohms

I = E/R
I = 10/10
I = 1 ohm

Case Two
Let's raise the Resistance to 100 ohms
E = 10
R = 100
I = 10/100 = 0.1

Conclusion
As the Resistance goes up, the current goes down. Answer: A

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3 years ago

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Yuki888 [10]

Answer:

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3 years ago

A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr

valina [46]

The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
$T=2 t = 2 \cdot 0.100 s = 0.200 s$
Which means that the frequency is
$f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz$
and the angular frequency is
$\omega=2 \pi f = 2 \pi (5 Hz)=31.4 rad/s$

In a spring-mass system, the maximum velocity of the object is given by
$v_{max} = A \omega$
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
$v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s$

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3 years ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23

Burka [1]

Answer:

a) $v(2) = 3.9\,\frac{m}{s}$ , b) $v(4) = -15.7\,\frac{m}{s}$

Explanation:

a) The equation for vertical velocity is obtained by deriving the function with respect to time:

$v(t) = 23.5 -9.8\cdot t$

The velocities at given instants are, respectivelly:

$v(2) = 3.9\,\frac{m}{s}$

$v(4) = -15.7\,\frac{m}{s}$

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3 years ago

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VladimirAG [237]

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3 years ago