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natulia [17]
3 years ago
7

Block a, with a mass of 2.0kg, moves along the x axis with a velocity of 5.0m/s in the positive x direction. it su?ers an elasti

c collision with block b, initially at rest, and the blocks leave the collision along the x axis. if b is much more massive than a, the speed of a after the collision is:
Physics
1 answer:
Evgen [1.6K]3 years ago
7 0
<span>-5m/s is what I got

</span>
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A chimpanzee sitting against his favorite tree gets up and walks 70.9 m due east and 31.9 m due south to reach a termite mound,
DIA [1.3K]
A right triangle is formed by the 70.9 m walked east and 31.9 m walked south.
The legs of this right triangle are 70.9 and 31.9.
The shortest distance between two points is a straaight line. Therefore the hypotenuse of this triangle is going to be that shortest distance.
We can use the Pythagorean Theorem to find the hypotenuse of this triangle.
a^2+b^2=c^2\\70.9^2+31.9^2=c^2\\5026.81+1017.61=c^2\\6044.42=c^2\\c=\sqrt{6044.42}

c=\sqrt{6044.42}\approx\boxed{77.7458681}\ (decimal\ form)\\\\c=\sqrt{6044.42}=\sqrt{\frac{604442}{100}}=\boxed{\frac{\sqrt{604442}}{10}}\ (exact\ form)

As for the second part of the question, we want to find the angle formed by the hypotenuse and the 31.9 walked east.
We could use any of the three trigonometric ratios here since we know all 3 sides.
sine = opposite / hypotenuse
cosine = adjacent / hypotenuse
tangent = opposite / adjacent
I am going to use tangent, because then I won't have to deal with the hypotenuse and so the answer will be more accurate.

If you haven't already drawn yourself a diagram, now is a good time to.
The side opposite our angle is the 31.9, and the adjacent is 70.9.
Therefore, \tan(m\angle)=\frac{31.9}{70.9}.

We can use inverse trig ratios here to find the measure of our angle.
\tan^{-1}(\tan(m\angle))=\tan^{-1}(\frac{31.9}{70.9})\\\\m\angle=\tan^{-1}(\frac{31.9}{70.9})\approx\boxed{24.2243851\°\ or\ 0.422795279\ rad}
4 0
3 years ago
If a 1.3 kg mass stretches a spring 4 cm, how much will a 5.8 kg mass stretch the
Likurg_2 [28]

Answer:

17.8cm

Explanation:

1.3kg --> 4cm

1kg --> 3, 1/13cm

5.8kg --> 18.8cm

6 0
2 years ago
A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele
Fantom [35]
The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, \Delta t:
I= \frac{Q}{\Delta t}
First we need to find the net charge flowing at a certain point of the wire in one second, \Delta t=1.0 s. Using I=0.92 A and re-arranging the previous equation, we find
Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C

Now we know that each electron carries a charge of e=1.6 \cdot 10^{-19} C, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}
3 0
3 years ago
A 100-lb child stands on a scale while riding in an elevator. What does the scale read while the elevator slows to stop at the l
Lelechka [254]

Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor

Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.

<h3>What is the apparent weight of a body in a lift?</h3>
  • Consider a body of mass m kept on a weighing machine in a lift.
  • The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
  • The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
<h3>How to solve the question?</h3>
  • Here we have given with the actual weight of the body as 100lbs.
  • This 100lb child is standing on the scale or the weighing machine, when it is riding .
  • During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
  • There is also<em> mg </em>downwards and a normal reaction in the upward direction.
  • when we equate both the upward force and downward force, we get,

                             ma=mg-N\\N=mg-ma    i.e. during riding the scale reads a weight less than that of actual weight.

  • When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.

Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.

Learn more about the apparent weight of the body in a lift here:

brainly.com/question/28045397

#SPJ4

7 0
1 year ago
A circuit has a resistance of 10ohmsband a current of 42 amps.caculate the voltage
Anon25 [30]
R=U/I so
U=RxI
U= 10 x 42
U= 420 volts
3 0
3 years ago
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