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solniwko [45]
3 years ago
13

Object A of mass M is moving east at speed v. It collides with object B of mass 2M that was initially at rest. The motion of the

objects before and after the collision is along the same line. After the collision, object A is moving west at a speed of v/3. What is the speed of object B immediately after the collision?
Physics
1 answer:
Firlakuza [10]3 years ago
3 0

Answer:

v_B=\frac{v}{3}

Explanation:

Given that:

mass of object A, m_A=M

mass of object B, m_B=2M

speed of object A, v_A=v

So, according to the conservation of momentum, the momentum before collision is equal to the momentum after conservation.

m_A.v+m_B\times 0=m_A\times v_A +m_B\times v_B

M\times v+0 = M\times \frac{v}{3}+2M\times v_B

2M\times v_B= \frac{2M\times v}{3}

v_B=\frac{v}{3}

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Arthur conducts a controlled experiment several times. The results of each test show that his hypothesis is not supported. Which
sladkih [1.3K]

Hello. You did not inform the experiment that Arthur is conducting, which makes it impossible for your question to be answered accurately. However, I will try to help you in the best possible way.

The hypothesis is an assumption that is made before the experiment is carried out. This hypothesis is formed with the observation of some phenomenon of nature where the researcher believes that two or more elements interact to form a result. In this case, the experiment is carried out to determine whether the assumption, that is, the hypothesis is false or true. In the event that an experiment determines that the hypothesis is false, two things may have occurred: (a) the experiment was set up, or analyzed incorrectly, (b) the elements tested have no relation to the observed phenomenon.

3 0
3 years ago
Consider a uniform electric field of 50 N/C directed toward the east. If the voltage measured relative to ground at a given poin
qaws [65]

Answer:

30 V

Explanation:

Given that:

The uniform electric field = 50 N/C

Voltage = 80 V

distance = 1.0 m

The potential difference of the electric field = Δ V

E_d = V₁ - V₂

50 × 1 = 80V - V₂

50 - 80 V = - V₂

-30 V = - V₂

V₂ = 30 V

5 0
3 years ago
hello friends,i need your help my home work now in physics,topic:motion.40 marks +brainliest if correct .​
fenix001 [56]

Answer:

See below

Explanation:

Vertical position is given by

df = do + vo t - 1/2 a t^2      df = final position = 0 (on the ground)

                                           do =original position = 2 m

                                            vo = original <u>VERTICAL</u> velocity = 0

                                            a = acceleration of gravity = 9.81 m/s^2

THIS BECOMES

0 = 2 + 0 * t  - 1/2 ( 9.81)t^2

  to show t =<u> .639 seconds to hit the ground </u>

During this .639 seconds it flies horizontally at 10 m/s for a distance of

      10 m/s * .639 s =<u> 6.39 m </u>

5 0
1 year ago
A man starts walking from home and walks 2 miles at 20° north of west, then 4 miles at 10° west of south, then 3 miles at 15° no
Rzqust [24]

Answer:

a)  R = 2.5 mi   b)  To return to your case you must walk in the opposite direction or θ = 98º

This is 8º north west

Explanation:

This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately

To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.

First vector A = 2 to 20º north west

Measured from the positive x axis is θ = 180 -20 = 160º

We use trigonometry to find the components

     Cos 20 = Aₓ / A

     sin 20 = A_{y} / A

    Aₓ = A cos 160 = 2 cos 160

    A_{y}  = A sin160 = 2 sin160

    Aₓ = -1,879 mi

    A_{y}  = 0.684 mi

Second vector B = 4 mi 10º west of the south

Angle θ = 270 - 10 = 260º

    cos 2600 = Bₓ / B

    sin 260 = B_{y} / B

    Bₓ = B cos 260

     B_{y}  = B sin 260

    Bₓ = 4 cos 260

     B_{y}  = 4 sin 260

     Bₓ = -0.6946mi

     B_{y}  = - 3,939 mi

Third vector C = 3 mi to 15 north east

     cos 15 = Cₓ / C

     sin15 = C_{y} / C

     Cₓ = C cos 15

     C_{y} = C sin15

     Cₓ = 3 cos 15

    C_{y} = 3 sin 15

     Cₓ = 2,898 mi

    C_{y} = 0.7765 mi

Now we can find the final position of the person

    X = Aₓ + Bₓ + Cₓ

    X = -1.879 -0.6949 + 2.898

    X = 0.3241 mi

    Y = A_{y} +  B_{y} + C_{y}

    Y = 0.684 - 3.939 +0.7765

    Y = -2.4785 mi

a) We use Pythagoras' theorem

     R = √ (x2 + y2)

     R = √ (0.3241 2 + (-2.4785) 2)

     R = 2.4996 mi

     R = 2.5 mi

b) let's use trigonometry

     Tan θ = y / x

     Tanθ  = -2.4785 / 0.3241

     θ = tan⁻¹ (-7,647)

     θ = -82

Measured from the positive side of the x axis is Te = 360 - 82 = 278º

(90-82) south east

To return to your case you must walk in the opposite direction or Te = 98º

This is 8º north west

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