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malfutka [58]
3 years ago
11

An object change in position according to its reference point is called

Physics
2 answers:
IRINA_888 [86]3 years ago
6 0
Motionnnnnnnnnnnnnn !
Flura [38]3 years ago
5 0

Answer:

motion

Explanation:

i had an assignment on it!

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One Newton is equivalent to<br> A. 1 kg/s2<br> B. 1 kg*m/s<br> C. 1 kg*m/s2<br> D. 1 kg/s
tiny-mole [99]

Answer:

B

Explanation:

7 0
3 years ago
Ship A is located 4.2 km north and 2.7 km east of ship B. Ship A has a velocity of 22 km/h toward the south and ship B has a vel
kotykmax [81]

Answer:

(a) The x-component of velocity is 31.55 km/h

(b) The y-component of velocity is 44.92 km/hr

Solution:

As per the solution:

The relative position of ship A relative to ship B is 4.2 km north and 2.7 km east.

Velocity of ship A, \vec{u_{A}} = 22 km/h towards South = - 22\hat{j}

Velocity of ship B, \vec{u_{B}} = 39 km/h Towards North east at an angle of 36^{\circ} = \vec{u_{B}} = 39sin36^{\circ} \hat{j}

Now, the velocity of ship A relative to ship B:

\vec{u_{AB}} = \vec{u_{A}} - \vec{u_{B}}

\vec{u_{A}} = - 22\hat{j}

\vec{u_{B}} = 39cos36^{\circ} \hat{i} - 39sin36^{\circ} \hat{j}

Now,

\vec{u_{AB}} = - 22\hat{j} +39cos36^{\circ} \hat{i} - 39sin36^{\circ} \hat{j}

\vec{u_{AB}} = 31.55\hat{i} - 44.92\hat{j}

4 0
3 years ago
A proton enters a uniform magnetic field of strength 2 T at 300 m/s. The magnetic field is oriented perpendicular to the proton’
sattari [20]

Answer:

Magnetic force, F=9.6\times 10^{-17}\ N

Explanation:

It is given that,

Magnetic field, B = 2 T

Velocity of the proton, v = 300 m/s

Charge on the proton, q=1.6\times 10^{-19}\ C

The magnetic field is oriented perpendicular to the proton’s velocity. The magnetic force on the charged particle is given by :

F=qvB\ sin\theta

The magnetic field is oriented perpendicular to the proton’s velocity, \theta=90^{\circ}

F=1.6\times 10^{-19}\times 300\times 2

F=9.6\times 10^{-17}\ N

So, the magnitude of the force that the proton experiences while it moves through the magnetic field is 9.6\times 10^{-17}\ N. Hence, this is the required solution.

7 0
3 years ago
Which statement is true about voice?
Ksju [112]
The answer is A. voice uses a wider range of pitch and volume as compared to speaking
3 0
4 years ago
Read 2 more answers
Car A of mass 1200kg traveling at 10m/s , collided in too the back of the car B, which is stationary. Following the collision,.
Nina [5.8K]

Answer:

Car B has a mass of 800 kg.

General Formulas and Concepts:

<u>Momentum</u>

Law of Conservation of Momentum: \displaystyle m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_{f}

Explanation:

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] m₁ = 1200 kg

[Given] v₁i = 10 m/s

[Solve] m₂

[Given] v₂i = 0 m/s

[Given] vf = 6 m/s

<u>Step 2: Solve for m₂</u>

  1. Substitute in variables [Law of Conservation of Momentum]:                       (1200 kg)(10 m/s) + m₂(0 m/s) = (1200 kg + m₂)(6 m/s)
  2. Multiply:                                                                                                             12000 kg · m/s = (1200 kg + m₂)(6 m/s)
  3. Isolate m₂ term:                                                                                                2000 kg = 1200 kg + m₂
  4. Isolate m₂:                                                                                                         800 kg = m₂
4 0
3 years ago
Read 2 more answers
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