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3 years ago

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 A. 1 kg/s2 B. 1 kg*m/s C. 1 kg*m/s2 D. 1 kg/s

tiny-mole [99]

Answer:

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:

Momentum

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

Explanation:

Step 1: Define

Identify variables

[Given] m₁ = 1200 kg

[Given] v₁i = 10 m/s

[Solve] m₂

[Given] v₂i = 0 m/s

[Given] vf = 6 m/s

Step 2: Solve for m₂

Substitute in variables [Law of Conservation of Momentum]: (1200 kg)(10 m/s) + m₂(0 m/s) = (1200 kg + m₂)(6 m/s)
Multiply: 12000 kg · m/s = (1200 kg + m₂)(6 m/s)
Isolate m₂ term: 2000 kg = 1200 kg + m₂
Isolate m₂: 800 kg = m₂

4 0

3 years ago

Read 2 more answers