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4 years ago

An object has a KE of 120 J. It has a mass of 6kg. What is its velocity?

Physics

1 answer:

Yuri [45]4 years ago

6 0

$120 = \frac{1}{2} \times 6 \times v {}^{2} \\ v {}^{2} = 40 \\ v = 2 \sqrt{10}$

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Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

4 years ago

A proton moves with a velocity of v with arrow = (4î − 6ĵ + k) m/s in a region in which the magnetic field is B with arrow = (î

nalin [4]

Answer:

F = [(6.4 × 10⁻¹⁹)î + (8.0 × 10⁻¹⁹)ĵ + (22.4 × 10⁻¹⁹)k] N

Magnitude of F = (2.466 × 10⁻¹⁸) N

Explanation:

The magnetic force, F, on a given charge, q, moving with velocity, v, in a magnetic field, B, is given as the vector product

F = qv × B

where v = (4î − 6ĵ + k) m/s

B = (î + 2ĵ − k) T

The particle is a proton, hence,

q = (1.602 × 10⁻¹⁹) C

F = qv × B = q (v × B)

(v × B) is given as (4î − 6ĵ + k) × (î + 2ĵ − k)

The cross product is evaluated as a determinant of

| î ĵ k |

|4 -6 1 |

|1 2 -1 |

î [(-6)(-1) - (2)(1)] - ĵ [(4)(-1) - (1)(1)] + k [(4)(2) - (-6)(1)]

î (6 - 2) - ĵ (-4 - 1) + k (8 + 6) = (4î + 5ĵ + 14k)

(v × B) = (4î + 5ĵ + 14k)

F = q (v × B) = (1.6 × 10⁻¹⁹) (4î + 5ĵ + 14k)

F = [(6.408 × 10⁻¹⁹)î + (8.01 × 10⁻¹⁹)ĵ + (22.428 × 10⁻¹⁹)k] N

Magnitude of F =

√[(6.408 × 10⁻¹⁹)² + (8.01 × 10⁻¹⁹)² + (22.428 × 10⁻¹⁹)²]

Magnitude of F = (2.466 × 10⁻¹⁸) N

Hope this Helps!!!

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