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3 years ago

15

if somebody swim to the right for 30 meters then swims to left for 15 meters and then swims back to the right for 50 meters what

s is the person displacement

1 answer:

zubka84 [21]3 years ago

8 0

50m

Explanation:

Displacement is the length of path traveled which is measured from start to the finishing of the path.

Analysis of the journey;

Starts from:

0 30m from right

15m to left

50m to right

The displacement is 50m from the starting point.

Distance is total path traveled and for this problem it is 30+ 15 + 50 = 95m

learn more:

displacement brainly.com/question/5461768

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No, resolve in both directions separately

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4 years ago

A ball of plasticine is released from rest at height of 2.2 m above the ground. After touching the ground, the plasticine ball c

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The magnitude of the acceleration of the ball while coming to rest is 477.43 m/s²

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height above the ground, h = 2.2 m

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2 years ago

1. An excited lithium atom emits a red light with wavelength a = 671nm. What is the corresponding photon energy? hc (6.63 x 10-3

Rudiy27

Answer:

E = 2,964 10⁻¹⁹ J

Explanation:

The energy of the photons is given by the Planck relation

E = h f

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let's reduce the magnitude to the SI system

λ = 671 nm = 671 10⁻⁹ m

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6 0

3 years ago

A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

Artist 52 [7]

Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

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3 years ago