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3 years ago

At which point on the image to the right would the ball have the greatest velocity if it moved from A to G.

Physics

1 answer:

marusya05 [52]3 years ago

4 0

Answer:

It would be A, because it is has more height in which the potential energy would be greater.

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A basketball star covers 2.70 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled pr

Likurg_2 [28]

Answer:

Part a)

$T = 0.81 s$

Part b)

$v_x = 3.33 m/s$

Part c)

$v_y = 3.91 m/s$

Part d)

$\theta = 49.55 degree$

Part e)

$T = 1.11 s$

Explanation:

Part a)

initial vertical position = 1.02 m

maximum height = 1.80 m

$\Delta y = 1.80 - 1.02$

$\Delta y = 0.78 m$

$v_f^2 - v_y^2 = 2a \Delta y$

$0 - v_y^2 = 2(-9.81)(0.78)$

$v_y = 3.91 m/s$

time taken by it to reach this height

$v_y = v_i + at$

$0 = 3.91 - 9.81 t_1$

$t_1 = 0.39 s$

Now when it again touch the ground then its speed is given as

$v_f^2 - v_y^2 = 2a \Delta y$

$v_f^2 - 0 = 2(9.81)(1.80 - 0.95)$

$v_y = 4.08 m/s$

time taken by it to reach this height

$4.08 = v_i + at$

$4.08 = 0 + 9.81 t_2$

$t_2 = 0.42 s$

$T = t_1 + t_2$

$T = 0.81 s$

Part b)

Horizontal velocity

$v_x = \frac{x}{t}$

$v_x = \frac{2.70}{0.81}$

$v_x = 3.33 m/s$

Part c)

vertical velocity is the intial y direction velocity

$v_y = 3.91 m/s$

Part d)

Take off angle is given as

$tan\theta = \frac{3.91}{3.33}$

$\theta = 49.55 degree$

Part e)

initial vertical position = 1.20 m

maximum height = 2.50 m

$\Delta y = 2.50 - 1.20$

$\Delta y = 1.30 m$

$v_f^2 - v_y^2 = 2a \Delta y$

$0 - v_y^2 = 2(-9.81)(1.30)$

$v_y = 5.05 m/s$

time taken by it to reach this height

$v_y = v_i + at$

$0 = 5.05 - 9.81 t_1$

$t_1 = 0.51 s$

Now when it again touch the ground then its speed is given as

$v_f^2 - v_y^2 = 2a \Delta y$

$v_f^2 - 0 = 2(9.81)(2.50 - 0.72)$

$v_y = 5.9 m/s$

time taken by it to reach this height

$5.9 = v_i + at$

$5.9 = 0 + 9.81 t_2$

$t_2 = 0.60 s$

$T = t_1 + t_2$

$T = 1.11 s$

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3 years ago

If a circle was flattened by pushing down on it, it would most likely form into the shape of an(1) ___ which has(2)___focal poin

worty [1.4K]

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Change in v= 9.8 m/s2xt. The diagram shows a ball falling toward Earth in<br> a vacuum.

Tresset [83]

Answer:

Option A. 39.2 m/s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 4 s

Final velocity (v) =?

v = u + gt

Since the initial velocity (u) is 0, the above equation becomes:

v = gt

Thus, inputting the value of g and t, we can obtain the value of v as shown below:

v = 9.8 × 4

v = 39.2 m/s

Therefore, the velocity of the ball at 4 s is 39.2 m/s.

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A particle with charge q = 10-9 C and mass m = 5.0 x 10-9 kg is moving in a magnetic field whose magnitude is 0.003 T. The speed

Marina86 [1]

Answer:

a=0.212 m/s²

Explanation:

Given that

q= 10⁻⁹ C

m = 5 x 10⁻⁹ kg

Magnetic filed ,B= 0.003 T

Speed ,V= 500 m/s

θ= 45°

Lets take acceleration of the mass is a m/s²

The force on the charge due to magnetic filed B

F= q V B sinθ

Also F= m a ( from Newton's law)

By balancing these above two forces

m a= q V B sinθ

$a=\dfrac{qVB\ sin\theta}{m}$

$a=\dfrac{10^{-9}\times 500\times 0.003\times\ sin45^{\circ}}{5\times 10^{-9}}\ m/s^2$

$a=\dfrac{10^{-9}\times 500\times 0.003\times\dfrac{1}{\sqrt2}}{5\times 10^{-9}}\ m/s^2$

a=0.212 m/s²

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3 years ago

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Answer:

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