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vesna_86 [32]
3 years ago
14

A 64.0 kg pole vaulter running at 10.2 m/s vaults over the bar. If the vaulter's horizontal component of velocity over the bar i

s 1.0 m/s and air resistance is disregarded, how high was the jump?
Physics
1 answer:
Lubov Fominskaja [6]3 years ago
4 0

Answer:

h = 5.05 m

Explanation:

given,

mass of pole vaulter, m = 64 Kg

speed of the vaulter,V = 10.2 m/s

horizontal component of velocity in air, v = 1 m/s

height of the jump,h = ?

using energy conservation

     E_i = E_f

\dfrac{1}{2}mv_i^2 + m g h_i= \dfrac{1}{2}mv_f^2+m g h_f

initial height of the vaulter is equal to zero.

\dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2+gh_f

h =\dfrac{v_i^2-v_f^2}{2g}

h =\dfrac{10^2-1^2}{2\times 9.8}

 h = 5.05 m

height of the jump is equal to 5.05 m.

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car was moving in a straight road of length 320 km it covered 240 km with an average velocity 75 km/hr then it ran out of fuel a
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The average velocity of the car for the whole journey is 69.57 km/h.

The given parameters:

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  • <em>Distance covered = 240 km at 75 km/h</em>
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The time spent by the before refueling is calculated as follows;

t = \frac{d}{v} \\\\t_1 = \frac{240}{75} \\\\t_1 = 3.2 \ hours

The time spent by the car for the remaining journey;

t_3 = \frac{320 - 240}{100} \\\\t_3 = 0.8 \ hr

The total time of the journey is calculated as follows;

t = t_1 + t_2 + t_3\\\\t = 3.2 \ hr \ + \ 0.6 \ hr \ + \ 0.8 \ hr\\\\t = 4.6 \ hours

The average velocity of the car for the whole journey is calculated as follows;

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3 years ago
An electron with a speed of 1.2 × 107 m/s moves horizontally into a region where a constant vertical force of 5.2 × 10-16 N acts
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Answer: 0.642mm

Explanation: F= force = 5.2×10^-16 N,

v = velocity of electron = 1.2×10^7 m/s,

m = mass of electron = 9.11×10^-31 kg.

We will assume the motion of the object to be of a constant acceleration, hence newton's laws of motion is applicable.

Recall that f = ma.

Where a = acceleration

This acceleration of vertical because it occurred when the object deflected.

5.2×10^-16 = 9.11×10^-31 (ay)

ay = 5.2×10^-16 / 9.11×10^-31

ay = 5.71×10^14 m/s²

For the horizontal motion, x = vt

Where x = horizontal distance = 0.019m and v is the velocity = 1.2×10^7 m/s,

By substituting the parameters, we have that

0.019 = 1.27×10^7 × t

t = 0.019 / 1.27 × 10^7

t = 1.5×10^-9 s

The vertical distance (y) is gotten by using the formulae below

y = ut + at²/2

but u = 0

y = at²/2

y = 5.71×10^14 × (1.5×10^-9)²/2

y = 0.00128475/2

y = 0.000642m = 0.642mm

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