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3 years ago

An athlete executing a long jump leaves the ground at a 28.0º angle and travels 7.80 m. (A) What was the takeoff speed?

Physics

1 answer:

Naddika [18.5K]3 years ago

7 0

Answer:

9.6 m/s

Explanation:

Angle of projection, θ = 28°

Horizontal distance, R = 7.8 m

Let the velocity of projection is given by u.

The formula used to find the velocity of projection is given by

$R=\frac{u^{2}Sin2\theta }{g}$

$7.8=\frac{u^{2}Sin\left (2\times 28 \right ) }{9.8}$

$7.8=\frac{u^{2}\times 0.829}{9.8}$

u = 9.6 m/s

Thus, the velocity of projection is 9.6 m/s.

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A 300 g walking stick is placed on a pivot and balanced by a 100 g mass, as shown in

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Answer:

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A .5kg bird is perched on its nest so that it has 50J of potential energy. how far is it off the of the ground?

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It is 10.20 m from the ground.

Explanation:

Given:

m = 0.5 kg

PE = 50 J

We know that the Potential energy is calculated by the formula:

$P. E = m \times g \times h$

where m is the is mass in kg; g is acceleration due to gravity which is 9.8 m/s and h is height in meters.

PE is the Potential Energy.

Potential Energy is the amount of energy stored when an object is stationary.

Here, if we substitute the values in the formula, we get

$P. E = m \times g \times h$

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3 years ago

Two particles are fixed to an x axis: particle 1 of charge −1.50 ✕ 10−7 c at x = 6.00 cm, and particle 2 of charge +1.50 ✕ 10−7

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Answer : $\underset{E_{R}}{\rightarrow} =-2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Explanation :

Given that,

Charge of particle 1 = $-1.50\times10^{-7} c$

Distance x = 6 cm

Charge of particle 2 = $1.50\times10^{-7} c$

Distance x = 27 cm

Total distance = $\dfrac{6+27}{2}$

$r = 16.5\ cm$

Particle 1 is at (6,0) and particle 2 is at (27,0) .

Therefore, midway (16.5, 0)

Now, $r = \dfrac{|6-16.5|}{2} = \dfrac{|27-16.5|}{2} = 10.5\ cm$

Formula of electric field

$E = \dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q}{r^{2}}$

Now, the the electric field due to particle 1

$\underset{E}{\rightarrow}\ = -\dfrac{9\times10^{9}\times1.50\times10^{-7 }}{10.5}\ \widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = \dfrac{13.5\times10^{2}}{(10.5\times10^{-2})^{2}}\widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Similarly, the electric field due to particle 2

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Resultant Electric field

$\underset{E_{R}}{\rightarrow} = \underset{E_{1}}{\rightarrow} + \underset{E_{2}}{\rightarrow}$

$\underset{E_{R}}{\rightarrow} = -2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Hence, this is the required answer.

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3 years ago

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Allied Forces. they became the allies.

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