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Firlakuza [10]
3 years ago
13

In Fig. 4-41, a ball is thrown up onto a roof, landing 4.00 s later at height h ???? 20.0 m above the release level. The ball’s

path just before landing is angled at u ???? 60.0° with the roof. (a) Find the horizontal dis- tance d it travels. (See the hint to Problem 39.) What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball’s initial velocity?

Physics
1 answer:
Yanka [14]3 years ago
3 0

"Fig is attacted with answer"

Answer:

a) d = 33.72 m

b) v_{i} = 26 m/s

c) β = 71.08°

Explanation:

a)

When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.

Given data:

time = t = 4.00 s

Height = h = 20 m

Angle = θ = 60°

Horizontal distance = d = ?

Using 2nd  equation of motion

h = v_{y_{f}}t + \frac{1}{2}gt^{2}

-20 = v_{y_{f}} (4) + 0.5(-9.8)(4)²

v_{y_{f}} (4) = 58.4

v_{y_{f}}  = 14.6 m/s

This is vertical component of velocity when the ball is on the roof. To calculate the Final velocity and horizontal component, we use

v_{f} = v_{y_{f}} / sinθ

v_{f} = 14.6 / sin 60

v_{f} = 16.86 m/s

v_{x_{f}} = v_{f}cosθ

v_{x_{f}} = 16.86 cos 60

v_{x_{f}} = 8.43 m/s

To calculate the horizontal distance

d = v_{y_{f}} t

d = (8.43)(4)

d = 33.72 m

b)

We know the values of Landing angle, height of roof, time of flight. In part a, We calculate the landing velocity of the ball and also its horizontal and vertical component. As the ball followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

So,

v_{x_{f}} = v_{x_{i}}

but the vertical component of velocity vary with and there is an acceleration along vertical direction which is equal to gravitation acceleration g.

So,

g = (v_{y_{f}} - v_{y_{i}} ) / t

9.8 =  14.6 - v_{y_{i}}) / 4

v_{y_{i}} = 24.6 m/s

v_{i} = \sqrt{v_{x_{i}}^{2}+v_{y_{i}}^{2} }

v_{i} = \sqrt{8.43^{2}+24.6^{2}}

v_{i} = 26 m/s

c)

cos β = v_{x_{i}} / v_{i}

β = cos⁻¹ (8.43 / 26)

β = 71.08°

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We can get the length of the pendulums likely to oscillate with the formula;

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Uma partícula com carga “Q”, no vácuo, gera um potencial elétrico de 600 volts a uma distância de 6,0 m, determine o valor dessa
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Answer:

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the correct answer is b

Explanation:

The expression for the electric potential of a point charge is

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let's calculate

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Traslate

La expresión para el potencial eléctrico de una carga puntual es

            V = k q/r

nos piden la carga eléctrica

          q= V r /k

          calculemos

         Q= 600  6,0 / 9 10⁹

         Q=  4 10⁻⁷ C

reduzcamos a nC

          Q = 4 10⁻⁷ C(10⁹ nC/1C )  

          q = 4 10² nC = 400 nC

la respuesta correcta es b

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